Fields and their consequences (A-level only)

3.7 Fields and their consequences (A-level only) — revision question pack

18 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.7. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.7.1 · Fields

Explanation

A force field is a region in which a body experiences a non-contact force. Gravitational fields arise from masses, electric fields from static charges, and magnetic fields from interactions involving moving charges.
Field strength is a vector whose direction is found by inspecting the force on a test mass or positive test charge.
Gravitational and electrostatic fields both obey inverse-square laws and can be represented using field lines, potential and equipotential surfaces.
Their key difference is that masses always attract, while charges may attract or repel.
Examiners expect comparisons to distinguish vector field strength from scalar potential and to determine direction from the interaction, not merely from the sign of a potential value.

Field directions around a mass and a positive charge, showing gravitational attraction and electric repulsion.

Worked example

Two identical positive charges are fixed symmetrically about a midpoint. Compare the resultant electric field and electric potential at the midpoint.

1.The two equal electric-field vectors point in opposite directions.
2.Vector addition gives zero resultant field.
3.The two positive potential contributions are scalars and add.

Answer: The electric field is zero, but the electric potential is positive and non-zero.

Common mistakes

Don't assume zero field at a point must also mean zero potential.
Don't state that gravitational interaction can repel like an electric interaction.
Don't use the force direction on an electron to define electric field direction.

Exam tip

In a comparison, state the shared inverse-square behaviour before contrasting attraction-only gravity with electric attraction or repulsion.

Tier 1 · Easy

State one difference between a gravitational force field and an electric force field.
[1 mark]
Total for this question: 1
State what the tangent to a field line represents.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A small positive charge and a small mass are placed separately at the same point near a negatively charged metal sphere and a planet. Describe the direction of each force and explain how a field-line diagram shows field strength.
[3 marks]
Total for this question: 3
A student claims that the field must be zero along an equipotential surface because no work is done moving along it. Explain why this claim is incorrect.
[3 marks]
Total for this question: 3
In a gravitational field, the potentials at A and B are $-900\,\text{J kg}^{-1}$ and $-720\,\text{J kg}^{-1}$ . In an electric field, the potentials at A and B are $-250\,\text{V}$ and $-70.0\,\text{V}$ . A $2.50\,\text{kg}$ mass and a $+4.00\,\text{mC}$ charge are each moved slowly from A to B in their respective fields. Calculate the external work done in each case and compare the work done per unit mass with the work done per unit charge.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two identical positive point charges are fixed a short distance apart. Discuss the electric field and electric potential at the midpoint, and describe the required relationship between field lines and equipotential surfaces near the charges.
[5 marks]
Total for this question: 5
In one arrangement, equal masses are fixed at $x=-d$ and $x=+d$ . In a second arrangement, charge $+Q$ is fixed at $x=-d$ and charge $-Q$ at $x=+d$ . Compare the field strength and potential at $x=0$ in the two arrangements.
[5 marks]
Total for this question: 5
At point P outside a positively charged massive sphere, an object of mass $m$ and charge $+q$ experiences a resultant force $3.00\,\mu\text{N}$ away from the sphere. An object of mass $m$ and charge $-q$ at P experiences $7.00\,\mu\text{N}$ towards the sphere. Deduce the separate gravitational and electric force magnitudes on either object, and the resultant force on an object of mass $2m$ and charge $+2q$ at P.
[5 marks]
Total for this question: 5
At one point, the local gravitational field is directed west and the local electric field is directed east. Compare the initial force directions on a neutral dust grain, a positively charged dust grain and an otherwise identical negatively charged dust grain. Explain why the direction of either field cannot be deduced from the sign of its potential alone.
[5 marks]
Total for this question: 5
Describe and explain the similarities and differences between field-line and equipotential patterns around an isolated spherical mass and around an isolated positively charged conducting sphere. Include direction, spacing, intersections and the work done when a suitable test object moves along one equipotential.
[6 marks]
Total for this question: 6

3.7.2.1 · Newton's law

Explanation

Newton’s law of gravitation gives the magnitude of the universal attractive force between point masses: $F=Gm_1m_2/r^2$ , where $G$ is the gravitational constant and $r$ is their centre-to-centre separation.
Each mass experiences an equal force directed towards the other mass.
Outside a spherically symmetric body, its mass may be treated as concentrated at its centre.
The inverse-square dependence means multiplying separation by a factor $k$ divides force by $k^2$ .
Examiners expect masses in kilograms, separation from centre to centre, a direction as well as a magnitude, and sensible estimation of the extremely small force between ordinary laboratory objects compared with astronomical bodies.

Two point masses separated centre to centre, with equal attractive forces directed towards each other.

Worked example

Masses $5.0\times10^{24}\,\text{kg}$ and $8.0\times10^{22}\,\text{kg}$ are $4.0\times10^8\,\text{m}$ apart. Calculate their gravitational force.

1.Use centre-to-centre separation in $F=Gm_1m_2/r^2$ .
2.Substitute $F=(6.67\times10^{-11})(5.0\times10^{24})(8.0\times10^{22})/(4.0\times10^8)^2$ .
3.Evaluate and give the attractive direction.

Answer: $1.7\times10^{20}\,\text{N}$ , attractive.

Common mistakes

Don't use the gap between two surfaces rather than centre-to-centre separation.
Don't apply an inverse $1/r$ relationship instead of the inverse square $1/r^2$ .
Don't call the force on the second mass a cancelling force on the first mass.

Exam tip

Write the separation definition beside $r$ before substituting values from a radius or altitude diagram.

Tier 1 · Easy

Calculate the gravitational force between masses $6.0\times10^{22}\,\text{kg}$ and $4.0\times10^{20}\,\text{kg}$ whose centres are $3.0\times10^7\,\text{m}$ apart.
[2 marks]
Total for this question: 2
Two bodies are $3.50\times10^7\,\text{m}$ apart. One has mass $4.50\times10^{22}\,\text{kg}$ and the gravitational force between them is $2.94\times10^{17}\,\text{N}$ . Calculate the mass of the second body.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An $850\,\text{kg}$ satellite is $7.20\times10^6\,\text{m}$ from the centre of a planet of mass $5.97\times10^{24}\,\text{kg}$ . Calculate the gravitational force on the satellite and its acceleration. Explain why the acceleration would be unchanged for a satellite of different mass at the same point.
[4 marks]
Total for this question: 4
In a laboratory experiment, masses $4.00\,\text{kg}$ and $6.00\,\text{kg}$ have centres $0.180\,\text{m}$ apart and attract with force $5.08\times10^{-8}\,\text{N}$ . Calculate the percentage difference between the experimental value of the gravitational constant and the accepted value $6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ .
[3 marks]
Total for this question: 3
A body of mass $8.00\times10^3\,\text{kg}$ , modelled as a point mass, is $2.00\,\text{m}$ from a second point mass. Their gravitational attraction is $4.00\times10^{-2}\,\text{N}$ . Determine the second mass and the acceleration of the first body.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two approximately spherical bodies have masses $4.0\times10^{24}\,\text{kg}$ and $7.5\times10^{23}\,\text{kg}$ . Their mutual gravitational force is $8.0\times10^{20}\,\text{N}$ . Determine their centre-to-centre separation and the new force if that separation increases by $20\%$ .
[5 marks]
Total for this question: 5
Two spherical, iron-rich asteroids have masses $3.20\times10^{15}\,\text{kg}$ and $5.50\times10^{15}\,\text{kg}$ and radii $5.00\,\text{km}$ and $6.00\,\text{km}$ . Their surfaces are $9.00\,\text{km}$ apart. Determine the mutual force and the acceleration of the smaller asteroid.
[5 marks]
Total for this question: 5
Two small asteroids, modelled as point masses of $2.00\times10^{12}\,\text{kg}$ and $3.00\times10^{12}\,\text{kg}$ , are initially at rest with their centres $3.00\times10^3\,\text{m}$ apart. Determine an approximate time for their separation to decrease by $10.0\,\text{m}$ , treating the gravitational force as constant at its initial value during this small movement.
[5 marks]
Total for this question: 5
A $4.00\,\text{kg}$ point mass is at the origin. Point masses $6.00\times10^9\,\text{kg}$ and $8.00\times10^9\,\text{kg}$ are fixed $3.00\,\text{m}$ east and $4.00\,\text{m}$ north of it respectively. Determine the magnitude and direction of the resultant gravitational force on the mass at the origin and its initial acceleration.
[6 marks]
Total for this question: 6
Two laboratory masses attract with force $2.90\times10^{-7}\,\text{N}$ when their centres are $0.240\,\text{m}$ apart and $1.29\times10^{-7}\,\text{N}$ when they are $0.360\,\text{m}$ apart. Use the measurements to determine the power $n$ in $F\propto r^{-n}$ . One mass is $12.5\,\text{kg}$ ; use the first reading to determine the other mass and comment on agreement with Newton's law.
[5 marks]
Total for this question: 5

3.7.2.2 · Gravitational field strength

Explanation

Gravitational field strength is the gravitational force per unit mass on a small test mass, $g=F/m$ , with units $\text{N kg}^{-1}$ , equivalent to $\text{m s}^{-2}$ .
Field lines point in the direction of force on the test mass.
Around a spherical mass they are radial and directed inward; their increasing separation with radius represents decreasing strength.
The magnitude outside the mass is $g=GM/r^2$ , where $r$ is measured from the centre.
Examiners expect altitude to be converted to radial distance, the inward vector direction to be stated, and field-line density to be interpreted qualitatively rather than treated as a literal count of physical lines.

A radial gravitational field with arrows directed towards the central mass.

Worked example

Determine the gravitational field strength $8.0\times10^6\,\text{m}$ from the centre of a planet of mass $6.0\times10^{24}\,\text{kg}$ .

1.Use $g=GM/r^2$ with distance measured from the centre.
2. $g=(6.67\times10^{-11})(6.0\times10^{24})/(8.0\times10^6)^2$ .
3.Evaluate and state the radial direction.

Answer: $6.3\,\text{N kg}^{-1}$ towards the planet.

Common mistakes

Don't use altitude above the surface directly as $r$ .
Don't quote only the magnitude although gravitational field strength is a vector.
Don't draw radial gravitational arrows pointing away from the mass.

Exam tip

If a radius and altitude are both given, write $r=R+h$ before using $g=GM/r^2$ .

Tier 1 · Easy

Define gravitational field strength at a point.
[1 mark]
Total for this question: 1
A $75\,\text{kg}$ object is at a point where gravitational field strength is $1.6\,\text{N kg}^{-1}$ directed vertically downwards. Calculate the gravitational force on the object due to the field.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Determine the gravitational field strength $2.40\times10^7\,\text{m}$ from the centre of a planet of mass $4.80\times10^{24}\,\text{kg}$ .
[3 marks]
Total for this question: 3
The gravitational field strength is $1.25\,\text{N kg}^{-1}$ at a distance $1.80\times10^7\,\text{m}$ from the centre of a spherical planet. Determine the planet's mass.
[3 marks]
Total for this question: 3
A $400\,\text{kg}$ probe at radial distance $r$ from a spherical planet experiences gravitational force $900\,\text{N}$ . Determine the force on a $600\,\text{kg}$ probe at radial distance $1.50r$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A planet has mass $6.40\times10^{24}\,\text{kg}$ and radius $7.00\times10^6\,\text{m}$ . Calculate the altitude above its surface where the gravitational field strength is $2.40\,\text{N kg}^{-1}$ .
[5 marks]
Total for this question: 5
Two spherical bodies of masses $4.8\times10^{24}\,\text{kg}$ and $1.2\times10^{24}\,\text{kg}$ have centres $6.0\times10^7\,\text{m}$ apart. Determine the point between them where the resultant gravitational field strength is zero.
[4 marks]
Total for this question: 4
A gravimeter reads $8.00\,\text{N kg}^{-1}$ on the surface of a spherical planet and $4.50\,\text{N kg}^{-1}$ after being raised vertically by $2.00\times10^6\,\text{m}$ . Use the readings to find the planet's radius and mass. Predict the reading after a total vertical rise of $4.00\times10^6\,\text{m}$ .
[5 marks]
Total for this question: 5
A point between a planet and its moon is $1.20\times10^7\,\text{m}$ from the planet's centre and $4.00\times10^6\,\text{m}$ from the moon's centre. Their masses are $5.00\times10^{24}\,\text{kg}$ and $8.00\times10^{22}\,\text{kg}$ respectively. Determine the resultant gravitational field strength at the point and the force on a $250\,\text{kg}$ probe placed there.
[5 marks]
Total for this question: 5
A spherical planet has mass $(3.74\pm0.11)\times10^{23}\,\text{kg}$ and radius $(2.86\pm0.06)\times10^6\,\text{m}$ . Calculate its surface gravitational field strength and the force on a $73.0\,\text{kg}$ probe. By adding the percentage uncertainty in mass to twice the percentage uncertainty in radius, determine the percentage and absolute uncertainties in the field strength.
[6 marks]
Total for this question: 6

3.7.2.3 · Gravitational potential

Explanation

Gravitational potential is work done per unit mass in bringing a small mass from infinity to a point, with zero potential defined at infinity. For a spherical mass, $V=-GM/r$ ; the negative sign shows that a mass at finite radius is gravitationally bound.
Potential difference determines external work in a slow transfer, $\Delta W=m\Delta V$ .
No work is done moving along an equipotential surface, which is perpendicular to field lines.
The radial graphs satisfy $g=-\Delta V/\Delta r$, so field strength is the magnitude of the potential gradient, while potential difference is obtained from the signed area under a $g$ – $r$ graph.
Examiners expect signs, the zero reference and the distinction between scalar potential and vector field strength.

Gravitational potential outside a spherical mass, remaining negative and approaching zero as radius increases.

Worked example

A $200\,\text{kg}$ probe moves slowly from $8.0\times10^6\,\text{m}$ to $1.6\times10^7\,\text{m}$ from a mass of $6.0\times10^{24}\,\text{kg}$ . Find the external work done.

1.Calculate $V_i=-GM/r_i=-5.00\times10^7\,\text{J kg}^{-1}$ .
2.Calculate $V_f=-GM/r_f=-2.50\times10^7\,\text{J kg}^{-1}$ .
3. $\Delta W=m(V_f-V_i)=200(2.50\times10^7)=5.0\times10^9\,\text{J}$ .

Answer: The external work done is $+5.0\times10^9\,\text{J}$ .

Common mistakes

Don't drop the negative sign from $V=-GM/r$ .
Don't use $mV_i$ rather than $m(V_f-V_i)$ for work between two finite points.
Don't claim that work is done while moving along one equipotential surface.

Exam tip

Write both potentials with signs before forming $\Delta V=V_f-V_i$ .

Tier 1 · Easy

State why gravitational potential near an isolated planet is negative when potential is defined as zero at infinity.
[1 mark]
Total for this question: 1
Calculate the gravitational potential $1.6\times10^7\,\text{m}$ from the centre of a spherical body of mass $8.0\times10^{23}\,\text{kg}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $320\,\text{kg}$ probe moves slowly from $9.0\times10^6\,\text{m}$ to $1.8\times10^7\,\text{m}$ from the centre of a planet of mass $7.2\times10^{24}\,\text{kg}$ . Calculate the change in gravitational potential and the work done by the external force.
[4 marks]
Total for this question: 4
Gravitational potential changes from $-2.10\times10^7\,\text{J kg}^{-1}$ to $-1.80\times10^7\,\text{J kg}^{-1}$ over an outward radial distance of $2.50\times10^5\,\text{m}$ . Estimate the gravitational field strength in this interval.
[3 marks]
Total for this question: 3
At the surface of a spherical planet, gravitational potential is $-4.50\times10^7\,\text{J kg}^{-1}$ and gravitational field strength is $10.0\,\text{N kg}^{-1}$ . Determine the planet's radius and mass.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $450\,\text{kg}$ craft is moved slowly from radius $8.0\times10^6\,\text{m}$ to radius $2.4\times10^7\,\text{m}$ from a body of mass $5.5\times10^{24}\,\text{kg}$ . Determine the external work done. The craft is then released from rest at the outer point; calculate its speed when it returns to the inner point.
[6 marks]
Total for this question: 6
A point is $2.0\times10^7\,\text{m}$ from a mass $3.0\times10^{24}\,\text{kg}$ and $1.0\times10^7\,\text{m}$ from a second mass $8.0\times10^{23}\,\text{kg}$ . Determine the gravitational potential at the point and the external work needed to move a $250\,\text{kg}$ craft slowly from there to infinity.
[5 marks]
Total for this question: 5
A graph of gravitational field strength against outward radial distance is approximated by straight-line segments. From A to B, field strength falls from $16.0$ to $10.0\,\text{N kg}^{-1}$ over $2.50\times10^6\,\text{m}$ . From B to C it falls from $10.0$ to $7.00\,\text{N kg}^{-1}$ over $1.50\times10^6\,\text{m}$ . Determine the minimum speed at A for an unpowered probe to reach C with zero speed.
[5 marks]
Total for this question: 5
A $300\,\text{kg}$ craft falls from rest from radial distance $1.50\times10^7\,\text{m}$ to $9.00\times10^6\,\text{m}$ from a spherical body of mass $4.50\times10^{24}\,\text{kg}$ . During the fall, resistive forces in the body's outer atmosphere transfer $30.0\%$ of the decrease in gravitational potential energy to thermal energy. Determine the gravitational potential-energy decrease and the craft's speed at the inner point.
[5 marks]
Total for this question: 5
Measurements outside a spherical planet give gravitational potential $-2.50\times10^7\,\text{J kg}^{-1}$ at $1.20\times10^7\,\text{m}$ and $-1.25\times10^7\,\text{J kg}^{-1}$ at $2.40\times10^7\,\text{m}$ . Explain how the data test the expected radial relationship, estimate the planet's mass from both readings, and calculate the external work needed to move a $500\,\text{kg}$ probe slowly between these points.
[6 marks]
Total for this question: 6

3.7.2.4 · Orbits of planets and satellites

Explanation

For a circular orbit, gravity supplies the centripetal force, giving $v=\sqrt{GM/r}$ . Combining this with $v=2\pi r/T$ gives $T^2=4\pi^2r^3/(GM)$ and therefore $T^2\propto r^3$ for one central mass.
A circular satellite’s total energy is $-GMm/(2r)$ ; escape speed follows by setting total energy at infinity to zero. A synchronous satellite matches the body’s rotation period.
A geostationary satellite is a special synchronous satellite in a circular equatorial orbit, moving in the rotation direction, with orbital radius measured from the planet’s centre.
Low orbits give shorter periods and different coverage.
Examiners expect energy, period, speed, orbit plane and radius to be connected to the stated satellite use.

Circular satellite orbits at different radii around a planet.

Worked example

A satellite orbits a planet of mass $6.0\times10^{24}\,\text{kg}$ at radius $4.0\times10^7\,\text{m}$ . Determine its speed and period.

1. $v=\sqrt{GM/r}=\sqrt{(6.67\times10^{-11})(6.0\times10^{24})/(4.0\times10^7)}$ .
2.Evaluate $v=3.16\times10^3\,\text{m s}^{-1}$ .
3. $T=2\pi r/v=7.95\times10^4\,\text{s}=22.1\,\text{h}$ .

Answer: $v=3.16\times10^3\,\text{m s}^{-1}$ and $T=22.1\,\text{h}$ .

Common mistakes

Don't use altitude above the surface instead of orbital radius from the centre.
Don't state that every synchronous orbit is geostationary.
Don't make total orbital energy positive for a bound circular orbit.

Exam tip

For a derivation, start by equating gravitational force to centripetal force before substituting $v=2\pi r/T$ .

Tier 1 · Easy

A satellite completes a circular orbit of radius $7.5\times10^6\,\text{m}$ in $6.0\times10^3\,\text{s}$ . Calculate its orbital speed.
[2 marks]
Total for this question: 2
Explain why a satellite in a circular polar orbit with period equal to the planet's rotation period is not geostationary.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Starting from Newton's law of gravitation, show that $T^2\propto r^3$ for circular orbits around one planet. Hence determine the factor by which the period changes when orbital radius is multiplied by $1.60$ .
[4 marks]
Total for this question: 4
A moon moves in a circular orbit of radius $2.20\times10^7\,\text{m}$ with period $1.80\times10^4\,\text{s}$ . Determine the mass of the body it orbits.
[3 marks]
Total for this question: 3
Two satellites orbit the same planet in circular paths. Their orbital radii are $7.00\times10^6\,\text{m}$ and $2.80\times10^7\,\text{m}$ , and their periods are $5.80\times10^3\,\text{s}$ and $4.64\times10^4\,\text{s}$ respectively. Calculate the gradient obtained from these points on a graph of $\log_{10}T$ against $\log_{10}r$ , and hence state the power-law relationship between $T$ and $r$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A planet of mass $6.0\times10^{24}\,\text{kg}$ and radius $7.0\times10^6\,\text{m}$ rotates once every $30\,\text{h}$ . Determine the radius and altitude of a synchronous circular orbit. Calculate the total energy of a $500\,\text{kg}$ satellite in that orbit.
[6 marks]
Total for this question: 6
A moon travels in a circular orbit of radius $8.00\times10^6\,\text{m}$ around an exoplanet once every $2.40\,\text{h}$ . The exoplanet's radius is $6.00\times10^6\,\text{m}$ . Determine the mean density of the exoplanet.
[5 marks]
Total for this question: 5
An $800\,\text{kg}$ satellite is initially in a circular orbit of radius $1.20\times10^7\,\text{m}$ . The central planet has mass $4.80\times10^{24}\,\text{kg}$ . A tangential impulse makes the satellite just able to escape. Determine the required increase in speed and the energy transferred to the satellite.
[5 marks]
Total for this question: 5
A $650\,\text{kg}$ satellite is moved from a circular orbit of radius $1.00\times10^7\,\text{m}$ to a circular orbit of radius $2.50\times10^7\,\text{m}$ around a planet of mass $5.20\times10^{24}\,\text{kg}$ . Determine the increase in the satellite's total energy and the factor by which its orbital period changes.
[5 marks]
Total for this question: 5
A moon has a circular orbit of radius $3.60\times10^8\,\text{m}$ and period $9.00\,\text{days}$ . Determine the mass of the planet, the moon's orbital speed and the escape speed from the orbit. Calculate the moon's total energy per unit mass while in the circular orbit.
[6 marks]
Total for this question: 6

3.7.3.1 · Coulomb's law

Explanation

For point charges in a vacuum, Coulomb’s law gives $F=|Q_1Q_2|/(4\pi\varepsilon_0r^2)$ , where $\varepsilon_0$ is the permittivity of free space and $r$ is the separation.
Air may be treated as a vacuum, and a charged sphere’s external effect may be modelled as charge concentrated at its centre.
Like charges repel and unlike charges attract, so the equation gives magnitude while signs and geometry determine direction.
Electrostatic and gravitational forces between subatomic particles both follow inverse-square laws, allowing their magnitude ratio to be compared because $r^2$ cancels.
Examiners expect charge conversion from nanocoulombs, vector addition for several charges, and a direction justified from charge signs.

Unlike point charges separated by distance r, with equal attractive forces.

Worked example

Charges $+3.0\,\text{nC}$ and $-6.0\,\text{nC}$ are $0.20\,\text{m}$ apart in air. Calculate the force.

1.Treat air as a vacuum and convert both charges to coulombs.
2. $F=(8.99\times10^9)(3.0\times10^{-9})(6.0\times10^{-9})/(0.20)^2$ .
3.Evaluate the magnitude and use opposite signs for the direction.

Answer: $4.0\times10^{-6}\,\text{N}$ , attractive.

Common mistakes

Don't leave nanocoulomb values unconverted before substitution.
Don't use the signs inside a magnitude calculation and report a negative force magnitude.
Don't add non-collinear force magnitudes without resolving vectors.

Exam tip

Calculate the positive magnitude first, then write a separate attraction-or-repulsion statement from the charge signs.

Tier 1 · Easy

Point charges $+3.0\,\text{nC}$ and $-8.0\,\text{nC}$ are $0.120\,\text{m}$ apart in air. Calculate the force between them.
[2 marks]
Total for this question: 2
Two point charges exert force $F$ on each other at separation $r$ . Express the new force in terms of $F$ when their separation becomes $2r$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

Determine the ratio of the electrostatic force to the gravitational force between a proton and an electron. Use $m_p=1.67\times10^{-27}\,\text{kg}$ and $m_e=9.11\times10^{-31}\,\text{kg}$ , and explain why their separation is not needed.
[4 marks]
Total for this question: 4
A point charge of magnitude $6.0\,\text{nC}$ exerts a force of $3.0\times10^{-5}\,\text{N}$ on a second point charge $0.060\,\text{m}$ away. Determine the magnitude of the second charge.
[3 marks]
Total for this question: 3
At a separation of $0.120\,\text{m}$ , two point charges exert a force of magnitude $5.40\times10^{-4}\,\text{N}$ on each other. Use Coulomb's law to predict the force magnitude at $0.200\,\text{m}$ . A student measures $1.90\times10^{-4}\,\text{N}$ at this separation. Calculate the percentage difference relative to the predicted value.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A charge $+2.0\,\text{nC}$ is at the origin. A charge $+5.0\,\text{nC}$ is at $(0.300\,\text{m},0)$ and a charge $-4.0\,\text{nC}$ is at $(0,0.400\,\text{m})$ . Determine the magnitude and direction of the resultant electrostatic force on the charge at the origin.
[6 marks]
Total for this question: 6
A charge $+2.0\,\text{nC}$ lies between two other positive charges on a straight line. A $+7.0\,\text{nC}$ charge is $0.180\,\text{m}$ to its left and a $+5.0\,\text{nC}$ charge is $0.420\,\text{m}$ to its right. Determine the resultant force on the middle charge.
[4 marks]
Total for this question: 4
Two fixed positive point charges are $0.600\,\text{m}$ apart. A $+2.00\,\text{nC}$ test charge placed between them experiences zero resultant force when it is $0.200\,\text{m}$ from the left-hand charge. At the midpoint, the resultant force on the test charge has magnitude $1.50\times10^{-5}\,\text{N}$ . Deduce the values of the two fixed charges and the direction of the midpoint force.
[5 marks]
Total for this question: 5
Charges $+6.00\,\text{nC}$ and $-4.00\,\text{nC}$ occupy the lower-left and lower-right vertices of an equilateral triangle of side $0.250\,\text{m}$ . A $+3.00\,\text{nC}$ charge is at the upper vertex. Determine the magnitude and direction of the resultant force on the upper charge.
[6 marks]
Total for this question: 6
Two identical small spheres, each of mass $2.50\,\text{g}$ , hang from the same point on insulating threads of length $0.300\,\text{m}$ . Equal charges make each thread settle at $8.00^\circ$ to the vertical. Treat the spheres as point charges. Determine the charge magnitude on each sphere and the number of elementary charges this represents.
[5 marks]
Total for this question: 5

3.7.3.2 · Electric field strength

Explanation

Electric field strength is force per unit positive test charge, $E=F/Q$ , measured in $\text{N C}^{-1}$ or $\text{V m}^{-1}$ . Field lines point away from positive charge and towards negative charge.
A point charge produces radial magnitude $E=Q/(4\pi\varepsilon_0r^2)$ .
Between parallel plates the field is approximately uniform: draw or describe straight, parallel and evenly spaced field lines, and use $E=V/d$ , derived from $Fd=Q\Delta V$ .
A charged particle entering perpendicular to a uniform field keeps constant velocity across the field but accelerates parallel or antiparallel to it, giving a parabolic trajectory.
Examiners expect the field direction to follow force on positive charge, so an electron accelerates opposite to the field, and may require field patterns investigated using conducting paper or an electrolytic tank.

A uniform electric field between parallel plates and the parabolic path of a positive particle entering perpendicular to it.

Worked example

Parallel plates have potential difference $1.20\,\text{kV}$ and separation $30\,\text{mm}$ . Find the field strength and force on an electron.

1.Convert $V=1.20\times10^3\,\text{V}$ and $d=0.030\,\text{m}$ .
2. $E=V/d=4.0\times10^4\,\text{V m}^{-1}$ .
3. $F=eE=(1.60\times10^{-19})(4.0\times10^4)=6.4\times10^{-15}\,\text{N}$ opposite to the field.

Answer: $E=4.0\times10^4\,\text{V m}^{-1}$ and $F=6.4\times10^{-15}\,\text{N}$ towards the positive plate.

Common mistakes

Don't use plate separation in millimetres in $E=V/d$ .
Don't draw the electron force in the electric-field direction.
Don't treat a charged particle’s path in a uniform field as circular.

Exam tip

For a uniform parallel-plate field, say the field lines are parallel and evenly spaced. Mark the plate signs and field direction before deciding the force direction for a positive or negative particle.

Tier 1 · Easy

Define electric field strength at a point.
[1 mark]
Total for this question: 1
A proton experiences an electric force of $4.80\times10^{-15}\,\text{N}$ to the left. Calculate the electric field strength and state its direction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Parallel plates have a potential difference of $1.80\,\text{kV}$ and separation $45.0\,\text{mm}$ . Calculate the uniform electric field strength, the force magnitude on an electron, and its acceleration magnitude.
[4 marks]
Total for this question: 4
Calculate the distance from a point charge of $+5.5\,\text{nC}$ at which the electric field strength is $2.0\times10^3\,\text{N C}^{-1}$ .
[3 marks]
Total for this question: 3
An external force does $4.32\,\mu\text{J}$ of work when a $+2.40\,\text{nC}$ test charge is moved slowly from one parallel plate to the other, opposite to the field direction. The plates are $18.0\,\text{mm}$ apart. Determine the potential difference and electric field strength, and state whether the destination plate is at higher or lower potential.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A proton enters at $3.00\times10^6\,\text{m s}^{-1}$ perpendicular to a uniform electric field of strength $2.50\times10^4\,\text{V m}^{-1}$ . The field region is $80.0\,\text{mm}$ long in the initial direction of travel. Determine the proton's deflection and the angle of its velocity to its original direction as it leaves. Use $m_p=1.67\times10^{-27}\,\text{kg}$ .
[6 marks]
Total for this question: 6
An oil drop of mass $6.00\times10^{-15}\,\text{kg}$ is stationary in a uniform electric field of strength $1.84\times10^5\,\text{N C}^{-1}$ directed vertically downwards. Determine the drop's charge and the number of excess electrons on it.
[5 marks]
Total for this question: 5
On conducting paper, two probes on an east-west line through P are spaced $8.00\,\text{mm}$ apart. The western probe reads $130\,\text{V}$ and the eastern probe reads $118\,\text{V}$ . Determine the electric field at P, the force on an electron there, and its acceleration. Use $m_e=9.11\times10^{-31}\,\text{kg}$ .
[5 marks]
Total for this question: 5
A negatively charged oil drop of mass $4.50\times10^{-15}\,\text{kg}$ carries three excess electrons and is initially stationary between horizontal plates $12.0\,\text{mm}$ apart. Determine the potential difference that balances its weight. The plate polarity is then reversed and the potential difference is set to $60.0\%$ of this value. Determine the drop's initial acceleration and the time taken to fall $4.00\,\text{mm}$ from rest.
[6 marks]
Total for this question: 6
A small charged bead of mass $3.26\times10^{-7}\,\text{kg}$ hangs from an insulating thread between vertical parallel plates $37.0\,\text{mm}$ apart. The potential difference is $2.65\,\text{kV}$ and the bead is deflected $13.5^\circ$ from the vertical towards the lower-potential plate. Determine the uniform electric field strength using $E=V/d$ , the magnitude and sign of the bead's charge, and the tension in the thread.
[6 marks]
Total for this question: 6

3.7.3.3 · Electric potential

Explanation

Absolute electric potential is work done per unit positive charge in bringing a small test charge from infinity, where potential is defined as zero. For a point charge, $V=Q/(4\pi\varepsilon_0r)$ ; potential is a scalar and carries the source charge’s sign.
External work in a slow transfer is $\Delta W=q\Delta V$ .
No work is done moving along an equipotential surface, which meets field lines at right angles.
The field points towards decreasing potential and has magnitude given by the potential gradient, while potential difference is the signed area under an $E$ – $r$ graph.
Examiners expect scalar potentials to be added algebraically, vector fields to be added separately, and recognition that zero potential need not mean zero field.

Radial electric field lines crossing circular equipotentials at right angles around a positive point charge.

Worked example

Find the electric potential $0.25\,\text{m}$ from a $+5.0\,\text{nC}$ point charge and the external work to bring $+2.0\,\text{nC}$ from infinity.

1. $V=(8.99\times10^9)(5.0\times10^{-9})/0.25=180\,\text{V}$ .
2.The initial potential at infinity is zero, so $\Delta V=+180\,\text{V}$ .
3. $\Delta W=q\Delta V=(2.0\times10^{-9})(180)=3.6\times10^{-7}\,\text{J}$ .

Answer: $V=+180\,\text{V}$ and external work $=+3.6\times10^{-7}\,\text{J}$ .

Common mistakes

Don't drop the sign of the source charge when calculating potential.
Don't add electric-field magnitudes as scalars when several charges are present.
Don't claim that a charge moving along an equipotential changes potential energy.

Exam tip

Calculate scalar potential first; only then use $q\Delta V$ with the moved charge’s sign.

Tier 1 · Easy

Calculate the electric potential $0.200\,\text{m}$ from an isolated point charge of $+4.00\,\text{nC}$ .
[2 marks]
Total for this question: 2
Calculate the external work needed to move a $+2.5\,\text{nC}$ charge slowly from infinity to a point at electric potential $+320\,\text{V}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A charge of $-3.0\,\text{nC}$ is moved slowly from a point at $+120\,\text{V}$ to a point at $-80\,\text{V}$ . Calculate the work done by the external force and state whether the charge's electric potential energy increases or decreases.
[4 marks]
Total for this question: 4
A point is $0.300\,\text{m}$ from a charge $+6.0\,\text{nC}$ and $0.400\,\text{m}$ from a charge $-2.0\,\text{nC}$ . Determine the electric potential at the point.
[3 marks]
Total for this question: 3
A fixed point charge of $+7.00\,\text{nC}$ produces a radial electric field. Calculate the electric potential at radial distances $0.150\,\text{m}$ and $0.300\,\text{m}$ . Hence determine the external work done when a $+2.00\,\text{nC}$ charge is moved slowly from the first point to the second.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Charges $+8.0\,\text{nC}$ and $-2.0\,\text{nC}$ are fixed $0.600\,\text{m}$ apart. Find the point between them where the electric potential is zero. Determine the electric field strength there and the external work needed to bring a $+3.0\,\text{nC}$ charge slowly from infinity to that point.
[6 marks]
Total for this question: 6
Charges $+9.0\,\text{nC}$ and $+4.0\,\text{nC}$ are fixed $0.650\,\text{m}$ apart. Determine the point between them where their contributions to electric potential are equal. Calculate the total potential there and the work done by the external force in bringing an electron slowly from infinity to that point.
[5 marks]
Total for this question: 5
Charges $+5.00\,\text{nC}$ and $-3.00\,\text{nC}$ are fixed at $x=0$ and $x=0.500\,\text{m}$ . A proton is released from rest at $x=0.100\,\text{m}$ . Determine its speed when it reaches $x=0.300\,\text{m}$ , taking the proton mass as $1.67\times10^{-27}\,\text{kg}$ .
[5 marks]
Total for this question: 5
Point charges $+8.00\,\text{nC}$ , $+3.00\,\text{nC}$ , $-5.00\,\text{nC}$ and $-2.00\,\text{nC}$ are fixed at the four corners of a square of side $0.400\,\text{m}$ . Determine the electric potential at the centre. An electron reaches the centre after starting from rest at infinity; determine its speed there and state the energy assumption used.
[5 marks]
Total for this question: 5
The radial electric field strength $0.200\,\text{m}$ from an isolated positive point charge is $3.40\times10^3\,\text{N C}^{-1}$ . Determine the source charge and the potentials at $0.200\,\text{m}$ and $0.450\,\text{m}$ . Calculate the external work done when a $-2.50\,\text{nC}$ charge is moved slowly between these radii in the outward direction.
[6 marks]
Total for this question: 6

3.7.4.1 · Capacitance

Explanation

Capacitance is charge stored per unit potential difference, $C=Q/V$ , measured in farads, where $1\,\text{F}=1\,\text{C V}^{-1}$ .
The two plates carry equal and opposite charges, and $Q$ denotes the magnitude on either plate rather than a net charge on the whole capacitor.
For a fixed capacitor, $Q$ is proportional to $V$ , so a graph of charge against potential difference is a straight line through the origin with gradient $C$ .
Examiners expect microfarads and nanofarads to be converted to farads, the correct plate-charge interpretation, and a gradient taken from axes in the stated order.

Worked example

A capacitor stores charge $3.0\,\text{mC}$ at $12\,\text{V}$ . Determine its capacitance.

1.Convert charge: $Q=3.0\times10^{-3}\,\text{C}$ .
2.Use $C=Q/V=(3.0\times10^{-3})/12$ .
3.Convert the result to microfarads.

Answer: $C=2.5\times10^{-4}\,\text{F}=250\,\mu\text{F}$ .

Common mistakes

Don't use the sum of the positive and negative plate-charge magnitudes as $Q$ .
Don't leave millcoulombs or microfarads unconverted in an SI calculation.
Don't call the gradient of a $V$ -against- $Q$ graph capacitance.

Exam tip

Write the graph axes beside $C=Q/V$ before interpreting a gradient.

Tier 1 · Easy

A capacitor stores charge of $3.6\,\text{mC}$ at a potential difference of $12\,\text{V}$ . Calculate its capacitance.
[2 marks]
Total for this question: 2
Define the capacitance of a capacitor.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A charge-potential-difference graph for a capacitor passes through the point $(16.0\,\text{V},4.80\,\text{mC})$ . Determine its capacitance and the charge stored at $27.0\,\text{V}$ .
[3 marks]
Total for this question: 3
The potential difference across a $220\,\mu\text{F}$ capacitor rises from $5.0\,\text{V}$ to $17.0\,\text{V}$ in $6.00\,\text{s}$ . Determine the additional charge stored and the mean charging current.
[3 marks]
Total for this question: 3
Capacitor A has capacitance $220\,\mu\text{F}$ and stores $1.32\,\text{mC}$ . Capacitor B stores $3.30\,\text{mC}$ at the same potential difference. Determine the potential difference and the capacitance of B.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $470\,\mu\text{F}$ capacitor is initially at $9.0\,\text{V}$ . A charge-transfer process then moves $1.2\times10^{16}$ electrons from its negatively charged plate to its positively charged plate while capacitance remains constant. Determine the new charge magnitude on each plate and the new potential difference.
[5 marks]
Total for this question: 5
A charge sensor has a constant zero offset. For a capacitor, it records charge $1.10\,\text{mC}$ at $2.00\,\text{V}$ and $3.50\,\text{mC}$ at $10.0\,\text{V}$ . Use the gradient of the charge–potential difference relationship to determine the capacitance and the sensor's zero offset.
[4 marks]
Total for this question: 4
A capacitor is charged by a measured current $(24.0\pm0.2)\,\mu\text{A}$ for $(150.0\pm0.5)\,\text{s}$ . Its potential difference is then $(9.60\pm0.05)\,\text{V}$ . Determine its capacitance and, by adding the percentage uncertainties in current, time and potential difference, calculate the percentage and absolute uncertainties to two significant figures.
[5 marks]
Total for this question: 5
For a fixed capacitor, a graph of potential difference against plate-charge magnitude is a straight line through the origin with gradient $3.20\times10^3\,\text{V C}^{-1}$ . Determine the capacitance and the charge magnitude at $19.2\,\text{V}$ . The charge magnitude is then reduced to $4.20\,\text{mC}$ ; calculate the new potential difference.
[5 marks]
Total for this question: 5
Capacitor A has capacitance $295\,\mu\text{F}$ and potential difference $21.4\,\text{V}$ . Capacitor B has capacitance $365\,\mu\text{F}$ and initially has the same plate-charge magnitude as A. Determine that charge and the initial potential difference across B. Calculate the additional charge required to raise B to $28.6\,\text{V}$ and the percentage increase in its plate-charge magnitude.
[5 marks]
Total for this question: 5

3.7.4.2 · Parallel plate capacitor

Explanation

A parallel-plate capacitor has $C=\varepsilon_0\varepsilon_rA/d$ , so capacitance increases with overlap area $A$ and relative permittivity $\varepsilon_r$ , but decreases with plate separation $d$ . In a dielectric, simple polar molecules rotate in the field and the resulting bound surface charges oppose the original field.
For fixed free charge this reduces potential difference and therefore increases capacitance.
If the capacitor remains connected to a fixed-voltage supply, extra charge flows onto the plates.
Experiments may determine relative permittivity or test the $C$ – $A$ and $C$ – $d$ relationships.
Examiners expect the overlap area, SI separation and whether charge or voltage is held constant to be identified.

Parallel capacitor plates separated by distance d with a dielectric polarised in the field.

Worked example

Plates of area $2.0\times10^{-2}\,\text{m}^2$ are separated by $1.0\,\text{mm}$ and filled by material of $\varepsilon_r=4.0$ . Find $C$ .

1.Convert separation: $d=1.0\times10^{-3}\,\text{m}$ .
2.Use $C=\varepsilon_0\varepsilon_rA/d$ .
3. $C=(8.85\times10^{-12})(4.0)(2.0\times10^{-2})/(1.0\times10^{-3})$ .

Answer: $C=7.1\times10^{-10}\,\text{F}$ .

Common mistakes

Don't use total plate area rather than the overlapping area.
Don't say a dielectric strengthens the field for fixed free charge.
Don't claim free charge increases for an isolated capacitor.

Exam tip

State whether the capacitor is isolated or remains connected before describing dielectric effects.

Tier 1 · Easy

Two parallel plates in air have overlap area $2.50\times10^{-2}\,\text{m}^2$ and separation $1.20\,\text{mm}$ . Calculate their capacitance.
[2 marks]
Total for this question: 2
A parallel-plate capacitor is rebuilt with twice the overlap area and three times the plate separation. Calculate the ratio $C_{\text{new}}/C_{\text{old}}$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

A parallel-plate capacitor has area $1.80\times10^{-2}\,\text{m}^2$ , separation $0.800\,\text{mm}$ and is connected to a $120\,\text{V}$ supply. A dielectric of relative permittivity $3.40$ completely fills the gap. Determine the new capacitance and charge stored.
[4 marks]
Total for this question: 4
A parallel-plate capacitor has capacitance $620\,\text{pF}$ , overlap area $2.40\times10^{-2}\,\text{m}^2$ and plate separation $0.900\,\text{mm}$ . Determine the relative permittivity of the material filling the gap.
[3 marks]
Total for this question: 3
Two parallel-plate capacitors have the same plate area and separation. The air-filled capacitor has capacitance $180\,\text{pF}$ ; the other is completely filled by a dielectric and has capacitance $630\,\text{pF}$ . Both are connected to $12.0\,\text{V}$ . Determine the relative permittivity, the charge on the dielectric-filled capacitor, and whether its electric field strength is greater than, equal to or less than that in the air-filled capacitor.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A designer needs a $2.00\,\text{nF}$ parallel-plate capacitor using a dielectric with relative permittivity $4.50$ and thickness $0.750\,\text{mm}$ . Calculate the required overlap area. Explain at the molecular level why the dielectric increases capacitance.
[6 marks]
Total for this question: 6
An isolated air-filled parallel-plate capacitor has area $2.50\times10^{-2}\,\text{m}^2$ , separation $1.00\,\text{mm}$ and potential difference $180\,\text{V}$ . The separation is increased to $1.50\,\text{mm}$ and a dielectric of relative permittivity $3.20$ fills the gap. Determine the new potential difference and explain its change.
[5 marks]
Total for this question: 5
A dielectric-filled parallel-plate capacitor has overlap area $2.00\times10^{-2}\,\text{m}^2$ . The micrometer used to set the separation has an unnoticed zero error, so every true plate separation exceeds its nominal value by a constant $d_0$ . The measured capacitances are $632\,\text{pF}$ at nominal separation $0.500\,\text{mm}$ and $295\,\text{pF}$ at $1.30\,\text{mm}$ . Determine the relative permittivity and $d_0$ .
[6 marks]
Total for this question: 6
A dielectric-filled parallel-plate capacitor remains connected to a $75.0\,\text{V}$ supply. Initially its overlap area is $1.50\times10^{-2}\,\text{m}^2$ , plate separation is $0.600\,\text{mm}$ and relative permittivity is $2.80$ . The overlap area is reduced by $20.0\%$ while the separation is increased by $25.0\%$ , the dielectric continuing to fill the gap completely,. Determine the initial and final capacitances and the charge that flows off either plate.
[6 marks]
Total for this question: 6
A variable-gap pressure sensor contains a dielectric-filled parallel-plate capacitor of overlap area $1.60\times10^{-2}\,\text{m}^2$ . Its readout circuit gives a graph of measured capacitance $C$ against reciprocal plate separation $1/d$ with gradient $4.25\times10^{-13}\,\text{F m}$ and vertical intercept $18.0\,\text{pF}$ . Determine the dielectric's relative permittivity, explain the intercept, and predict the measured capacitance and stored charge at separation $0.750\,\text{mm}$ and potential difference $90.0\,\text{V}$ .
[6 marks]
Total for this question: 6

3.7.4.3 · Energy stored by a capacitor

Explanation

Energy stored by a capacitor is the work done separating charge and is $E=\tfrac12QV=\tfrac12CV^2=Q^2/(2C)$ .
On a graph of potential difference against charge, energy is the area under the line; for a linear capacitor the triangular area gives the factor $1/2$ .
The formula should be selected to match the quantities held fixed.
If voltage changes, energy released is the difference between initial and final stored energies, not $Q\Delta V$ with one unchanged charge value.
Examiners expect the squared quantity to be retained, graph axes to be checked, and energy changes to be calculated from two complete energy values.

A potential-difference against charge graph whose triangular area is the capacitor’s stored energy.

Worked example

A $220\,\mu\text{F}$ capacitor is charged to $12\,\text{V}$ . Calculate its stored energy.

1.Convert $C=220\times10^{-6}\,\text{F}$ .
2.Use $E=\tfrac12CV^2$ .
3. $E=\tfrac12(220\times10^{-6})(12)^2$ .

Answer: $E=1.58\times10^{-2}\,\text{J}$ .

Common mistakes

Don't omit the factor $1/2$ in the capacitor-energy equation.
Don't forget to square voltage in $E=\tfrac12CV^2$ .
Don't use $Q\Delta V$ as the released energy while charge is changing.

Exam tip

For a discharge between two voltages, calculate $E_i-E_f$ using $\tfrac12CV^2$ .

Tier 1 · Easy

A capacitor stores $2.0\,\text{mC}$ at a potential difference of $9.0\,\text{V}$ . Calculate its stored energy.
[2 marks]
Total for this question: 2
A capacitor stores energy $E$ at potential difference $V$ . Its capacitance is unchanged when the potential difference is raised to $2V$ . Express the new stored energy in terms of $E$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

The potential difference across a $150\,\mu\text{F}$ capacitor falls from $20.0\,\text{V}$ to $8.0\,\text{V}$ . Determine the energy transferred from the capacitor.
[3 marks]
Total for this question: 3
A capacitor holds charge $4.00\,\text{mC}$ and stores $3.20\times10^{-2}\,\text{J}$ . Determine its capacitance and potential difference.
[3 marks]
Total for this question: 3
A graph of energy stored by a capacitor against $V^2$ is a straight line with gradient $4.00\times10^{-4}\,\text{J V}^{-2}$ . Determine the capacitance and the charge stored at $12.0\,\text{V}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $330\,\mu\text{F}$ capacitor is charged from zero to $24.0\,\text{V}$ through a resistor by an ideal constant-voltage supply. Calculate the work done by the supply, the final stored energy, and the mean power dissipated in the resistor if charging takes $0.800\,\text{s}$ .
[5 marks]
Total for this question: 5
A $680\,\mu\text{F}$ photoflash capacitor is initially at $320\,\text{V}$ . One flash transfers $29.0\,\text{J}$ from the capacitor. Determine the potential difference immediately after the flash and the percentage of the initial stored energy transferred.
[5 marks]
Total for this question: 5
The potential difference across a single $4.00\,\mu\text{F}$ capacitor falls from $600\,\text{V}$ to $200\,\text{V}$ . Calculate the initial and final stored energies, and hence determine the energy released and the percentage of the initial energy released.
[5 marks]
Total for this question: 5
For a fixed capacitor, a graph of potential difference against charge is a straight line through the origin and the point $(6.84\,\text{mC},38.0\,\text{V})$ . Use the graph to determine the capacitance and the work done in charging the capacitor to this point. Calculate the additional work required to increase its charge from $2.16\,\text{mC}$ to $6.84\,\text{mC}$ .
[6 marks]
Total for this question: 6
A $1.35\,\text{mF}$ capacitor initially holds charge $21.6\,\text{mC}$ . It is discharged completely through a small motor that raises a $0.160\,\text{kg}$ load through $64.0\,\text{mm}$ . Determine the initial potential difference, the initial stored energy and the efficiency of the lifting process.
[5 marks]
Total for this question: 5

3.7.4.4 · Capacitor charge and discharge

Explanation

The time constant of a resistor–capacitor circuit is $\tau=RC$ . During discharge, $Q=Q_0e^{-t/RC}$ , with corresponding equations for $V$ and current magnitude; after one time constant each is $0.368$ of its initial value and $T_{1/2}=0.69RC$ .
During charging, $Q=Q_0(1-e^{-t/RC})$ : charge and voltage rise to limiting values while current falls.
Graph gradients give rates and area under an $I$ – $t$ graph gives transferred charge.
Required practical 9 includes charge and discharge graphs and a log-linear plot whose gradient is $-1/RC$ .
Examiners expect a time constant from data, not a claim that discharge completes after one $RC$ .

Exponential charging and discharging curves with one time constant marked.

Worked example

A $220\,\mu\text{F}$ capacitor discharges through $47\,\text{k}\Omega$ . Find the time constant and half-life.

1. $\tau=RC=(47\times10^3)(220\times10^{-6})=10.34\,\text{s}$ .
2.Use $T_{1/2}=0.69RC$ .
3. $T_{1/2}=0.69(10.34)=7.13\,\text{s}$ .

Answer: $\tau=10.3\,\text{s}$ and $T_{1/2}=7.13\,\text{s}$ .

Common mistakes

Don't treat one time constant as complete charge or discharge.
Don't use $e^{-t/RC}$ for charging without the leading $1-$ .
Don't report the log-plot gradient as $RC$ instead of $-1/RC$ .

Exam tip

On a discharge graph, one time constant is the time to fall to $37\%$ of the initial value.

Tier 1 · Easy

A $220\,\mu\text{F}$ capacitor discharges through a $47\,\text{k}\Omega$ resistor. Calculate the time constant and state the fraction of initial charge remaining after this time.
[2 marks]
Total for this question: 2
Calculate the time taken for the charge to halve when a $150\,\mu\text{F}$ capacitor discharges through $33\,\text{k}\Omega$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

A capacitor initially stores $6.00\,\text{mC}$ and discharges through an $82.0\,\text{k}\Omega$ resistor. Its capacitance is $100\,\mu\text{F}$ . Determine the charge remaining after $12.0\,\text{s}$ .
[3 marks]
Total for this question: 3
During charging, a capacitor reaches $63\%$ of its final potential difference after $8.60\,\text{s}$ . It is connected through a $39.0\,\text{k}\Omega$ resistor. Estimate the capacitance.
[3 marks]
Total for this question: 3
During discharge through a $39.0\,\text{k}\Omega$ resistor, current falls from $3.00\,\text{mA}$ to $1.20\,\text{mA}$ in $7.20\,\text{s}$ . Determine the capacitance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

During discharge, charge falls from $8.0\,\text{mC}$ to $1.6\,\text{mC}$ in $14.0\,\text{s}$ . The capacitance is $120\,\mu\text{F}$ . Determine the time constant, the resistance, and the charge after one further time constant. State the gradient of a graph of $\ln Q$ against $t$ .
[6 marks]
Total for this question: 6
An initially uncharged $220\,\mu\text{F}$ capacitor charges through $56.0\,\text{k}\Omega$ from a $15.0\,\text{V}$ supply. Determine the time taken to reach $90.0\%$ of its final charge, the charge then and the charging current then.
[5 marks]
Total for this question: 5
At the start of a capacitor discharge, the current is $4.80\,\text{mA}$ and the tangent to the current-time graph has gradient $-0.960\,\text{mA s}^{-1}$ . The initial potential difference is $12.0\,\text{V}$ . The area under the $I$ – $t$ graph equals the charge that flows, $Q_0$ . Determine the resistance, capacitance and $Q_0$ .
[5 marks]
Total for this question: 5
For a capacitor discharge, a graph of $\ln(V/1\,\text{V})$ against time has gradient $-0.0860\,\text{s}^{-1}$ and vertical intercept $2.708$ . The capacitance is $330\,\mu\text{F}$ . Determine the initial potential difference, the resistance and the time at which the potential difference reaches $2.00\,\text{V}$ .
[5 marks]
Total for this question: 5
A capacitor charges from zero through a $47.0\,\text{k}\Omega$ resistor. Its potential differences are $5.00\,\text{V}$ at $4.00\,\text{s}$ and $8.00\,\text{V}$ at $8.00\,\text{s}$ . Determine the supply potential difference, the time constant, the capacitance and the initial charging current.
[6 marks]
Total for this question: 6

3.7.5.1 · Magnetic flux density

Explanation

A straight current-carrying wire perpendicular to a magnetic field experiences force $F=BIl$ .
Magnetic flux density $B$ is therefore force per unit current per unit length, and one tesla gives $1\,\text{N}$ on a $1\,\text{m}$ wire carrying $1\,\text{A}$ at right angles.
Fleming’s left-hand rule links field, conventional current and force directions; electron flow is opposite to conventional current.
Required practical 10 uses a top-pan balance to investigate how force varies with $B$ , $I$ and $l$ , converting a mass-reading change using $F=\Delta mg$ .
Examiners expect the perpendicular condition, active wire length and mass conversion to be stated.

A current-carrying wire perpendicular to a magnetic field, showing the magnetic-force direction.

Worked example

A $0.080\,\text{m}$ wire carries $3.2\,\text{A}$ perpendicular to a $0.45\,\text{T}$ field. Calculate the force.

1.The field is perpendicular, so use $F=BIl$ .
2. $F=(0.45)(3.2)(0.080)$ .
3.Round to two significant figures.

Answer: $F=0.12\,\text{N}$ .

Common mistakes

Don't use the total wire length rather than the length inside the field.
Don't substitute a balance change in grams directly as force.
Don't use electron-flow direction in Fleming’s left-hand rule.

Exam tip

State that the field and current are perpendicular before using the full $F=BIl$ expression.

Tier 1 · Easy

A $0.080\,\text{m}$ wire carries $3.2\,\text{A}$ perpendicular to a magnetic field of flux density $0.45\,\text{T}$ . Calculate the force on the wire.
[2 marks]
Total for this question: 2
Define the tesla using the force on a current-carrying wire.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A horizontal wire of length $0.120\,\text{m}$ carries $4.00\,\text{A}$ perpendicular to a magnetic field. Switching on the current changes the balance reading by $6.50\,\text{g}$ . Determine the magnetic flux density.
[3 marks]
Total for this question: 3
A straight wire lies in the plane of the page. Conventional current is vertically upwards and a uniform magnetic field is directed from left to right. Use Fleming's left-hand rule to determine the force direction. State the new direction if the current is reversed.
[3 marks]
Total for this question: 3
A horizontal wire has mass per unit length $0.0180\,\text{kg m}^{-1}$ and lies perpendicular to a uniform $0.400\,\text{T}$ magnetic field directed into the page. Determine the current that makes magnetic force support the wire, and state the conventional-current direction.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In a balance experiment, a perpendicular wire of active length $0.0850\,\text{m}$ gives a straight-line graph of balance-reading change against current with gradient $1.75\,\text{g A}^{-1}$ . Determine the flux density and predict the reading change at $5.20\,\text{A}$ . Explain what a non-zero vertical intercept would suggest.
[5 marks]
Total for this question: 5
A $32.0\,\text{mm}$ wire carries $2.00\,\text{A}$ perpendicular to a magnetic field. Reversing the current changes a balance reading from $18.60\,\text{g}$ to $16.20\,\text{g}$ . Determine the magnetic flux density and explain why using the difference removes a zero offset.
[4 marks]
Total for this question: 4
A horizontal wire of mass $6.00\,\text{g}$ carries $3.00\,\text{A}$ from left to right. A length $80.0\,\text{mm}$ is in a $0.300\,\text{T}$ field into the page, while an adjacent length $50.0\,\text{mm}$ is in a $0.500\,\text{T}$ field out of the page. Determine the resultant magnetic force and the total upward force required from the supports. Calculate the change in support force when the current is reversed.
[5 marks]
Total for this question: 5
A straight wire of active length $0.180\,\text{m}$ and mass per unit length $0.0120\,\text{kg m}^{-1}$ rests on a smooth plane inclined at $25.0^\circ$ to the horizontal. A uniform $0.280\,\text{T}$ magnetic field is perpendicular to the wire and arranged so that its magnetic force is up the slope. Determine the current needed for equilibrium and the normal contact force on the wire.
[5 marks]
Total for this question: 5
A bent wire lies in the plane of the page in a uniform $0.350\,\text{T}$ field directed into the page. It carries $2.50\,\text{A}$ first east through a $0.120\,\text{m}$ section and then north through a $0.0800\,\text{m}$ section. Determine the magnetic force on each section and the magnitude and direction of the resultant force on the bent wire.
[5 marks]
Total for this question: 5

3.7.5.2 · Moving charges in a magnetic field

Explanation

A charge moving perpendicular to a magnetic field experiences force $F=BQv$ , perpendicular to both velocity and field. Fleming’s left-hand rule gives the force on positive conventional current; the direction reverses for a negative particle.
Since magnetic force is always perpendicular to motion, it does no work and changes direction without changing speed or kinetic energy.
Equating magnetic and centripetal forces gives $r=mv/(BQ)$ for a circular path.
This underlies devices such as the cyclotron, whose non-relativistic frequency is $f=BQ/(2\pi m)$ and is independent of orbit radius and speed.
Examiners expect correct three-dimensional directions, charge magnitude and radius rather than diameter.

A positive charged particle following a circular path in a magnetic field into the page.

Worked example

A proton travels at $4.0\times10^6\,\text{m s}^{-1}$ perpendicular to a $0.25\,\text{T}$ field. Find the force and orbit radius.

1. $F=BQv=(0.25)(1.60\times10^{-19})(4.0\times10^6)=1.6\times10^{-13}\,\text{N}$ .
2.Use $r=mv/(BQ)$ .
3. $r=(1.67\times10^{-27})(4.0\times10^6)/[(0.25)(1.60\times10^{-19})]=0.167\,\text{m}$ .

Answer: $F=1.6\times10^{-13}\,\text{N}$ and $r=0.167\,\text{m}$ .

Common mistakes

Don't use the same force direction for positive and negative particles.
Don't claim magnetic force increases the particle’s kinetic energy.
Don't substitute path diameter for $r$ .

Exam tip

Find the force direction for a positive charge first, then reverse it only if the particle is negative.

Tier 1 · Easy

A proton moves at $4.0\times10^6\,\text{m s}^{-1}$ perpendicular to a $0.25\,\text{T}$ magnetic field. Calculate the magnetic force magnitude.
[2 marks]
Total for this question: 2
Explain why a magnetic field cannot change the speed of a charged particle when magnetic force is the only force acting.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A proton of speed $3.20\times10^6\,\text{m s}^{-1}$ enters a uniform $0.480\,\text{T}$ magnetic field perpendicular to the field. Determine the radius of its path. Use $m_p=1.67\times10^{-27}\,\text{kg}$ .
[3 marks]
Total for this question: 3
A charged particle moves at $4.50\times10^6\,\text{m s}^{-1}$ in a circular path of radius $84.0\,\text{mm}$ perpendicular to a $0.320\,\text{T}$ magnetic field. Determine its specific charge $Q/m$ .
[3 marks]
Total for this question: 3
Positive ions X and Y travel from left to right into the same uniform magnetic field directed into the page, with equal speeds perpendicular to the field. X has charge $+e$ and path radius $0.120\,\text{m}$ ; Y has charge $+2e$ and radius $0.180\,\text{m}$ . Determine $m_Y/m_X$ and state the initial direction in which both paths curve.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An alpha particle of mass $6.64\times10^{-27}\,\text{kg}$ and charge $+3.20\times10^{-19}\,\text{C}$ reaches radius $0.450\,\text{m}$ in a cyclotron with magnetic flux density $0.800\,\text{T}$ . Determine its speed, kinetic energy in $\text{MeV}$ , and cyclotron frequency.
[6 marks]
Total for this question: 6
A proton accelerates from rest through $1.80\,\text{kV}$ and then enters a $0.350\,\text{T}$ magnetic field perpendicular to its velocity. Determine the radius of its semicircular path and the time spent in the field. Use $m_p=1.67\times10^{-27}\,\text{kg}$ .
[5 marks]
Total for this question: 5
A proton and an alpha particle have the same non-relativistic kinetic energy. They enter the same uniform magnetic field at the same point and in the same direction, with each velocity perpendicular to the field. Deduce the ratios of their orbit radii and periods. Hence state their relative positions when the proton first completes one orbit. Take the alpha-particle mass as $4m_p$ and its charge as $+2e$ .
[5 marks]
Total for this question: 5
A singly charged positive ion is accelerated from rest through $2.50\,\text{kV}$ and enters a $0.450\,\text{T}$ magnetic field perpendicular to its velocity. Its circular path has radius $32.0\,\text{mm}$ . Determine the ion's mass, speed and period of revolution. Identify the likely ion using $1\,\text{u}=1.66\times10^{-27}\,\text{kg}$ .
[6 marks]
Total for this question: 6
A proton starts near the centre of a cyclotron with magnetic flux density $0.720\,\text{T}$ and leaves at orbit radius $0.420\,\text{m}$ . The potential difference across the gap is $3.50\,\text{kV}$ each time it crosses. Determine the exit speed, kinetic energy in $\text{MeV}$ , approximate number of gap crossings and acceleration time, neglecting the initial energy and time spent crossing the gap.
[6 marks]
Total for this question: 6

3.7.5.3 · Magnetic flux and flux linkage

Explanation

Magnetic flux is $\Phi=BA$ when flux density is normal to area $A$ , and is measured in webers. For a coil of $N$ turns, flux linkage counts the flux through every turn.
If $\theta$ is the angle between the field and the coil’s normal, $N\Phi=BAN\cos\theta$ . Flux linkage is maximum when the normal is parallel to the field and zero when the coil plane is parallel to the field.
A negative sign indicates flux opposite to the chosen positive normal.
Required practical 11 uses a search coil and oscilloscope while varying angle.
Examiners expect the angle to the normal, not the plane, and clear distinction between flux and flux linkage.

A tilted coil showing angle theta between its normal and the magnetic field.

Worked example

A $200$ -turn coil of area $1.5\times10^{-2}\,\text{m}^2$ is in a $0.32\,\text{T}$ field with its normal at $60^{\circ}$ to the field. Find the flux linkage.

1.Use $N\Phi=BAN\cos\theta$ .
2.Substitute $(0.32)(1.5\times10^{-2})(200)\cos60^{\circ}$ .
3.Evaluate in weber-turns.

Answer: $N\Phi=0.48\,\text{Wb-turn}$ .

Common mistakes

Don't use the angle between the field and the coil plane.
Don't omit the number of turns when calculating flux linkage.
Don't state that flux is maximum when the field lies in the coil plane.

Exam tip

Draw the coil normal explicitly before choosing the angle in $BAN\cos\theta$ .

Tier 1 · Easy

Magnetic flux passes normally through a flat surface of area $1.5\times10^{-2}\,\text{m}^2$ . The flux density is $0.32\,\text{T}$ . Calculate the flux.
[2 marks]
Total for this question: 2
State the difference between magnetic flux through one turn and magnetic flux linkage for an $N$ -turn coil.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $240$ -turn coil of area $3.50\times10^{-3}\,\text{m}^2$ is in a $0.180\,\text{T}$ field. The field makes an angle of $35.0^\circ$ with the normal to the coil. Determine the flux linkage.
[3 marks]
Total for this question: 3
A $420$ -turn coil of area $2.20\times10^{-3}\,\text{m}^2$ has flux linkage $0.0750\,\text{Wb turns}$ in a $0.280\,\text{T}$ field. Determine the angle between the field and the coil normal.
[3 marks]
Total for this question: 3
Coil A has $240$ turns and area $1.50\times10^{-3}\,\text{m}^2$ in a $0.320\,\text{T}$ field at $60.0^\circ$ to its normal. Coil B is beside it at the same angle and has $120$ turns and twice the area. Determine the flux linkage of coil A and the ratio of the flux linkages $(N\Phi)_B/(N\Phi)_A$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $500$ -turn search coil of area $8.0\times10^{-4}\,\text{m}^2$ is in a uniform $0.120\,\text{T}$ field. Its chosen normal is rotated from $20.0^\circ$ to $110^\circ$ relative to the field. Calculate the initial and final flux linkages and their signed change.
[5 marks]
Total for this question: 5
A $350$ -turn rectangular coil measures $60.0\,\text{mm}$ by $40.0\,\text{mm}$ . Initially $55\%$ of its area overlaps a uniform $0.420\,\text{T}$ field, with the coil normal at $25.0^\circ$ to the field. Determine the initial flux linkage and its increase when the whole coil enters the field without rotating.
[5 marks]
Total for this question: 5
A $250$ -turn search coil of area $2.40\times10^{-3}\,\text{m}^2$ is rotated through $180^\circ$ from one field-normal orientation to the opposite orientation. The measured change in flux linkage is $0.240\,\text{Wb turns}$ . Determine the magnetic flux density. Each flux-linkage reading has uncertainty $\pm0.003\,\text{Wb turns}$ . Compare the percentage uncertainty in the measured change with that for a $90^\circ$ rotation, and deduce which rotation is better for measuring the field.
[5 marks]
Total for this question: 5
A $320$ -turn coil of area $1.80\times10^{-3}\,\text{m}^2$ rotates uniformly in a magnetic field. Its flux-linkage graph is sinusoidal with amplitude $0.138\,\text{Wb turns}$ and period $40.0\,\text{ms}$ . Determine the magnetic flux density and the signed flux linkage $5.00\,\text{ms}$ after a positive maximum. Calculate the angle turned by the coil normal in this time.
[5 marks]
Total for this question: 5
A flat $180$ -turn coil of area $1.20\times10^{-2}\,\text{m}^2$ has its chosen normal perpendicular to two adjacent field regions. One region covers $60.0\%$ of the coil and has flux density $+0.350\,\text{T}$ ; the other covers the remaining area and has flux density $-0.200\,\text{T}$ . Determine the net flux linkage. The coil is then turned through $180^\circ$ without changing its position; determine the final linkage and signed change.
[5 marks]
Total for this question: 5

3.7.5.4 · Electromagnetic induction

Explanation

Faraday’s law states that induced emf magnitude equals the rate of change of flux linkage, $|\varepsilon|=|\Delta(N\Phi)/\Delta t|$ . Lenz’s law gives direction: the induced effect opposes the change in flux that produced it, not necessarily the original field.
A straight conductor cutting field lines develops an emf, while a stationary coil in a steady field does not.
A current requires a complete circuit.
For a coil rotating uniformly, $\varepsilon=BAN\omega\sin\omega t$ and peak emf is $BAN\omega$ .
Examiners expect the relevant change in linkage and time interval, a direction justified by the change, and qualitative interpretation of simple induction experiments.

A magnet moving towards a coil, changing its magnetic flux linkage and inducing an emf.

Worked example

A $250$ -turn coil’s flux per turn falls from $6.0\times10^{-4}\,\text{Wb}$ to $1.0\times10^{-4}\,\text{Wb}$ in $0.020\,\text{s}$ . Find the mean induced emf.

1.Find the flux change per turn: $|\Delta\Phi|=5.0\times10^{-4}\,\text{Wb}$ .
2.Use $|\varepsilon|=N|\Delta\Phi|/\Delta t$ .
3. $|\varepsilon|=250(5.0\times10^{-4})/0.020$ .

Answer: $|\varepsilon|=6.25\,\text{V}$ .

Common mistakes

Don't use flux change per turn but omit $N$ .
Don't say the induced field opposes the original field in every case.
Don't claim a steady flux through a stationary coil induces a continuous emf.

Exam tip

For Lenz’s law, name the change being opposed: increasing or decreasing flux linkage.

Tier 1 · Easy

State Faraday's law and Lenz's law for electromagnetic induction.
[2 marks]
Total for this question: 2
A bar magnet is held stationary inside a coil connected to a voltmeter. Explain why the voltmeter reads zero and state one change that would produce a reading.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The magnetic flux through each turn of a $250$ -turn coil decreases uniformly from $3.2\,\text{mWb}$ to $0.80\,\text{mWb}$ in $40\,\text{ms}$ . Determine the induced emf magnitude and state how Lenz's law fixes its polarity.
[3 marks]
Total for this question: 3
A straight conductor of length $0.280\,\text{m}$ moves at $4.50\,\text{m s}^{-1}$ perpendicular to a $0.750\,\text{T}$ magnetic field. Determine the induced emf across its ends.
[3 marks]
Total for this question: 3
A square $120$ -turn coil of side $80.0\,\text{mm}$ is pulled completely out of a uniform $0.350\,\text{T}$ field normal to its plane at constant speed $0.250\,\text{m s}^{-1}$ . One side remains parallel to the field boundary. Determine the mean induced emf while the coil leaves the field.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $180$ -turn coil of area $6.50\times10^{-3}\,\text{m}^2$ rotates at $50.0\,\text{Hz}$ in a uniform $0.240\,\text{T}$ field. At $t=0$ its flux linkage is maximum and positive. Calculate the peak emf and the emf magnitude at $t=2.50\,\text{ms}$ . Explain the origin and direction of the emf using Faraday's and Lenz's laws.
[5 marks]
Total for this question: 5
A conducting rod of length $0.350\,\text{m}$ slides along parallel rails at constant speed $3.20\,\text{m s}^{-1}$ through a perpendicular magnetic field of flux density $0.640\,\text{T}$ . The total circuit resistance is $2.80\,\Omega$ . Determine the charge that flows and the thermal energy transferred while the rod moves $0.480\,\text{m}$ .
[4 marks]
Total for this question: 4
The uniform field normal to a $200$ -turn coil of area $4.00\times10^{-3}\,\text{m}^2$ changes steadily from $+0.150\,\text{T}$ to $-0.0500\,\text{T}$ in $80.0\,\text{ms}$ . The coil forms a circuit of resistance $4.00\,\Omega$ . Determine the induced emf, current and charge transferred. State the direction of the field produced by the induced current relative to the original $0.150\,\text{T}$ field.
[5 marks]
Total for this question: 5
A $600$ -turn search coil of area $2.50\times10^{-3}\,\text{m}^2$ is pulled completely out of a field normal to the coil. The induced-emf pulse rises linearly from zero to $4.80\,\text{V}$ in $10.0\,\text{ms}$ and then falls linearly to zero over the next $20.0\,\text{ms}$ . Use the pulse to determine the initial magnetic flux density and explain how its polarity is set.
[5 marks]
Total for this question: 5
A conducting rod of length $0.240\,\text{m}$ starts from rest and accelerates at $2.50\,\text{m s}^{-2}$ along rails in a uniform $0.550\,\text{T}$ field perpendicular to the circuit. The total resistance is $1.80\,\Omega$ . At $0.800\,\text{s}$ , determine the induced emf, current, magnetic force and the mechanical power needed to overcome magnetic braking. Show that this power equals the rate of thermal energy transfer.
[6 marks]
Total for this question: 6

3.7.5.5 · Alternating currents

Explanation

For sinusoidal current and voltage, $I_{\mathrm{rms}}=I_0/\sqrt2$ and $V_{\mathrm{rms}}=V_0/\sqrt2$ , where subscript $0$ denotes peak value. Peak-to-peak is twice the peak.
An rms current has the same mean heating effect in a resistor as a direct current of that value; mains voltage is quoted as rms.
An oscilloscope can display ac waveforms, act as an ac or dc voltmeter, and measure period and frequency.
Amplitude is vertical divisions times volts per division; period is horizontal divisions times time per division.
Examiners expect the $\sqrt2$ relationships only for sinusoids and careful separation of peak, rms and peak-to-peak values.

A sinusoidal voltage waveform showing its peak value and period.

Worked example

A mains supply is $230\,\text{V rms}$ . Calculate its peak and peak-to-peak voltages.

1. $V_0=\sqrt2V_{\mathrm{rms}}=\sqrt2(230)=325\,\text{V}$ .
2.Peak-to-peak voltage is $2V_0$ .
3. $V_{\mathrm{p-p}}=650\,\text{V}$ .

Answer: Peak $=325\,\text{V}$ and peak-to-peak $=650\,\text{V}$ .

Common mistakes

Don't treat the quoted mains value as a peak voltage.
Don't divide rms by $\sqrt2$ when finding peak.
Don't use the sinusoidal rms formula for a non-sinusoidal trace.

Exam tip

Label the requested value as rms, peak or peak-to-peak before applying any factor.

Tier 1 · Easy

A sinusoidal voltage has peak value $12.0\,\text{V}$ . Calculate its rms value.
[1 mark]
Total for this question: 1
State what is meant by the rms value of an alternating current.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sinusoidal trace spans $6.4$ horizontal divisions per cycle at $0.50\,\text{ms div}^{-1}$ . Its peak-to-peak height is $5.6$ vertical divisions at $2.0\,\text{V div}^{-1}$ . Determine the frequency, peak voltage and rms voltage.
[4 marks]
Total for this question: 4
A sinusoidal voltage has peak value $18.0\,\text{V}$ across a $9.00\,\Omega$ resistor. Determine the rms current and show that the mean power is half the peak instantaneous power.
[3 marks]
Total for this question: 3
A sinusoidal current through a $12.0\,\Omega$ resistor has peak-to-peak value $6.40\,\text{A}$ and frequency $50.0\,\text{Hz}$ . Determine the mean power and the energy transferred in one cycle.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A resistive heater rated at $1.80\,\text{kW}$ operates from a sinusoidal $230\,\text{V rms}$ supply. Determine the peak and peak-to-peak supply voltages, the rms and peak currents, and the energy transferred in $12.0\,\text{min}$ .
[6 marks]
Total for this question: 6
A $110\,\text{V rms}$ , $50.0\,\text{Hz}$ sinusoidal signal is displayed on an oscilloscope set to $50.0\,\text{V div}^{-1}$ and $2.00\,\text{ms div}^{-1}$ . Determine the peak-to-peak height and the width of one cycle in divisions. A $+15.0\,\text{V}$ dc offset is then added; state the displacement of the trace centre.
[5 marks]
Total for this question: 5
A resistor has the same mean heating effect when connected to a $24.0\,\text{V}$ dc supply as when connected to a sinusoidal ac supply. Calculate the peak and peak-to-peak voltages of the ac supply. An oscilloscope screen is eight vertical divisions high and its available settings are $5.0$ , $10.0$ and $20.0\,\text{V div}^{-1}$ . Select the most sensitive setting that displays the complete ac trace and calculate its peak-to-peak height in divisions.
[5 marks]
Total for this question: 5
Oscilloscope time cursors show that a sinusoidal voltage crosses zero upwards at $2.15\,\text{ms}$ and reaches the next positive maximum at $3.60\,\text{ms}$ . A voltage cursor gives this maximum as $+14.8\,\text{V}$ . Determine the period, frequency, negative peak, peak-to-peak and rms voltages, and the time at which the trace next crosses zero downwards.
[6 marks]
Total for this question: 6
Two sinusoidal voltage signals are displayed on separate oscilloscope channels. Successive upward zero crossings of channel X occur at $1.45\,\text{ms}$ and $7.85\,\text{ms}$ . The corresponding upward zero crossing of channel Y occurs at $3.05\,\text{ms}$ . Voltage cursors give peak-to-peak values of $17.6\,\text{V}$ for X and $8.80\,\text{V}$ for Y. Determine the frequency, the time lag and phase angle by which Y lags X, and the rms voltage of each signal.
[6 marks]
Total for this question: 6

3.7.5.6 · The operation of a transformer

Explanation

For an ideal transformer, $N_s/N_p=V_s/V_p$ . Power is conserved, so increasing secondary voltage reduces secondary current.
Real efficiency is $I_sV_s/(I_pV_p)$ . Losses arise from winding resistance, eddy currents, hysteresis and flux leakage.
Laminating and insulating the core restricts eddy-current loops, while a suitable soft magnetic core reduces hysteresis.
In power transmission, transformers raise voltage so the same power uses a smaller current, reducing line loss $I^2R$ , then lower voltage for consumers.
Examiners expect turn and voltage ratios in matching order, separate input and output power calculations, named loss mechanisms and a transmission-loss calculation using line current and resistance.

A transformer with primary and secondary coils linked by a laminated magnetic core.

Worked example

A transformer has $N_p=500$ , $N_s=100$ and $V_p=230\,\text{V}$ . It supplies $10\,\text{A}$ at $92\%$ efficiency. Find $V_s$ and $I_p$ .

1. $V_s=V_pN_s/N_p=230(100/500)=46\,\text{V}$ .
2. $P_{out}=I_sV_s=10(46)=460\,\text{W}$ , so $P_{in}=460/0.92=500\,\text{W}$ .
3. $I_p=P_{in}/V_p=500/230=2.17\,\text{A}$ .

Answer: $V_s=46\,\text{V}$ and $I_p=2.17\,\text{A}$ .

Common mistakes

Don't reverse only one of the turns or voltage ratios.
Don't assume input and output powers are equal for a stated non-ideal efficiency.
Don't use transmission voltage rather than current in $I^2R$ line loss.

Exam tip

Write primary quantities on one side and secondary quantities on the other before forming either ratio.

Tier 1 · Easy

An ideal transformer has $300$ primary turns and $1200$ secondary turns. The primary voltage is $24\,\text{V}$ . Calculate the secondary voltage.
[2 marks]
Total for this question: 2
A transformer has $600$ primary turns and changes $240\,\text{V}$ to $12.0\,\text{V}$ . Calculate the number of secondary turns.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A transformer takes $1.80\,\text{A}$ from a $230\,\text{V}$ supply and delivers $24.0\,\text{V}$ at $84.0\%$ efficiency. Determine the secondary current and the power dissipated in the transformer.
[4 marks]
Total for this question: 4
Identify two energy-loss mechanisms in a transformer and state one design feature that reduces either loss.
[3 marks]
Total for this question: 3
For a transformer, a graph of secondary rms voltage against secondary turns has gradient $0.0400\,\text{V turn}^{-1}$ when the primary rms voltage is $12.0\,\text{V}$ . The primary takes $0.800\,\text{A}$ while the secondary supplies $1.20\,\text{A}$ at $6.00\,\text{V}$ . Determine the primary turns and efficiency.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A station transmits $2.50\,\text{MW}$ through cables of total resistance $3.20\,\Omega$ . Compare the cable power losses when transmission voltage is $25.0\,\text{kV}$ and $250\,\text{kV}$ . Explain the transformer's role and how core construction reduces eddy-current loss.
[6 marks]
Total for this question: 6
A student connects the primary of a transformer first to a steady $12\,\text{V}$ dc supply and then to a $12\,\text{V rms}$ ac supply. Explain why a sustained secondary potential difference occurs only with the ac supply. For ideal operation, state the effects of doubling the secondary turns on the secondary rms voltage and frequency.
[5 marks]
Total for this question: 5
An ideal transformer sends $600\,\text{kW}$ into transmission cables of total resistance $5.00\,\Omega$ . Cable loss must not exceed $1.50\%$ of the transmitted power. The primary has $320$ turns at $1.50\,\text{kV}$ . Determine the minimum whole number of secondary turns.
[5 marks]
Total for this question: 5
A station supplies $400\,\text{kW}$ at $8.00\,\text{kV}$ to a $96.0\%$ efficient step-up transformer whose primary has $250$ turns. Its secondary voltage is $160\,\text{kV}$ . Power then passes through cables of total resistance $6.00\,\Omega$ and a $94.0\%$ efficient step-down transformer. Determine the step-up secondary turns, cable current and loss, power delivered after the second transformer and overall efficiency.
[6 marks]
Total for this question: 6
A loaded transformer has $870$ primary turns connected to a $225\,\text{V rms}$ supply and draws $2.68\,\text{A}$ . The secondary potential difference is $37.5\,\text{V rms}$ and its current is $13.7\,\text{A}$ . Determine the number of secondary turns and the loaded efficiency using $I_sV_s/(I_pV_p)$ .
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.7.1 · Fields

Tier 1 · Easy

Mark scheme for 3.7.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	A gravitational field always produces attraction between masses, whereas an electric field can produce attraction or repulsion between charges.	Award the mark for a valid contrast: gravitational interaction between masses is always attractive, but the electric interaction depends on the signs of the charges and may be attractive or repulsive.	1
02.1	The tangent gives the direction of the force field at that point.	Award the mark for identifying the tangent as the local field direction, which is also the force direction on a positive test charge or a test mass.	1

Tier 2 · Standard

Mark scheme for 3.7.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Both forces are towards the source; the electric force is towards the negative sphere and the gravitational force is towards the planet. A greater field-line density represents a stronger field.	A positive test charge is attracted to the negative sphere, so its force is directed towards the sphere. A test mass is attracted to the planet, so its force is directed towards the planet. In either diagram, closer field lines indicate a larger field-strength magnitude; the arrows give the vector direction.	3
02.1	No work is done because displacement along the surface is perpendicular to the field, not because the field is zero. The field can be non-zero and is normal to the equipotential surface.	Use $W=Fs\cos\theta$ : along an equipotential, $W=0$ because $\theta=90^\circ$ . Therefore a non-zero field may exist, but its field lines must cross the equipotential surface at right angles.	3
03.1	Gravitational work $+450\,\text{J}$ ; electric work $+0.720\,\text{J}$ ; the work per unit mass and work per unit charge are numerically equal at $+180\,\text{J kg}^{-1}$ and $+180\,\text{J C}^{-1}$ respectively.	For the gravitational transfer, $\Delta V_g=-720-(-900)=+180\,\text{J kg}^{-1}$ , so $W_{\text{external}}=m\Delta V_g=(2.50)(180)=+450\,\text{J}$ . For the electric transfer, $\Delta V=-70.0-(-250)=+180\,\text{V}=+180\,\text{J C}^{-1}$ , so $W_{\text{external}}=q\Delta V=(4.00\times10^{-3})(180)=+0.720\,\text{J}$ . Therefore the two transfers require equal work per unit mass and per unit charge numerically, and both external-work values are positive because potential energy increases.	4

Tier 3 · Hard

Mark scheme for 3.7.1 Tier 3 · Hard
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01.1	At the midpoint the two electric-field vectors cancel, so the resultant field is zero, but the two positive potentials add, so the potential is positive and non-zero. Field lines leave each positive charge, never cross, and meet equipotential surfaces at right angles; moving along an equipotential requires no work.	Treat electric field as a vector: the equal fields at the midpoint point in opposite directions and cancel. Treat potential as a scalar: each positive charge contributes a positive potential, so the contributions add rather than cancel. Field lines point away from positive charges and cannot cross because the field has one direction at a point. Since no work is done along an equipotential, the electric field must be perpendicular to every equipotential surface.	5
02.1	For the masses, the gravitational field vectors cancel but the two negative gravitational potentials add. For the opposite charges, the electric field vectors point from the positive charge towards the negative charge and add, while the equal positive and negative electric potentials cancel.	At the midpoint, each source in a pair is equally distant. Gravitational field is vector and the two attractions are opposite, so resultant $g=0$ ; gravitational potential is scalar and both contributions are negative, so $V_g<0$ . For the charge pair, both electric-field contributions point from the positive charge to the negative charge, so $E\ne0$ . Their scalar potentials have equal magnitudes and opposite signs, so $V=0$ .	5
03.1	Gravitational force $2.00\,\mu\text{N}$ towards the sphere; electric force $5.00\,\mu\text{N}$ , away for $+q$ and towards for $-q$ ; force on the $2m$ , $+2q$ object is $6.00\,\mu\text{N}$ away from the sphere.	Let the gravitational magnitude on mass $m$ be $F_g$ and the electric magnitude on charge $q$ be $F_e$ . Gravity is always towards the sphere. The positive charge is repelled, so $F_e-F_g=3.00\,\mu\text{N}$ . The negative charge is attracted, so $F_e+F_g=7.00\,\mu\text{N}$ . Solving gives $F_e=5.00\,\mu\text{N}$ and $F_g=2.00\,\mu\text{N}$ . Doubling charge doubles the outward electric force to $10.0\,\mu\text{N}$ , while doubling mass doubles the inward gravitational force to $4.00\,\mu\text{N}$ . The resultant is therefore $6.00\,\mu\text{N}$ away from the sphere.	5
04.1	The neutral grain is forced west by gravity. The positive grain has an electric force east and a gravitational force west, so its resultant depends on their magnitudes. The negative grain has both electric and gravitational forces west. Potential is scalar; field direction is towards decreasing potential and is set by the potential gradient, not by whether the potential value is positive or negative.	Gravity acts on all three masses in the gravitational-field direction, west. A positive charge is forced in the electric-field direction, east, so its electric and gravitational forces oppose and their magnitudes must be compared. A negative charge is forced opposite to the electric field, west, so both forces act west. The sign of a scalar potential gives a reference-dependent value at one point; a field is a vector determined by how potential changes with position, with field lines normal to equipotentials towards lower potential.	5
05.1	Both patterns are radial outside the sphere, weaken with distance and have concentric spherical equipotentials crossed at right angles by field lines. Gravitational lines point inwards, whereas electric lines point outwards from the positive sphere. Field lines never cross, and wider spacing represents a weaker field. No work is done by either field when a test mass or positive test charge moves along one equipotential.	Outside either spherical source the field behaves as though the source were concentrated at the centre, so field lines are radial and their density falls with distance. Gravity attracts a test mass, giving inward arrows; the positive sphere repels a positive test charge, giving outward arrows. One point cannot have two field directions, so field lines cannot cross. Equipotentials are concentric spheres and are perpendicular to the field because displacement along one has no field component: $W=Fs\cos90^\circ=0$ . Both strengths decrease with distance, shown by increasing line spacing.	6

3.7.2.1 · Newton's law

Tier 1 · Easy

Mark scheme for 3.7.2.1 Tier 1 · Easy
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01.1	$1.8\times10^{18}\,\text{N}$ , attractive	Use $F=\dfrac{Gm_1m_2}{r^2}$ . Hence $F=\dfrac{(6.67\times10^{-11})(6.0\times10^{22})(4.0\times10^{20})}{(3.0\times10^7)^2}=1.78\times10^{18}\,\text{N}$ . To two significant figures this is $1.8\times10^{18}\,\text{N}$ , and gravity makes the force attractive.	2
02.1	$1.20\times10^{20}\,\text{kg}$	Rearrange $F=Gm_1m_2/r^2$ to $m_2=Fr^2/(Gm_1)$ . Hence $m_2=(2.94\times10^{17})(3.50\times10^7)^2/[(6.67\times10^{-11})(4.50\times10^{22})]=1.20\times10^{20}\,\text{kg}$ to three significant figures.	2

Tier 2 · Standard

Mark scheme for 3.7.2.1 Tier 2 · Standard
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01.1	Force $6.53\times10^3\,\text{N}$ ; acceleration $7.68\,\text{m s}^{-2}$ , independent of satellite mass.	The force is $F=\dfrac{GMm}{r^2}=\dfrac{(6.67\times10^{-11})(5.97\times10^{24})(850)}{(7.20\times10^6)^2}=6.53\times10^3\,\text{N}$ . Then $a=F/m=(6.53\times10^3)/850=7.68\,\text{m s}^{-2}$ . Substituting Newton's law into $a=F/m$ cancels the satellite mass: $a=GM/r^2$ .	4
02.1	$2.82\%$	Rearrange Newton's law to $G_{\text{exp}}=Fr^2/(m_1m_2)$ . This gives $G_{\text{exp}}=(5.08\times10^{-8})(0.180)^2/[(4.00)(6.00)]=6.858\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$ . Therefore the percentage difference is $\|6.858-6.67\|/6.67\times100=2.82\%$ .	3
03.1	Second mass $3.00\times10^5\,\text{kg}$ ; acceleration of the first body $5.00\times10^{-6}\,\text{m s}^{-2}$ towards the second point mass.	Newton's law gives $m_2=Fr^2/(Gm_1)=(4.00\times10^{-2})(2.00)^2/[(6.67\times10^{-11})(8.00\times10^3)]=3.00\times10^5\,\text{kg}$ . For the first body, $a_1=F/m_1=(4.00\times10^{-2})/(8.00\times10^3)=5.00\times10^{-6}\,\text{m s}^{-2}$ , directed towards the second point mass.	4

Tier 3 · Hard

Mark scheme for 3.7.2.1 Tier 3 · Hard
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01.1	Separation $5.0\times10^8\,\text{m}$ ; new force $5.6\times10^{20}\,\text{N}$ .	Rearrange Newton's law to $r=\sqrt{Gm_1m_2/F}$ . Thus $r=\sqrt{(6.67\times10^{-11})(4.0\times10^{24})(7.5\times10^{23})/(8.0\times10^{20})}=5.00\times10^8\,\text{m}$ . The new separation is $1.20r$ . Since $F\propto1/r^2$ , $F_{\text{new}}=8.0\times10^{20}/(1.20)^2=5.56\times10^{20}\,\text{N}$ , or $5.6\times10^{20}\,\text{N}$ to two significant figures.	5
02.1	Force $2.93\times10^{12}\,\text{N}$ ; acceleration of the smaller asteroid $9.17\times10^{-4}\,\text{m s}^{-2}$ .	The centre separation is $r=(5.00+9.00+6.00)\,\text{km}=2.00\times10^4\,\text{m}$ . Treating each spherical asteroid as a point mass at its centre, $F=(6.67\times10^{-11})(3.20\times10^{15})(5.50\times10^{15})/(2.00\times10^4)^2=2.93\times10^{12}\,\text{N}$ . For the smaller asteroid, $a=F/m=(2.9348\times10^{12})/(3.20\times10^{15})=9.17\times10^{-4}\,\text{m s}^{-2}$ .	5
03.1	$7.35\times10^2\,\text{s}$	Initially $F=Gm_1m_2/r^2=(6.67\times10^{-11})(2.00\times10^{12})(3.00\times10^{12})/(3.00\times10^3)^2=4.45\times10^7\,\text{N}$ . Both asteroids accelerate towards each other, so the separation closes at the sum of their accelerations: $a_{\text{rel}}=F/m_1+F/m_2=3.71\times10^{-5}\,\text{m s}^{-2}$ . From $10.0=\tfrac12a_{\text{rel}}t^2$ , $t=\sqrt{20.0/(3.71\times10^{-5})}=7.35\times10^2\,\text{s}$ .	5
04.1	Resultant force $0.222\,\text{N}$ at $36.9^\circ$ north of east; initial acceleration $5.56\times10^{-2}\,\text{m s}^{-2}$ in the same direction.	The eastern mass produces $F_E=GmM_E/r_E^2=(6.67\times10^{-11})(4.00)(6.00\times10^9)/(3.00)^2=0.1778\,\text{N}$ east. The northern mass produces $F_N=(6.67\times10^{-11})(4.00)(8.00\times10^9)/(4.00)^2=0.1334\,\text{N}$ north. Hence $F=\sqrt{(0.1778)^2+(0.1334)^2}=0.2223\,\text{N}$ and $\theta=\tan^{-1}(0.1334/0.1778)=36.9^\circ$ north of east. Finally $a=F/m=0.2223/4.00=5.56\times10^{-2}\,\text{m s}^{-2}$ .	6
05.1	$n=2.00$ (accept $n=1.99$ to $2.01$ ); second mass $20.0\,\text{kg}$ ; the results agree with the inverse-square form of Newton's law within the precision of the data.	For $F\propto r^{-n}$ , $F_1/F_2=(r_2/r_1)^n$ , so $n=\ln(2.90/1.29)/\ln(0.360/0.240)=2.00$ to three significant figures. Newton's law gives $m_2=F_1r_1^2/(Gm_1)=(2.90\times10^{-7})(0.240)^2/[(6.67\times10^{-11})(12.5)]=20.0\,\text{kg}$ . The measured exponent is the predicted value $2$ ; allow the conclusion that the rounded readings are consistent with an inverse-square relationship.	5

3.7.2.2 · Gravitational field strength

Tier 1 · Easy

Mark scheme for 3.7.2.2 Tier 1 · Easy
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01.1	Gravitational field strength is the gravitational force per unit mass on a small test mass at that point.	State both the force-per-unit-mass ratio and the test mass at the point: $g=F/m$ . Gravitational field strength is a vector directed as the force on that test mass.	1
02.1	$1.2\times10^2\,\text{N}$ in the field direction	From $g=F/m$ , $F=mg=(75)(1.6)=120\,\text{N}=1.2\times10^2\,\text{N}$ to two significant figures. The force has the same direction as the gravitational field.	2

Tier 2 · Standard

Mark scheme for 3.7.2.2 Tier 2 · Standard
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01.1	$0.556\,\text{N kg}^{-1}$ towards the planet	Use $g=GM/r^2$ . Substitution gives $g=(6.67\times10^{-11})(4.80\times10^{24})/(2.40\times10^7)^2=0.5558\,\text{N kg}^{-1}$ . Therefore $g=0.556\,\text{N kg}^{-1}$ to three significant figures, directed radially towards the planet.	3
02.1	$6.07\times10^{24}\,\text{kg}$	Use $g=GM/r^2$ and rearrange to $M=gr^2/G$ . Therefore $M=(1.25)(1.80\times10^7)^2/(6.67\times10^{-11})=6.07\times10^{24}\,\text{kg}$ to three significant figures.	3
03.1	$600\,\text{N}$ towards the planet	At $r$ , $g_1=F_1/m_1=900/400=2.25\,\text{N kg}^{-1}$ . Since radial field strength follows $g\propto1/r^2$ , at $1.50r$ the field is $g_2=2.25/(1.50)^2=1.00\,\text{N kg}^{-1}$ . Hence $F_2=m_2g_2=(600)(1.00)=600\,\text{N}$ , directed towards the planet.	3

Tier 3 · Hard

Mark scheme for 3.7.2.2 Tier 3 · Hard
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01.1	$6.34\times10^6\,\text{m}$ above the surface	From $g=GM/r^2$ , the distance from the centre is $r=\sqrt{GM/g}$ . Hence $r=\sqrt{(6.67\times10^{-11})(6.40\times10^{24})/2.40}=1.3337\times10^7\,\text{m}$ . Altitude is not this radial distance: $h=r-R=1.3337\times10^7-7.00\times10^6=6.34\times10^6\,\text{m}$ to three significant figures.	5
02.1	The point is $4.0\times10^7\,\text{m}$ from the larger mass and $2.0\times10^7\,\text{m}$ from the smaller mass.	Let the distance from the larger mass be $x$ , so the distance from the smaller is $6.0\times10^7-x$ . Between the bodies the fields oppose. Set their magnitudes equal: $G(4.8\times10^{24})/x^2=G(1.2\times10^{24})/(6.0\times10^7-x)^2$ . Taking positive square roots gives $2/x=1/(6.0\times10^7-x)$ , so $x=4.0\times10^7\,\text{m}$ .	4
03.1	Radius $6.00\times10^6\,\text{m}$ ; mass $4.32\times10^{24}\,\text{kg}$ ; field strength at $4.00\times10^6\,\text{m}$ altitude $2.88\,\text{N kg}^{-1}$ .	Let the radius be $R$ . Since $g\propto1/r^2$ , $8.00/4.50=(R+2.00\times10^6)^2/R^2$ . Taking the positive square root gives $4/3=(R+2.00\times10^6)/R$ , so $R=6.00\times10^6\,\text{m}$ . Then $M=gR^2/G=(8.00)(6.00\times10^6)^2/(6.67\times10^{-11})=4.32\times10^{24}\,\text{kg}$ . At altitude $4.00\times10^6\,\text{m}$ , the radial distance is $1.00\times10^7\,\text{m}$ , so $g=8.00(6.00/10.0)^2=2.88\,\text{N kg}^{-1}$ .	5
04.1	Resultant field strength $1.98\,\text{N kg}^{-1}$ towards the planet; force on the probe $4.96\times10^2\,\text{N}$ towards the planet.	Between the bodies their fields oppose. The planet produces $g_P=(6.67\times10^{-11})(5.00\times10^{24})/(1.20\times10^7)^2=2.316\,\text{N kg}^{-1}$ . The moon produces $g_M=(6.67\times10^{-11})(8.00\times10^{22})/(4.00\times10^6)^2=0.3335\,\text{N kg}^{-1}$ . Therefore $g=2.316-0.3335=1.982\,\text{N kg}^{-1}$ towards the planet. The probe force is $F=mg=(250)(1.982)=4.96\times10^2\,\text{N}$ .	5
05.1	Surface field strength $3.05\,\text{N kg}^{-1}$ ; probe force $2.23\times10^2\,\text{N}$ ; field-strength uncertainty $7.14\%$ , or $\pm0.218\,\text{N kg}^{-1}$ .	At the surface, $g=GM/R^2=(6.67\times10^{-11})(3.74\times10^{23})/(2.86\times10^6)^2=3.05\,\text{N kg}^{-1}$ . The probe force is $F=mg=(73.0)(3.0498)=2.23\times10^2\,\text{N}$ towards the planet. With the stated addition rule, the percentage uncertainty is $[(0.11/3.74)+2(0.06/2.86)]\times100=7.14\%$ . Hence the absolute uncertainty in $g$ is $(0.0714)(3.0498)=0.218\,\text{N kg}^{-1}$ .	6

3.7.2.3 · Gravitational potential

Tier 1 · Easy

Mark scheme for 3.7.2.3 Tier 1 · Easy
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01.1	The gravitational force attracts a mass, so positive external work is required to move the mass from the point to infinity; the point therefore has lower potential than infinity.	Infinity is assigned $V_g=0$ . A mass at finite distance is gravitationally bound and energy must be added to remove it to infinity, so its potential energy per unit mass, and hence $V_g$ , is negative.	1
02.1	$-3.3\times10^6\,\text{J kg}^{-1}$	Use $V_g=-GM/r$ . Hence $V_g=-(6.67\times10^{-11})(8.0\times10^{23})/(1.6\times10^7)=-3.335\times10^6\,\text{J kg}^{-1}$ , which is $-3.3\times10^6\,\text{J kg}^{-1}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.7.2.3 Tier 2 · Standard
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01.1	$\Delta V_g=+2.7\times10^7\,\text{J kg}^{-1}$ ; external work $+8.5\times10^9\,\text{J}$ .	The initial potential is $V_i=-GM/r_i=-(6.67\times10^{-11})(7.2\times10^{24})/(9.0\times10^6)=-5.34\times10^7\,\text{J kg}^{-1}$ . The final potential is $V_f=-2.67\times10^7\,\text{J kg}^{-1}$ . Therefore $\Delta V_g=V_f-V_i=+2.67\times10^7\,\text{J kg}^{-1}$ and $\Delta W=m\Delta V_g=320(2.67\times10^7)=+8.54\times10^9\,\text{J}$ .	4
02.1	$12.0\,\text{N kg}^{-1}$ directed inwards	The outward potential gradient is $\Delta V_g/\Delta r=[-1.80\times10^7-(-2.10\times10^7)]/(2.50\times10^5)=+12.0\,\text{J kg}^{-1}\text{m}^{-1}$ . Since $g=-\Delta V_g/\Delta r$ , the field is $12.0\,\text{N kg}^{-1}$ in the inward direction.	3
03.1	Radius $4.50\times10^6\,\text{m}$ ; mass $3.04\times10^{24}\,\text{kg}$ .	At radius $r$ , $\|V_g\|=GM/r$ and $g=GM/r^2$ . Dividing gives $\|V_g\|/g=r$ , so $r=(4.50\times10^7)/10.0=4.50\times10^6\,\text{m}$ . Then $M=gr^2/G=(10.0)(4.50\times10^6)^2/(6.67\times10^{-11})=3.04\times10^{24}\,\text{kg}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.7.2.3 Tier 3 · Hard
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01.1	External work $+1.38\times10^{10}\,\text{J}$ ; return speed $7.82\times10^3\,\text{m s}^{-1}$ .	At the inner point $V_A=-GM/r_A=-4.5856\times10^7\,\text{J kg}^{-1}$ ; at the outer point $V_B=-GM/r_B=-1.5285\times10^7\,\text{J kg}^{-1}$ . For the outward move, $\Delta V_g=V_B-V_A=+3.0571\times10^7\,\text{J kg}^{-1}$ , so external work is $m\Delta V_g=450(3.0571\times10^7)=+1.38\times10^{10}\,\text{J}$ . On return, $\Delta V_g=V_A-V_B=-3.0571\times10^7\,\text{J kg}^{-1}$ and conservation of energy gives $\tfrac12mv^2=-m\Delta V_g$ . Thus $v=\sqrt{-2\Delta V_g}=7.82\times10^3\,\text{m s}^{-1}$ .	6
02.1	Potential $-1.53\times10^7\,\text{J kg}^{-1}$ ; external work $+3.84\times10^9\,\text{J}$ .	Add the scalar potentials: $V_g=-G(M_1/r_1+M_2/r_2)=-(6.67\times10^{-11})[(3.0\times10^{24})/(2.0\times10^7)+(8.0\times10^{23})/(1.0\times10^7)]=-1.5341\times10^7\,\text{J kg}^{-1}$ . At infinity $V_g=0$ , so $\Delta V_g=+1.5341\times10^7\,\text{J kg}^{-1}$ . For a slow transfer the external work is $W=m\Delta V_g=(250)(1.5341\times10^7)=+3.84\times10^9\,\text{J}$ to three significant figures.	5
03.1	$9.51\times10^3\,\text{m s}^{-1}$	The increase in potential is the area under the $g$ – $r$ graph. From A to B, $\Delta V_{AB}=\tfrac12(16.0+10.0)(2.50\times10^6)=3.25\times10^7\,\text{J kg}^{-1}$ . From B to C, $\Delta V_{BC}=\tfrac12(10.0+7.00)(1.50\times10^6)=1.275\times10^7\,\text{J kg}^{-1}$ . Thus $\Delta V=4.525\times10^7\,\text{J kg}^{-1}$ . Conservation of energy for zero final speed gives $\tfrac12v^2=\Delta V$ , so $v=\sqrt{2(4.525\times10^7)}=9.51\times10^3\,\text{m s}^{-1}$ .	5
04.1	Gravitational potential energy decreases by $4.00\times10^9\,\text{J}$ ; speed $4.32\times10^3\,\text{m s}^{-1}$ .	The potential change per unit mass is $\Delta V_g=-GM(1/r_f-1/r_i)=-(6.67\times10^{-11})(4.50\times10^{24})[1/(9.00\times10^6)-1/(1.50\times10^7)]=-1.334\times10^7\,\text{J kg}^{-1}$ . Hence the potential-energy decrease has magnitude $m\|\Delta V_g\|=(300)(1.334\times10^7)=4.00\times10^9\,\text{J}$ . The kinetic-energy gain is $0.700m\|\Delta V_g\|$ , so $\tfrac12mv^2=0.700m\|\Delta V_g\|$ and $v=\sqrt{2(0.700)(1.334\times10^7)}=4.32\times10^3\,\text{m s}^{-1}$ .	5
05.1	The potential magnitude halves when radius doubles, supporting $V_g\propto-1/r$ ; each reading gives mass $4.50\times10^{24}\,\text{kg}$ ; external work $6.25\times10^9\,\text{J}$ .	For $V_g=-GM/r$ , doubling $r$ should halve $\|V_g\|$ , as the two readings do. From the inner reading, $M=\|V_g\|r/G=(2.50\times10^7)(1.20\times10^7)/(6.67\times10^{-11})=4.50\times10^{24}\,\text{kg}$ ; the outer reading gives the same value. The slow outward transfer changes potential by $\Delta V_g=-1.25\times10^7-(-2.50\times10^7)=+1.25\times10^7\,\text{J kg}^{-1}$ . Thus $W_{\text{external}}=m\Delta V_g=(500)(1.25\times10^7)=6.25\times10^9\,\text{J}$ .	6

3.7.2.4 · Orbits of planets and satellites

Tier 1 · Easy

Mark scheme for 3.7.2.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$7.9\times10^3\,\text{m s}^{-1}$	The distance travelled in one orbit is $2\pi r$ . Therefore $v=2\pi r/T=2\pi(7.5\times10^6)/(6.0\times10^3)=7.85\times10^3\,\text{m s}^{-1}$ , which is $7.9\times10^3\,\text{m s}^{-1}$ to two significant figures.	2
02.1	A polar orbit is not in the equatorial plane, so the satellite does not remain above one point on the planet's equator.	Award one mark for recognising that a geostationary orbit must be equatorial, which a polar orbit is not, and one mark for the consequence that the satellite does not remain above one fixed point on the equator.	2

Tier 2 · Standard

Mark scheme for 3.7.2.4 Tier 2 · Standard
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01.1	The period is multiplied by $2.02$ .	Newton's law provides the centripetal force: $GMm/r^2=mv^2/r$ . With $v=2\pi r/T$ , this becomes $GMm/r^2=m(4\pi^2r^2/T^2)/r$ . Cancelling $m$ and rearranging gives $T^2=4\pi^2r^3/(GM)$ , so $T^2\propto r^3$ . Therefore $T_{\text{new}}/T_{\text{old}}=(1.60)^{3/2}=2.02$ .	4
02.1	$1.95\times10^{25}\,\text{kg}$	For a circular orbit, $T^2=4\pi^2r^3/(GM)$ . Rearranging gives $M=4\pi^2r^3/(GT^2)$ . Hence $M=4\pi^2(2.20\times10^7)^3/[(6.67\times10^{-11})(1.80\times10^4)^2]=1.95\times10^{25}\,\text{kg}$ .	3
03.1	Gradient $1.50$ ; $T\propto r^{3/2}$ .	The gradient is $[\log_{10}(4.64\times10^4)-\log_{10}(5.80\times10^3)]/[\log_{10}(2.80\times10^7)-\log_{10}(7.00\times10^6)]=\log_{10}8/\log_{10}4=1.50$ . Therefore $\log T=1.50\log r+\text{constant}$ , so $T\propto r^{1.50}=r^{3/2}$ .	3

Tier 3 · Hard

Mark scheme for 3.7.2.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Orbital radius $4.91\times10^7\,\text{m}$ ; altitude $4.21\times10^7\,\text{m}$ ; total energy $-2.04\times10^9\,\text{J}$ .	Newton's gravitational force supplies centripetal force, giving $T^2=4\pi^2r^3/(GM)$ . Convert $T=30\times3600=1.08\times10^5\,\text{s}$ . Then $r=[GMT^2/(4\pi^2)]^{1/3}=4.908\times10^7\,\text{m}$ . The altitude is $r-R=4.908\times10^7-7.0\times10^6=4.21\times10^7\,\text{m}$ . For a circular orbit, $E=-GMm/(2r)=-(6.67\times10^{-11})(6.0\times10^{24})(500)/(2\times4.908\times10^7)=-2.04\times10^9\,\text{J}$ .	6
02.1	Mean density $4.49\times10^3\,\text{kg m}^{-3}$ .	Convert the period to $T=(2.40)(3600)=8.64\times10^3\,\text{s}$ . From $T^2=4\pi^2r^3/(GM)$ , $M=4\pi^2(8.00\times10^6)^3/[(6.67\times10^{-11})(8.64\times10^3)^2]=4.06\times10^{24}\,\text{kg}$ . The planet's volume is $4\pi R^3/3=4\pi(6.00\times10^6)^3/3=9.05\times10^{20}\,\text{m}^3$ . Hence $\rho=M/V=4.49\times10^3\,\text{kg m}^{-3}$ to three significant figures.	5
03.1	Speed increase $2.14\times10^3\,\text{m s}^{-1}$ ; energy transferred $1.07\times10^{10}\,\text{J}$ .	The circular speed is $v_c=\sqrt{GM/r}=\sqrt{(6.67\times10^{-11})(4.80\times10^{24})/(1.20\times10^7)}=5.17\times10^3\,\text{m s}^{-1}$ . The escape speed at the same point is $v_e=\sqrt{2GM/r}=7.30\times10^3\,\text{m s}^{-1}$ . Hence the impulse must add $\Delta v=v_e-v_c=2.14\times10^3\,\text{m s}^{-1}$ . The energy transferred is the kinetic-energy increase, $\Delta E=\tfrac12m(v_e^2-v_c^2)=1.07\times10^{10}\,\text{J}$ .	5
04.1	Total energy increases by $6.76\times10^9\,\text{J}$ ; the period is multiplied by $3.95$ .	For a circular orbit, $E=-GMm/(2r)$ . Hence $\Delta E=GMm(1/r_1-1/r_2)/2=[(6.67\times10^{-11})(5.20\times10^{24})(650)/2][1/(1.00\times10^7)-1/(2.50\times10^7)]=6.76\times10^9\,\text{J}$ . For the same planet, $T\propto r^{3/2}$ , so $T_2/T_1=(2.50)^{3/2}=3.95$ .	5
05.1	Planet mass $4.57\times10^{25}\,\text{kg}$ ; orbital speed $2.91\times10^3\,\text{m s}^{-1}$ ; escape speed $4.11\times10^3\,\text{m s}^{-1}$ (accept $4.11$ – $4.12\times10^3\,\text{m s}^{-1}$ ); total energy per unit mass $-4.23\times10^6\,\text{J kg}^{-1}$ .	Convert $T=(9.00)(86400)=7.776\times10^5\,\text{s}$ . From $T^2=4\pi^2r^3/(GM)$ , $M=4\pi^2(3.60\times10^8)^3/[(6.67\times10^{-11})(7.776\times10^5)^2]=4.57\times10^{25}\,\text{kg}$ . The orbital speed is $v=2\pi r/T=2.91\times10^3\,\text{m s}^{-1}$ . Escape speed at that radius is $v_e=\sqrt{2GM/r}=\sqrt2v=4.11\times10^3\,\text{m s}^{-1}$ . Circular-orbit energy per unit mass is $-GM/(2r)=-v^2/2=-4.23\times10^6\,\text{J kg}^{-1}$ .	6

3.7.3.1 · Coulomb's law

Tier 1 · Easy

Mark scheme for 3.7.3.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.50\times10^{-5}\,\text{N}$ , attractive	Treat air as a vacuum and use $F=k\|Q_1Q_2\|/r^2$ . Thus $F=(8.99\times10^9)(3.0\times10^{-9})(8.0\times10^{-9})/(0.120)^2=1.50\times10^{-5}\,\text{N}$ . The opposite signs make the force attractive.	2
02.1	The new force is $F/4$ .	Coulomb's law gives $F\propto1/r^2$ . Replacing $r$ by $2r$ gives $F_{\text{new}}/F_{\text{old}}=1/(2^2)=1/4$ .	1

Tier 2 · Standard

Mark scheme for 3.7.3.1 Tier 2 · Standard
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01.1	$F_E/F_G=2.27\times10^{39}$	The force magnitudes are $F_E=ke^2/r^2$ and $F_G=Gm_pm_e/r^2$ . Dividing cancels the common $r^2$ : $F_E/F_G=ke^2/(Gm_pm_e)$ . Substitution gives $(8.99\times10^9)(1.60\times10^{-19})^2/[(6.67\times10^{-11})(1.67\times10^{-27})(9.11\times10^{-31})]=2.27\times10^{39}$ .	4
02.1	$2.0\,\text{nC}$	From $F=k\|Q_1Q_2\|/r^2$ , $\|Q_2\|=Fr^2/(k\|Q_1\|)$ . Thus $\|Q_2\|=(3.0\times10^{-5})(0.060)^2/[(8.99\times10^9)(6.0\times10^{-9})]=2.00\times10^{-9}\,\text{C}=2.0\,\text{nC}$ .	3
03.1	Predicted force $1.94\times10^{-4}\,\text{N}$ ; percentage difference $2.26\%$ .	For unchanged charges, Coulomb's law gives $F\propto1/r^2$ . Hence $F_2=(5.40\times10^{-4})(0.120/0.200)^2=1.944\times10^{-4}\,\text{N}$ . The percentage difference relative to the predicted value is $\|1.90-1.944\|\times10^{-4}/(1.944\times10^{-4})\times100=2.26\%$ .	3

Tier 3 · Hard

Mark scheme for 3.7.3.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.10\times10^{-6}\,\text{N}$ , $24.2^\circ$ above the negative $x$ -axis.	The $+5.0\,\text{nC}$ charge repels the origin charge in the negative $x$ direction: $F_x=-(8.99\times10^9)(2.0\times10^{-9})(5.0\times10^{-9})/(0.300)^2=-9.99\times10^{-7}\,\text{N}$ . The negative charge attracts it in the positive $y$ direction: $F_y=(8.99\times10^9)(2.0\times10^{-9})(4.0\times10^{-9})/(0.400)^2=4.50\times10^{-7}\,\text{N}$ . Hence $F=\sqrt{F_x^2+F_y^2}=1.10\times10^{-6}\,\text{N}$ and $\theta=\tan^{-1}(\|F_y/F_x\|)=24.2^\circ$ above the negative $x$ -axis.	6
02.1	$3.37\times10^{-6}\,\text{N}$ to the right	Both interactions are repulsive. The left charge exerts a rightward force $F_L=(8.99\times10^9)(2.0\times10^{-9})(7.0\times10^{-9})/(0.180)^2=3.88\times10^{-6}\,\text{N}$ . The right charge exerts a leftward force $F_R=(8.99\times10^9)(2.0\times10^{-9})(5.0\times10^{-9})/(0.420)^2=5.10\times10^{-7}\,\text{N}$ . Therefore the resultant is $F_L-F_R=3.37\times10^{-6}\,\text{N}$ to the right.	4
03.1	Left-hand charge $+25.0\,\text{nC}$ ; right-hand charge $+1.00\times10^2\,\text{nC}$ ; midpoint force towards the left.	At the zero-force point the opposing force magnitudes are equal, so $kqQ_L/(0.200)^2=kqQ_R/(0.400)^2$ and therefore $Q_R=4Q_L$ . At the midpoint, both separations are $0.300\,\text{m}$ and the larger right-hand charge produces the larger force, so the resultant is towards the left. Its magnitude is $1.50\times10^{-5}=kq(Q_R-Q_L)/(0.300)^2=3kqQ_L/(0.300)^2$ . Thus $Q_L=(1.50\times10^{-5})(0.300)^2/[3(8.99\times10^9)(2.00\times10^{-9})]=2.50\times10^{-8}\,\text{C}=25.0\,\text{nC}$ and $Q_R=1.00\times10^2\,\text{nC}$ .	5
04.1	$2.28\times10^{-6}\,\text{N}$ , directed $19.1^\circ$ above the horizontal to the right, pointing away from the base of the triangle.	The positive lower-left charge repels the upper charge with $F_L=(8.99\times10^9)(6.00\times10^{-9})(3.00\times10^{-9})/(0.250)^2=2.589\times10^{-6}\,\text{N}$ along $60^\circ$ above the horizontal. The negative lower-right charge attracts it with $F_R=(8.99\times10^9)(4.00\times10^{-9})(3.00\times10^{-9})/(0.250)^2=1.726\times10^{-6}\,\text{N}$ along $60^\circ$ below the horizontal. Thus $F_x=(F_L+F_R)\cos60^\circ=2.157\times10^{-6}\,\text{N}$ and $F_y=(F_L-F_R)\sin60^\circ=7.47\times10^{-7}\,\text{N}$ . Therefore $F=2.28\times10^{-6}\,\text{N}$ and $\theta=\tan^{-1}(F_y/F_x)=19.1^\circ$ above the horizontal to the right.	6
05.1	Charge magnitude $5.17\times10^{-8}\,\text{C}$ on each sphere; $3.23\times10^{11}$ elementary charges.	The centre separation is $r=2L\sin\theta=2(0.300)\sin8.00^\circ=8.35\times10^{-2}\,\text{m}$ . For one sphere in equilibrium, $F_E=mg\tan\theta=(2.50\times10^{-3})(9.81)\tan8.00^\circ=3.45\times10^{-3}\,\text{N}$ . Coulomb's law gives $q=\sqrt{F_Er^2/k}=\sqrt{(3.45\times10^{-3})(8.35\times10^{-2})^2/(8.99\times10^9)}=5.17\times10^{-8}\,\text{C}$ . Hence $q/e=(5.17\times10^{-8})/(1.60\times10^{-19})=3.23\times10^{11}$ .	5

3.7.3.2 · Electric field strength

Tier 1 · Easy

Mark scheme for 3.7.3.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Electric field strength is the force per unit positive test charge at that point.	Use $E=F/Q$ and specify a positive test charge so that the definition also fixes the direction of the field vector.	1
02.1	$3.00\times10^4\,\text{N C}^{-1}$ to the left	For a proton, $Q=+e=+1.60\times10^{-19}\,\text{C}$ . From $E=F/Q$ , $E=(4.80\times10^{-15})/(1.60\times10^{-19})=3.00\times10^4\,\text{N C}^{-1}$ . A positive charge is forced in the field direction, so the field is to the left.	2

Tier 2 · Standard

Mark scheme for 3.7.3.2 Tier 2 · Standard
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01.1	$E=4.00\times10^4\,\text{V m}^{-1}$ ; $F=6.40\times10^{-15}\,\text{N}$ ; $a=7.03\times10^{15}\,\text{m s}^{-2}$ .	Convert $V=1.80\times10^3\,\text{V}$ and $d=45.0\times10^{-3}\,\text{m}$ . Then $E=V/d=4.00\times10^4\,\text{V m}^{-1}$ . The force magnitude is $F=eE=(1.60\times10^{-19})(4.00\times10^4)=6.40\times10^{-15}\,\text{N}$ . Using $m_e=9.11\times10^{-31}\,\text{kg}$ , $a=F/m_e=7.03\times10^{15}\,\text{m s}^{-2}$ ; its direction is opposite to the field.	4
02.1	$0.16\,\text{m}$	For a point charge, $E=kQ/r^2$ , so $r=\sqrt{kQ/E}$ . Therefore $r=\sqrt{(8.99\times10^9)(5.5\times10^{-9})/(2.0\times10^3)}=0.157\,\text{m}$ , which is $0.16\,\text{m}$ to two significant figures.	3
03.1	Potential difference $1.80\times10^3\,\text{V}$ ; field strength $1.00\times10^5\,\text{V m}^{-1}$ ; the destination plate is at higher potential.	For a slow transfer, $W=q\Delta V$ , so $\Delta V=(4.32\times10^{-6})/(2.40\times10^{-9})=1.80\times10^3\,\text{V}$ . The uniform field magnitude is $E=V/d=(1.80\times10^3)/(18.0\times10^{-3})=1.00\times10^5\,\text{V m}^{-1}$ . Electric potential increases opposite to the field direction, so the destination plate is at higher potential.	4

Tier 3 · Hard

Mark scheme for 3.7.3.2 Tier 3 · Hard
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01.1	Deflection $0.852\,\text{mm}$ ; angle $1.22^\circ$ .	The force is $F=eE$ , so the transverse acceleration is $a=eE/m_p=(1.60\times10^{-19})(2.50\times10^4)/(1.67\times10^{-27})=2.40\times10^{12}\,\text{m s}^{-2}$ . The transit time is fixed by horizontal motion: $t=L/v_x=0.0800/(3.00\times10^6)=2.67\times10^{-8}\,\text{s}$ . Thus $y=\tfrac12at^2=8.52\times10^{-4}\,\text{m}=0.852\,\text{mm}$ . The transverse exit speed is $v_y=at=6.39\times10^4\,\text{m s}^{-1}$ , so $\theta=\tan^{-1}(v_y/v_x)=1.22^\circ$ .	6
02.1	Charge $-3.20\times10^{-19}\,\text{C}$ ; two excess electrons.	The weight is downward, so the electric force must be upward. Because the field points downward, the drop is negative. Equilibrium gives $\|Q\|E=mg$ , hence $\|Q\|=(6.00\times10^{-15})(9.81)/(1.84\times10^5)=3.20\times10^{-19}\,\text{C}$ . Thus $Q=-3.20\times10^{-19}\,\text{C}$ and $\|Q\|/e=(3.20\times10^{-19})/(1.60\times10^{-19})=2$ excess electrons.	5
03.1	Field $1.50\times10^3\,\text{V m}^{-1}$ east; electron force $2.40\times10^{-16}\,\text{N}$ west; electron acceleration $2.63\times10^{14}\,\text{m s}^{-2}$ west.	The probes span a small gap with a $130-118=12\,\text{V}$ potential difference across it, so $E=V/d=12/(8.00\times10^{-3})=1.50\times10^3\,\text{V m}^{-1}$ , directed from high to low potential (east). The electron force has magnitude $F=eE=(1.60\times10^{-19})(1.50\times10^3)=2.40\times10^{-16}\,\text{N}$ and points west because the electron is negative. Its acceleration is $a=F/m_e=(2.40\times10^{-16})/(9.11\times10^{-31})=2.63\times10^{14}\,\text{m s}^{-2}$ west.	5
04.1	Balancing potential difference $1.10\times10^3\,\text{V}$ ; acceleration $15.7\,\text{m s}^{-2}$ downwards; fall time $2.26\times10^{-2}\,\text{s}$ .	The charge magnitude is $\|q\|=3e=4.80\times10^{-19}\,\text{C}$ . For equilibrium, $\|q\|E=mg$ , so $E=(4.50\times10^{-15})(9.81)/(4.80\times10^{-19})=9.20\times10^4\,\text{V m}^{-1}$ and $V=Ed=(9.20\times10^4)(12.0\times10^{-3})=1.10\times10^3\,\text{V}$ . Reversing polarity makes the electric force downward; at $0.600V$ it has magnitude $0.600mg$ . Therefore $a=(mg+0.600mg)/m=1.600g=15.7\,\text{m s}^{-2}$ downward. From $s=\tfrac12at^2$ , $t=\sqrt{2(4.00\times10^{-3})/15.696}=2.26\times10^{-2}\,\text{s}$ .	6
05.1	Field strength $7.16\times10^4\,\text{V m}^{-1}$ ; charge $+1.07\times10^{-11}\,\text{C}$ ; tension $3.29\times10^{-6}\,\text{N}$ .	The uniform field strength is $E=V/d=(2.65\times10^3)/(37.0\times10^{-3})=7.16\times10^4\,\text{V m}^{-1}$ . In equilibrium, the horizontal electric force and vertical weight give $qE=mg\tan\theta$ . Hence $q=(3.26\times10^{-7})(9.81)\tan13.5^\circ/(7.162\times10^4)=1.07\times10^{-11}\,\text{C}$ . The field points towards the lower-potential plate and the bead deflects in that direction, so its charge is positive. The tension is $T=mg/\cos13.5^\circ=3.29\times10^{-6}\,\text{N}$ .	6

3.7.3.3 · Electric potential

Tier 1 · Easy

Mark scheme for 3.7.3.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$+180\,\text{V}$	Use $V=kQ/r$ . Therefore $V=(8.99\times10^9)(4.00\times10^{-9})/0.200=179.8\,\text{V}$ . The source charge is positive, so $V=+180\,\text{V}$ to three significant figures.	2
02.1	$+8.0\times10^{-7}\,\text{J}$	At infinity $V_i=0$ , so $\Delta V=+320\,\text{V}$ . For a slow transfer, $W_{\text{external}}=q\Delta V=(2.5\times10^{-9})(320)=+8.0\times10^{-7}\,\text{J}$ .	2

Tier 2 · Standard

Mark scheme for 3.7.3.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	External work $+6.0\times10^{-7}\,\text{J}$ ; potential energy increases.	The potential change is $\Delta V=V_f-V_i=-80-120=-200\,\text{V}$ . For a slow transfer, external work equals the change in potential energy: $\Delta W=q\Delta V=(-3.0\times10^{-9})(-200)=+6.0\times10^{-7}\,\text{J}$ . The positive sign means the potential energy increases.	4
02.1	$+135\,\text{V}$	Electric potential is scalar, so add contributions with their signs: $V=k(Q_1/r_1+Q_2/r_2)=(8.99\times10^9)[(6.0\times10^{-9})/0.300-(2.0\times10^{-9})/0.400]=134.85\,\text{V}$ . To three significant figures, $V=+135\,\text{V}$ .	3
03.1	Potential at $0.150\,\text{m}$ is $+420\,\text{V}$ ; potential at $0.300\,\text{m}$ is $+210\,\text{V}$ ; external work $-4.20\times10^{-7}\,\text{J}$ (accept $-4.19\times10^{-7}\,\text{J}$ from intermediate rounding, or from an unrounded $1/4\pi\varepsilon_0=8.988\times10^9\,\text{m F}^{-1}$ ).	For a point charge, $V=kQ/r$ . Thus $V_1=(8.99\times10^9)(7.00\times10^{-9})/0.150=419.53\,\text{V}$ and $V_2=(8.99\times10^9)(7.00\times10^{-9})/0.300=209.77\,\text{V}$ . Therefore $\Delta V=209.77-419.53=-209.76\,\text{V}$ and, for a slow transfer, $W_{\text{external}}=q\Delta V=(2.00\times10^{-9})(-209.76)=-4.20\times10^{-7}\,\text{J}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.7.3.3 Tier 3 · Hard
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01.1	The point is $0.480\,\text{m}$ from the positive charge; $E=1.56\times10^3\,\text{N C}^{-1}$ towards the negative charge; external work $0\,\text{J}$ .	Let the point be distance $x$ from the positive charge, so it is $0.600-x$ from the negative charge. Set the scalar potential to zero: $k[8.0\times10^{-9}/x-2.0\times10^{-9}/(0.600-x)]=0$ . Hence $8(0.600-x)=2x$ , giving $x=0.480\,\text{m}$ . At this point both field vectors point from the positive charge towards the negative charge, so they add: $E=k[8.0\times10^{-9}/(0.480)^2+2.0\times10^{-9}/(0.120)^2]=1.56\times10^3\,\text{N C}^{-1}$ . Since $V_{\infty}=V_{\text{point}}=0$ , $\Delta W=q\Delta V=0$ even though the field is not zero.	6
02.1	The point is $0.450\,\text{m}$ from the $+9.0\,\text{nC}$ charge; total potential $+360\,\text{V}$ ; external work $-5.75\times10^{-17}\,\text{J}$ .	Let the distance from $+9.0\,\text{nC}$ be $x$ . Equal potential contributions require $9/x=4/(0.650-x)$ , giving $x=0.450\,\text{m}$ and the other distance $0.200\,\text{m}$ . Each contribution is $(8.99\times10^9)(9.0\times10^{-9})/0.450=179.8\,\text{V}$ , so $V=+359.6\,\text{V}=+360\,\text{V}$ to three significant figures. For an electron, $W_{\text{external}}=q\Delta V=(-1.60\times10^{-19})(359.6)=-5.75\times10^{-17}\,\text{J}$ .	5
03.1	$2.65\times10^5\,\text{m s}^{-1}$	At $x=0.100\,\text{m}$ , $V_A=(8.99\times10^9)[(5.00\times10^{-9})/0.100-(3.00\times10^{-9})/0.400]=382\,\text{V}$ . At $x=0.300\,\text{m}$ , $V_B=(8.99\times10^9)[(5.00\times10^{-9})/0.300-(3.00\times10^{-9})/0.200]=15.0\,\text{V}$ . The decrease in proton potential energy becomes kinetic energy: $\tfrac12m_pv^2=e(V_A-V_B)=(1.60\times10^{-19})(382.075-14.983)$ . Hence $v=2.65\times10^5\,\text{m s}^{-1}$ to three significant figures.	5
04.1	Potential at the centre $+127\,\text{V}$ ; electron speed $6.68\times10^6\,\text{m s}^{-1}$ , assuming no non-electrostatic work or energy loss.	Each corner is distance $r=0.400/\sqrt2=0.2828\,\text{m}$ from the centre. Potential is scalar, so $V=k\Sigma Q/r=(8.99\times10^9)[(8.00+3.00-5.00-2.00)\times10^{-9}]/0.2828=127.1\,\text{V}$ . For an electron, the potential-energy change from infinity is $\Delta U=qV=-eV$ , so the kinetic-energy gain is $eV$ . Hence $\tfrac12m_ev^2=eV$ and $v=\sqrt{2(1.60\times10^{-19})(127.1)/(9.11\times10^{-31})}=6.68\times10^6\,\text{m s}^{-1}$ . This assumes conservation of mechanical energy with no other work or losses.	5
05.1	Source charge $+15.1\,\text{nC}$ ; potentials $+680\,\text{V}$ and $+302\,\text{V}$ ; external work $+9.44\times10^{-7}\,\text{J}$ (accept $+9.45\times10^{-7}\,\text{J}$ from the rounded $302\,\text{V}$ ).	For a point charge, $E=kQ/r^2$ , so $Q=Er^2/k=(3.40\times10^3)(0.200)^2/(8.99\times10^9)=1.51\times10^{-8}\,\text{C}$ . Since $V=kQ/r=Er^2/r$ , $V_1=(3.40\times10^3)(0.200)=680\,\text{V}$ and $V_2=680(0.200/0.450)=302\,\text{V}$ . Thus $\Delta V=302.2-680=-377.8\,\text{V}$ . For a slow transfer, $W_{\text{external}}=q\Delta V=(-2.50\times10^{-9})(-377.8)=+9.44\times10^{-7}\,\text{J}$ .	6

3.7.4.1 · Capacitance

Tier 1 · Easy

Mark scheme for 3.7.4.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$300\,\mu\text{F}$	Use $C=Q/V$ . Convert $Q=3.6\times10^{-3}\,\text{C}$ , so $C=(3.6\times10^{-3})/12=3.0\times10^{-4}\,\text{F}=300\,\mu\text{F}$ .	2
02.1	Capacitance is the charge stored on either plate per unit potential difference across the capacitor.	State the ratio $C=Q/V$ , where $Q$ is the magnitude of charge on one plate and $V$ is the potential difference between the plates.	1

Tier 2 · Standard

Mark scheme for 3.7.4.1 Tier 2 · Standard
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01.1	Capacitance $300\,\mu\text{F}$ ; charge $8.10\,\text{mC}$ .	The gradient of a $Q$ - $V$ graph is $C$ . Therefore $C=(4.80\times10^{-3})/16.0=3.00\times10^{-4}\,\text{F}=300\,\mu\text{F}$ . At $27.0\,\text{V}$ , $Q=CV=(3.00\times10^{-4})(27.0)=8.10\times10^{-3}\,\text{C}=8.10\,\text{mC}$ .	3
02.1	Additional charge $2.64\,\text{mC}$ ; mean current $0.440\,\text{mA}$ .	The extra charge is $\Delta Q=C\Delta V=(220\times10^{-6})(17.0-5.0)=2.64\times10^{-3}\,\text{C}=2.64\,\text{mC}$ . The mean current is $I=\Delta Q/\Delta t=(2.64\times10^{-3})/6.00=4.40\times10^{-4}\,\text{A}=0.440\,\text{mA}$ .	3
03.1	Potential difference $6.00\,\text{V}$ ; capacitance of B $550\,\mu\text{F}$ .	For A, $V=Q_A/C_A=(1.32\times10^{-3})/(220\times10^{-6})=6.00\,\text{V}$ . The capacitors have the same potential difference, so $C_B=Q_B/V=(3.30\times10^{-3})/6.00=5.50\times10^{-4}\,\text{F}=550\,\mu\text{F}$ .	3

Tier 3 · Hard

Mark scheme for 3.7.4.1 Tier 3 · Hard
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01.1	Charge $2.31\,\text{mC}$ ; potential difference $4.91\,\text{V}$ .	Initially $Q_i=CV=(470\times10^{-6})(9.0)=4.23\times10^{-3}\,\text{C}$ . The removed electron charge magnitude is $\Delta Q=ne=(1.2\times10^{16})(1.60\times10^{-19})=1.92\times10^{-3}\,\text{C}$ . Hence $Q_f=Q_i-\Delta Q=2.31\times10^{-3}\,\text{C}$ . Since $C$ is unchanged, $V_f=Q_f/C=(2.31\times10^{-3})/(470\times10^{-6})=4.91\,\text{V}$ .	5
02.1	Capacitance $300\,\mu\text{F}$ ; sensor zero offset $+0.500\,\text{mC}$ .	For a graph of measured charge against potential difference, the gradient is the capacitance: $C=\Delta Q/\Delta V=(3.50-1.10)\,\text{mC}/(10.0-2.00)\,\text{V}=0.300\,\text{mC V}^{-1}=300\,\mu\text{F}$ . Writing $Q_{\text{measured}}=CV+Q_0$ and using the first pair gives $Q_0=1.10-(0.300)(2.00)=+0.500\,\text{mC}$ .	4
03.1	Capacitance $375\,\mu\text{F}$ ; percentage uncertainty $1.7\%$ ; absolute uncertainty $6.3\,\mu\text{F}$ .	The stored charge is $Q=It$ , so $C=It/V=(24.0\times10^{-6})(150.0)/9.60=3.75\times10^{-4}\,\text{F}=375\,\mu\text{F}$ . Using the stated convention, the percentage uncertainty is $[(0.2/24.0)+(0.5/150.0)+(0.05/9.60)]\times100=1.69\%$ , or $1.7\%$ to two significant figures. The absolute uncertainty is $(0.016875)(375)=6.33\,\mu\text{F}$ , or $6.3\,\mu\text{F}$ .	5
04.1	Capacitance $3.13\times10^{-4}\,\text{F}=313\,\mu\text{F}$ ; charge magnitude $6.00\,\text{mC}$ at $19.2\,\text{V}$ ; new potential difference $13.4\,\text{V}$ .	On a graph of $V$ against $Q$ , the gradient is $V/Q=1/C$ . Thus $C=1/(3.20\times10^3)=3.125\times10^{-4}\,\text{F}=313\,\mu\text{F}$ . At $19.2\,\text{V}$ , $Q=V/(V/Q)=19.2/(3.20\times10^3)=6.00\times10^{-3}\,\text{C}$ . When $Q=4.20\times10^{-3}\,\text{C}$ , $V=Q/C=(4.20\times10^{-3})(3.20\times10^3)=13.44\,\text{V}=13.4\,\text{V}$ .	5
05.1	Initial plate-charge magnitude $6.31\,\text{mC}$ ; initial potential difference across B $17.3\,\text{V}$ ; additional charge $4.13\,\text{mC}$ ; charge increase $65.4\%$ (accept $65.5\%$ from the rounded charges).	For A, $Q_A=C_AV_A=(295\times10^{-6})(21.4)=6.313\times10^{-3}\,\text{C}$ . Initially B has this same charge magnitude, so $V_B=Q_A/C_B=(6.313\times10^{-3})/(365\times10^{-6})=17.3\,\text{V}$ . At $28.6\,\text{V}$ , B must hold $Q_f=C_BV_f=(365\times10^{-6})(28.6)=10.439\times10^{-3}\,\text{C}$ . Therefore $\Delta Q=10.439-6.313=4.126\,\text{mC}$ and the increase relative to B's initial charge is $(4.126/6.313)\times100=65.4\%$ .	5

3.7.4.2 · Parallel plate capacitor

Tier 1 · Easy

Mark scheme for 3.7.4.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.84\times10^{-10}\,\text{F}$	For air take $\varepsilon_r=1$ . Convert $d=1.20\times10^{-3}\,\text{m}$ and use $C=\varepsilon_0A/d$ . Thus $C=(8.85\times10^{-12})(2.50\times10^{-2})/(1.20\times10^{-3})=1.84\times10^{-10}\,\text{F}$ .	2
02.1	$C_{\text{new}}/C_{\text{old}}=2/3$	For the same dielectric, $C\propto A/d$ . The change factor is therefore $2/3$ .	1

Tier 2 · Standard

Mark scheme for 3.7.4.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Capacitance $6.77\times10^{-10}\,\text{F}$ ; charge $8.12\times10^{-8}\,\text{C}$ .	Convert $d=8.00\times10^{-4}\,\text{m}$ . With the dielectric, $C=\varepsilon_0\varepsilon_rA/d=(8.85\times10^{-12})(3.40)(1.80\times10^{-2})/(8.00\times10^{-4})=6.77\times10^{-10}\,\text{F}$ . The supply keeps $V$ fixed, so $Q=CV=(6.77\times10^{-10})(120)=8.12\times10^{-8}\,\text{C}$ .	4
02.1	$2.63$	Rearrange $C=\varepsilon_0\varepsilon_rA/d$ to $\varepsilon_r=Cd/(\varepsilon_0A)$ . Hence $\varepsilon_r=(620\times10^{-12})(0.900\times10^{-3})/[(8.85\times10^{-12})(2.40\times10^{-2})]=2.63$ .	3
03.1	Relative permittivity $3.50$ ; charge $7.56\,\text{nC}$ ; the electric field strengths are equal.	For unchanged geometry, $C$ is proportional to relative permittivity, so $\varepsilon_r=630/180=3.50$ . The charge is $Q=CV=(630\times10^{-12})(12.0)=7.56\times10^{-9}\,\text{C}=7.56\,\text{nC}$ . Both capacitors have the same potential difference and separation, so $E=V/d$ is the same in each.	4

Tier 3 · Hard

Mark scheme for 3.7.4.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Required area $3.77\times10^{-2}\,\text{m}^2$ ; polarisation reduces the field and potential difference for a given free charge.	Rearrange $C=\varepsilon_0\varepsilon_rA/d$ to $A=Cd/(\varepsilon_0\varepsilon_r)$ . Hence $A=(2.00\times10^{-9})(0.750\times10^{-3})/[(8.85\times10^{-12})(4.50)]=3.77\times10^{-2}\,\text{m}^2$ . In the dielectric, permanent polar molecules rotate, or induced dipoles align, so bound surface charges form. Their field opposes the field due to the free plate charges. For a fixed free charge this lowers $V$ , and because $C=Q/V$ , the capacitance is larger.	6
02.1	New potential difference $84.4\,\text{V}$ ; the capacitance becomes $2.13$ times larger because the relative permittivity rises by a factor of $3.20$ while the separation rises by a factor of $1.50$ .	Initially $C_i=\varepsilon_0A/d_i=(8.85\times10^{-12})(2.50\times10^{-2})/(1.00\times10^{-3})=2.2125\times10^{-10}\,\text{F}$ , so $Q=C_iV_i=3.9825\times10^{-8}\,\text{C}$ . The net capacitance ratio is $C_f/C_i=\varepsilon_r(d_i/d_f)=3.20/1.50=2.13$ , not $3.20$ , because the plate separation also increases. Thus $C_f=4.72\times10^{-10}\,\text{F}$ . The isolated capacitor keeps the same free charge, so $V_f=Q/C_f=84.4\,\text{V}$ .	5
03.1	Relative permittivity $2.50$ ; zero error $d_0=0.200\,\text{mm}$ .	The model is $C=\varepsilon A/(d+d_0)$ , so $1/C=(d+d_0)/(\varepsilon A)$ . Subtracting the two equations eliminates $d_0$ : $1/C_2-1/C_1=(d_2-d_1)/(\varepsilon A)$ . Hence $\varepsilon=(1.30-0.500)\times10^{-3}/\{(2.00\times10^{-2})[(295\times10^{-12})^{-1}-(632\times10^{-12})^{-1}]\}=2.21\times10^{-11}\,\text{F m}^{-1}$ . Thus $\varepsilon_r=\varepsilon/\varepsilon_0=2.50$ . Finally, $d_0=\varepsilon A/C_1-d_1=2.00\times10^{-4}\,\text{m}=0.200\,\text{mm}$ .	6
04.1	Initial capacitance $6.20\times10^{-10}\,\text{F}$ ; final capacitance $3.96\times10^{-10}\,\text{F}$ ; charge flowing off either plate $1.67\times10^{-8}\,\text{C}$ .	Initially $C_i=\varepsilon_0\varepsilon_rA/d=(8.85\times10^{-12})(2.80)(1.50\times10^{-2})/(0.600\times10^{-3})=6.195\times10^{-10}\,\text{F}$ . The geometry gives $C_f/C_i=0.800/1.250=0.640$ , so $C_f=3.965\times10^{-10}\,\text{F}$ . The supply holds $V$ constant, hence the decrease in plate-charge magnitude is $\Delta Q=(C_i-C_f)V=(6.195-3.965)\times10^{-10}(75.0)=1.67\times10^{-8}\,\text{C}$ . This charge flows off each plate through the supply circuit.	6
05.1	Relative permittivity $3.00$ ; the intercept represents stray or parasitic capacitance in the sensor and leads; measured capacitance $585\,\text{pF}$ ; charge $52.6\,\text{nC}$ .	From $C=\varepsilon_0\varepsilon_rA(1/d)$ , the ideal gradient is $\varepsilon_0\varepsilon_rA$ . Therefore $\varepsilon_r=(4.25\times10^{-13})/[(8.85\times10^{-12})(1.60\times10^{-2})]=3.00$ . A non-zero intercept is a separation-independent parasitic capacitance from the sensor's leads and readout circuit. At $d=0.750\times10^{-3}\,\text{m}$ , $C_{\text{measured}}=(4.25\times10^{-13})/(0.750\times10^{-3})+18.0\times10^{-12}=5.847\times10^{-10}\,\text{F}=585\,\text{pF}$ . Thus $Q=CV=(5.847\times10^{-10})(90.0)=5.26\times10^{-8}\,\text{C}=52.6\,\text{nC}$ .	6

3.7.4.3 · Energy stored by a capacitor

Tier 1 · Easy

Mark scheme for 3.7.4.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$9.0\,\text{mJ}$	Use $E=\tfrac12QV$ . Thus $E=\tfrac12(2.0\times10^{-3})(9.0)=9.0\times10^{-3}\,\text{J}=9.0\,\text{mJ}$ .	2
02.1	$4E$	For constant capacitance, $E=\tfrac12CV^2$ . Doubling $V$ multiplies $E$ by $2^2=4$ .	1

Tier 2 · Standard

Mark scheme for 3.7.4.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.52\times10^{-2}\,\text{J}$	Energy transferred is the decrease in stored energy: $\Delta E=\tfrac12C(V_i^2-V_f^2)$ . Therefore $\Delta E=\tfrac12(150\times10^{-6})[(20.0)^2-(8.0)^2]=2.52\times10^{-2}\,\text{J}$ .	3
02.1	Capacitance $250\,\mu\text{F}$ ; potential difference $16.0\,\text{V}$ .	Use $E=Q^2/(2C)$ , so $C=Q^2/(2E)=(4.00\times10^{-3})^2/[2(3.20\times10^{-2})]=2.50\times10^{-4}\,\text{F}=250\,\mu\text{F}$ . Then $V=Q/C=(4.00\times10^{-3})/(2.50\times10^{-4})=16.0\,\text{V}$ .	3
03.1	Capacitance $800\,\mu\text{F}$ ; charge $9.60\,\text{mC}$ .	From $E=\tfrac12CV^2$ , the gradient of $E$ against $V^2$ is $C/2$ . Hence $C=2(4.00\times10^{-4})=8.00\times10^{-4}\,\text{F}=800\,\mu\text{F}$ . The charge is $Q=CV=(8.00\times10^{-4})(12.0)=9.60\times10^{-3}\,\text{C}=9.60\,\text{mC}$ .	3

Tier 3 · Hard

Mark scheme for 3.7.4.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Supply work $0.190\,\text{J}$ ; stored energy $0.0950\,\text{J}$ ; mean resistor power $0.119\,\text{W}$ .	The final charge is $Q=CV=(330\times10^{-6})(24.0)=7.92\times10^{-3}\,\text{C}$ . The supply works at fixed voltage, so $W_{\text{supply}}=QV=CV^2=(330\times10^{-6})(24.0)^2=0.19008\,\text{J}$ . The capacitor stores $E=\tfrac12CV^2=0.09504\,\text{J}$ , equal to the triangular area under its $V$ - $Q$ graph. The remainder $0.19008-0.09504=0.09504\,\text{J}$ is dissipated, so mean power is $0.09504/0.800=0.1188\,\text{W}$ .	5
02.1	Final potential difference $131\,\text{V}$ ; $83.3\%$ of the initial energy is transferred.	Initially $E_i=\tfrac12CV_i^2=\tfrac12(680\times10^{-6})(320)^2=34.816\,\text{J}$ . The final energy is $E_f=34.816-29.0=5.816\,\text{J}$ . From $E_f=\tfrac12CV_f^2$ , $V_f=\sqrt{2E_f/C}=130.8\,\text{V}$ , or $131\,\text{V}$ to three significant figures. The transferred percentage is $(29.0/34.816)\times100=83.3\%$ .	5
03.1	Initial energy $0.720\,\text{J}$ ; final energy $0.0800\,\text{J}$ ; energy released $0.640\,\text{J}$ , which is $88.9\%$ of the initial energy.	Initially $E_i=\tfrac12CV_i^2=\tfrac12(4.00\times10^{-6})(600)^2=0.720\,\text{J}$ . Finally $E_f=\tfrac12CV_f^2=\tfrac12(4.00\times10^{-6})(200)^2=0.0800\,\text{J}$ . The energy released is $E_i-E_f=0.720-0.0800=0.640\,\text{J}$ . The percentage released is $(0.640/0.720)\times100=88.9\%$ .	5
04.1	Capacitance $180\,\mu\text{F}$ ; total work $0.130\,\text{J}$ ; additional work $0.117\,\text{J}$ .	For a fixed capacitor, $C=Q/V=(6.84\times10^{-3})/38.0=1.80\times10^{-4}\,\text{F}=180\,\mu\text{F}$ . The work done is the triangular area under the straight $V$ – $Q$ graph: $E_f=\tfrac12Q_fV_f=\tfrac12(6.84\times10^{-3})(38.0)=0.12996\,\text{J}$ . At $Q_i=2.16\,\text{mC}$ , $V_i=Q_i/C=12.0\,\text{V}$ and $E_i=\tfrac12(2.16\times10^{-3})(12.0)=0.01296\,\text{J}$ . The additional work is $E_f-E_i=0.117\,\text{J}$ .	6
05.1	Initial potential difference $16.0\,\text{V}$ ; initial stored energy $0.173\,\text{J}$ ; lifting efficiency $58.1\%$ .	Initially $V=Q/C=(21.6\times10^{-3})/(1.35\times10^{-3})=16.0\,\text{V}$ . The stored energy is $E=Q^2/(2C)=(21.6\times10^{-3})^2/[2(1.35\times10^{-3})]=0.1728\,\text{J}$ . The useful gravitational-potential-energy gain is $mgh=(0.160)(9.81)(64.0\times10^{-3})=0.10045\,\text{J}$ . Therefore $\eta=(0.10045/0.1728)\times100=58.1\%$ .	5

3.7.4.4 · Capacitor charge and discharge

Tier 1 · Easy

Mark scheme for 3.7.4.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Time constant $10.3\,\text{s}$ ; fraction remaining $0.368$ .	Convert $R=47\times10^3\,\Omega$ and $C=220\times10^{-6}\,\text{F}$ . Then $\tau=RC=(47\times10^3)(220\times10^{-6})=10.34\,\text{s}$ . At $t=\tau$ , $Q/Q_0=e^{-1}=0.368$ .	2
02.1	Time to halve $=3.4\,\text{s}$	Use $T_{1/2}=0.69RC$ . Thus $T_{1/2}=0.69(33\times10^3)(150\times10^{-6})=3.42\,\text{s}$ , which is $3.4\,\text{s}$ to two significant figures.	2

Tier 2 · Standard

Mark scheme for 3.7.4.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.39\,\text{mC}$	The time constant is $RC=(82.0\times10^3)(100\times10^{-6})=8.20\,\text{s}$ . Use $Q=Q_0e^{-t/RC}$ : $Q=(6.00\times10^{-3})e^{-12.0/8.20}=1.3887\times10^{-3}\,\text{C}=1.39\,\text{mC}$ .	3
02.1	$221\,\mu\text{F}$	A charging capacitor reaches $1-e^{-1}=0.632$ of its final potential difference after one time constant, so $RC\approx8.60\,\text{s}$ . Therefore $C=8.60/(39.0\times10^3)=2.21\times10^{-4}\,\text{F}=221\,\mu\text{F}$ to three significant figures.	3
03.1	$201\,\mu\text{F}$	For discharge, $I/I_0=e^{-t/RC}$ . Thus $RC=-t/\ln(I/I_0)=-7.20/\ln(1.20/3.00)=7.8578\,\text{s}$ . Therefore $C=7.8578/(39.0\times10^3)=2.01\times10^{-4}\,\text{F}=201\,\mu\text{F}$ .	3

Tier 3 · Hard

Mark scheme for 3.7.4.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$\tau=8.70\,\text{s}$ ; $R=72.5\,\text{k}\Omega$ ; later charge $0.589\,\text{mC}$ ; gradient $-0.115\,\text{s}^{-1}$ .	Use $Q/Q_0=e^{-t/\tau}$ . Here $1.6/8.0=0.200$ , so $\ln(0.200)=-14.0/\tau$ and $\tau=14.0/\ln 5=8.70\,\text{s}$ . Since $\tau=RC$ , $R=8.70/(120\times10^{-6})=7.25\times10^4\,\Omega=72.5\,\text{k}\Omega$ . One further time constant multiplies charge by $e^{-1}$ , giving $1.6/e=0.589\,\text{mC}$ . Taking logs gives $\ln Q=\ln Q_0-t/RC$ , so the gradient is $-1/\tau=-0.115\,\text{s}^{-1}$ .	6
02.1	Time $28.4\,\text{s}$ ; charge $2.97\,\text{mC}$ ; current $26.8\,\mu\text{A}$ .	For charging, $Q/Q_0=1-e^{-t/RC}$ . At $90.0\%$ , $e^{-t/RC}=0.100$ , so $t=-RC\ln0.100=-(56.0\times10^3)(220\times10^{-6})\ln0.100=28.4\,\text{s}$ . The final charge is $Q_0=CV=(220\times10^{-6})(15.0)=3.30\,\text{mC}$ , hence $Q=0.900Q_0=2.97\,\text{mC}$ . Current is $I=(V/R)e^{-t/RC}=[15.0/(56.0\times10^3)](0.100)=26.8\,\mu\text{A}$ .	5
03.1	Resistance $2.50\,\text{k}\Omega$ ; capacitance $2.00\times10^{-3}\,\text{F}$ ( $2000\,\mu\text{F}$ ); initial charge $24.0\,\text{mC}$ , equal to the total area under the current-time graph.	For $I=I_0e^{-t/RC}$ , the initial gradient is $-I_0/RC$ . Hence $RC=(4.80\,\text{mA})/(0.960\,\text{mA s}^{-1})=5.00\,\text{s}$ . Initially $I_0=V_0/R$ , so $R=12.0/(4.80\times10^{-3})=2.50\times10^3\,\Omega$ . Then $C=RC/R=5.00/(2.50\times10^3)=2.00\times10^{-3}\,\text{F}$ . Using the stated graph-area relation, $Q_0=I_0RC=(4.80\times10^{-3})(5.00)=2.40\times10^{-2}\,\text{C}=24.0\,\text{mC}$ .	5
04.1	Initial potential difference $15.0\,\text{V}$ ; resistance $35.2\,\text{k}\Omega$ ; time $23.4\,\text{s}$ .	For discharge, $\ln(V/1\,\text{V})=\ln(V_0/1\,\text{V})-t/RC$ . Hence $V_0=e^{2.708}\,\text{V}=15.0\,\text{V}$ and $1/RC=0.0860\,\text{s}^{-1}$ . Therefore $R=1/[(0.0860)(330\times10^{-6})]=3.52\times10^4\,\Omega=35.2\,\text{k}\Omega$ . Using $V/V_0=e^{-0.0860t}$ , $t=\ln(15.0/2.00)/0.0860=23.4\,\text{s}$ .	5
05.1	Supply potential difference $12.5\,\text{V}$ ; time constant $7.83\,\text{s}$ ; capacitance $167\,\mu\text{F}$ ; initial current $266\,\mu\text{A}$ .	Let $x=e^{-4.00/RC}$ . Then $5.00=V_s(1-x)$ and $8.00=V_s(1-x^2)=V_s(1-x)(1+x)$ . Dividing gives $8.00/5.00=1+x$ , so $x=0.600$ . Hence $V_s=5.00/(1-0.600)=12.5\,\text{V}$ and $RC=-4.00/\ln0.600=7.83\,\text{s}$ . Thus $C=7.83/(47.0\times10^3)=1.67\times10^{-4}\,\text{F}=167\,\mu\text{F}$ . At the start the uncharged capacitor has zero potential difference, so $I_0=V_s/R=12.5/(47.0\times10^3)=266\,\mu\text{A}$ .	6

3.7.5.1 · Magnetic flux density

Tier 1 · Easy

Mark scheme for 3.7.5.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.12\,\text{N}$	Use $F=BIl=(0.45)(3.2)(0.080)=0.1152\,\text{N}$ . To two significant figures the force is $0.12\,\text{N}$ .	2
02.1	One tesla is the flux density that produces a force of $1\,\text{N}$ per metre on a wire carrying $1\,\text{A}$ perpendicular to the field.	Award one mark for a force of $1\,\text{N}$ on a $1\,\text{m}$ wire carrying $1\,\text{A}$ , and one mark for stating that the wire is perpendicular to the field. This follows from $B=F/(Il)$ .	2

Tier 2 · Standard

Mark scheme for 3.7.5.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.133\,\text{T}$	The stated change from the zero-current reading corresponds to magnetic force $F=\Delta mg=(6.50\times10^{-3})(9.81)=0.0638\,\text{N}$ . Then $B=F/(Il)=0.0638/[(4.00)(0.120)]=0.133\,\text{T}$ .	3
02.1	The force is into the page; reversing the current makes the force act out of the page.	Point the first finger to the right for the field and the second finger upwards for conventional current; the thumb then points into the page. Reversing current reverses the magnetic force, so it points out of the page.	3
03.1	Current $0.441\,\text{A}$ from left to right.	For wire length $l$ , weight is $(\lambda l)g$ and magnetic force is $BIl$ . Equating them cancels $l$ : $I=\lambda g/B=(0.0180)(9.81)/0.400=0.441\,\text{A}$ . Fleming's left-hand rule gives an upward force for conventional current from left to right when the field is into the page.	3

Tier 3 · Hard

Mark scheme for 3.7.5.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$B=0.202\,\text{T}$ ; reading change $9.10\,\text{g}$ ; a non-zero intercept indicates a zero offset or other systematic force.	Convert the gradient to force per current: $(1.75\times10^{-3})(9.81)=1.71675\times10^{-2}\,\text{N A}^{-1}$ . Since $F/I=Bl$ , $B=(1.71675\times10^{-2})/0.0850=0.202\,\text{T}$ . At $5.20\,\text{A}$ the mass-reading change is $(1.75)(5.20)=9.10\,\text{g}$ . The magnetic relationship predicts zero force at zero current, so a non-zero intercept indicates a systematic zero error or a current-independent force, not a change in $B$ .	5
02.1	Magnetic flux density $0.184\,\text{T}$ ; the two readings contain the same baseline but opposite magnetic-force contributions.	Reversing current reverses the magnetic force, so the $2.40\,\text{g}$ reading difference corresponds to twice the magnetic force. Thus $F=\tfrac12(2.40\times10^{-3})(9.81)=1.177\times10^{-2}\,\text{N}$ . With $l=0.0320\,\text{m}$ , $B=F/(Il)=1.177\times10^{-2}/[(2.00)(0.0320)]=0.1839\,\text{T}=0.184\,\text{T}$ to three significant figures. Subtracting cancels any constant balance load or zero offset present in both readings.	4
03.1	Magnetic force $3.00\times10^{-3}\,\text{N}$ downwards; initial support force $6.19\times10^{-2}\,\text{N}$ upwards; reversing the current reduces the support force by $6.00\times10^{-3}\,\text{N}$ .	For the first region, $F_1=BIl=(0.300)(3.00)(80.0\times10^{-3})=0.0720\,\text{N}$ upwards. In the reversed field, $F_2=(0.500)(3.00)(50.0\times10^{-3})=0.0750\,\text{N}$ downwards. The resultant magnetic force is $3.00\times10^{-3}\,\text{N}$ downwards. The weight is $(6.00\times10^{-3})(9.81)=5.886\times10^{-2}\,\text{N}$ , so the supports initially exert $5.886\times10^{-2}+3.00\times10^{-3}=6.19\times10^{-2}\,\text{N}$ upwards. Reversing the current reverses the magnetic force, so the support force becomes $5.586\times10^{-2}\,\text{N}$ and decreases by $6.00\times10^{-3}\,\text{N}$ .	5
04.1	Current $0.178\,\text{A}$ ; normal contact force $1.92\times10^{-2}\,\text{N}$ .	For length $l$ , the downslope weight component is $(\lambda l)g\sin\theta$ and the magnetic force is $BIl$ . Equilibrium gives $I=\lambda g\sin\theta/B=(0.0120)(9.81)\sin25.0^\circ/0.280=0.178\,\text{A}$ . The magnetic force is parallel to the plane, so it has no normal component. Thus the contact force is $N=(\lambda l)g\cos\theta=(0.0120)(0.180)(9.81)\cos25.0^\circ=1.92\times10^{-2}\,\text{N}$ .	5
05.1	Forces $0.105\,\text{N}$ north and $0.0700\,\text{N}$ west; resultant $0.126\,\text{N}$ at $33.7^\circ$ west of north.	Both wire sections are perpendicular to the field. Fleming's left-hand rule gives a northward force on the eastward-current section: $F_1=BIl=(0.350)(2.50)(0.120)=0.105\,\text{N}$ . It gives a westward force on the northward-current section: $F_2=(0.350)(2.50)(0.0800)=0.0700\,\text{N}$ . These forces are perpendicular, so $F=\sqrt{(0.105)^2+(0.0700)^2}=0.126\,\text{N}$ and $\theta=\tan^{-1}(0.0700/0.105)=33.7^\circ$ west of north.	5

3.7.5.2 · Moving charges in a magnetic field

Tier 1 · Easy

Mark scheme for 3.7.5.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.6\times10^{-13}\,\text{N}$	Use $F=BQv=(0.25)(1.60\times10^{-19})(4.0\times10^6)=1.6\times10^{-13}\,\text{N}$ .	2
02.1	The magnetic force is always perpendicular to the particle's velocity, so it does no work and cannot change kinetic energy or speed.	Use $F=BQv$ with the direction perpendicular to motion. Since work requires a force component along displacement, the magnetic force changes only the velocity direction, leaving kinetic energy and speed constant.	2

Tier 2 · Standard

Mark scheme for 3.7.5.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.96\times10^{-2}\,\text{m}$	Magnetic force supplies centripetal force: $BQv=mv^2/r$ . Hence $r=mv/(BQ)=(1.67\times10^{-27})(3.20\times10^6)/[(0.480)(1.60\times10^{-19})]=6.96\times10^{-2}\,\text{m}$ .	3
02.1	$1.67\times10^8\,\text{C kg}^{-1}$	Equate magnetic and centripetal forces: $BQv=mv^2/r$ . Therefore $Q/m=v/(Br)=(4.50\times10^6)/[(0.320)(84.0\times10^{-3})]=1.67\times10^8\,\text{C kg}^{-1}$ .	3
03.1	$m_Y/m_X=3.00$ ; both paths initially curve upwards.	For circular motion, $r=mv/(BQ)$ , so with the same $v$ and $B$ , mass is proportional to $rQ$ . Therefore $m_Y/m_X=(r_YQ_Y)/(r_XQ_X)=[(0.180)(2e)]/[(0.120)(e)]=3.00$ . The force on a positive charge moving right in a field into the page is upwards, so both paths initially curve upwards.	3

Tier 3 · Hard

Mark scheme for 3.7.5.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Speed $1.73\times10^7\,\text{m s}^{-1}$ ; kinetic energy $6.25\,\text{MeV}$ ; frequency $6.14\,\text{MHz}$ .	For circular motion, $BQv=mv^2/r$ , so $v=BQr/m=(0.800)(3.20\times10^{-19})(0.450)/(6.64\times10^{-27})=1.73\times10^7\,\text{m s}^{-1}$ . Then $E_k=\tfrac12mv^2=9.99\times10^{-13}\,\text{J}$ . Since $1\,\text{MeV}=1.60\times10^{-13}\,\text{J}$ , this is $6.25\,\text{MeV}$ . The period follows from $2\pi r/v$ , giving $f=BQ/(2\pi m)=(0.800)(3.20\times10^{-19})/[2\pi(6.64\times10^{-27})]=6.14\times10^6\,\text{Hz}$ .	6
02.1	Radius $1.75\times10^{-2}\,\text{m}$ ; time $9.37\times10^{-8}\,\text{s}$ .	Electrical work becomes kinetic energy: $eV=\tfrac12m_pv^2$ , so $v=\sqrt{2eV/m_p}=5.87\times10^5\,\text{m s}^{-1}$ . The magnetic orbit radius is $r=m_pv/(Be)=(1.67\times10^{-27})(5.87\times10^5)/[(0.350)(1.60\times10^{-19})]=1.75\times10^{-2}\,\text{m}$ . A semicircle takes half a cyclotron period: $t=\pi m_p/(Be)=9.37\times10^{-8}\,\text{s}$ .	5
03.1	$r_\alpha/r_p=1$ and $T_\alpha/T_p=2$ ; when the proton returns to the entry point, the alpha particle is at the diametrically opposite point of the same-radius path.	Using $E_k=\tfrac12mv^2$ in $r=mv/(BQ)$ gives $r=\sqrt{2mE_k}/(BQ)$ . Therefore $r_\alpha/r_p=\sqrt{4m_p/m_p}/2=1$ , so their circular paths have the same radius and curve in the same sense because both charges are positive. The period is $T=2\pi m/(BQ)$ , giving $T_\alpha/T_p=(4m_p/2e)/(m_p/e)=2$ . During one proton period the alpha particle therefore completes half an orbit and is at the point diametrically opposite the common entry point.	5
04.1	Mass $6.64\times10^{-27}\,\text{kg}=4.00\,\text{u}$ ; speed $3.47\times10^5\,\text{m s}^{-1}$ ; period $5.79\times10^{-7}\,\text{s}$ ; likely a singly ionised helium-4 ion.	Use $qV=\tfrac12mv^2$ and $r=mv/(Bq)$ . Eliminating $v$ gives $m=qB^2r^2/(2V)=(1.60\times10^{-19})(0.450)^2(32.0\times10^{-3})^2/[2(2.50\times10^3)]=6.64\times10^{-27}\,\text{kg}$ . This is $6.64\times10^{-27}/(1.66\times10^{-27})=4.00\,\text{u}$ , consistent with helium-4. Then $v=Bqr/m=(0.450)(1.60\times10^{-19})(32.0\times10^{-3})/(6.64\times10^{-27})=3.47\times10^5\,\text{m s}^{-1}$ . Finally $T=2\pi m/(Bq)=5.79\times10^{-7}\,\text{s}$ .	6
05.1	Exit speed $2.90\times10^7\,\text{m s}^{-1}$ ; kinetic energy $4.38\,\text{MeV}$ ; approximately $1.25\times10^3$ gap crossings; acceleration time $5.70\times10^{-5}\,\text{s}$ (accept $5.68$ – $5.70\times10^{-5}\,\text{s}$ ).	At the exit, $v=Bqr/m=(0.720)(1.60\times10^{-19})(0.420)/(1.67\times10^{-27})=2.90\times10^7\,\text{m s}^{-1}$ . Thus $E_k=\tfrac12mv^2=7.01\times10^{-13}\,\text{J}=4.38\,\text{MeV}$ . Each gap crossing adds $qV=3.50\,\text{keV}$ , so the number is $1.25\times10^3$ . The cyclotron frequency is $f=Bq/(2\pi m)=1.10\times10^7\,\text{Hz}$ . There are two crossings per orbit, so $t=n/(2f)=1251.4/[2(1.0979\times10^7)]=5.70\times10^{-5}\,\text{s}$ .	6

3.7.5.3 · Magnetic flux and flux linkage

Tier 1 · Easy

Mark scheme for 3.7.5.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.8\times10^{-3}\,\text{Wb}$	With the field normal to the area, $\Phi=BA=(0.32)(1.5\times10^{-2})=4.8\times10^{-3}\,\text{Wb}$ .	2
02.1	Magnetic flux is $\Phi$ for one turn, whereas flux linkage is $N\Phi$ for the whole coil.	Award one mark for identifying $\Phi$ as the flux through one turn and one mark for multiplying by the number of turns to obtain the linkage $N\Phi$ .	2

Tier 2 · Standard

Mark scheme for 3.7.5.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.124\,\text{Wb turns}$	Use $N\Phi=BAN\cos\theta$ . Thus $N\Phi=(0.180)(3.50\times10^{-3})(240)\cos35.0^\circ=0.1239\,\text{Wb turns}$ , which is $0.124\,\text{Wb turns}$ .	3
02.1	$73.1^\circ$	Use $N\Phi=BAN\cos\theta$ . Thus $\cos\theta=0.0750/[(0.280)(2.20\times10^{-3})(420)]=0.290$ . Therefore $\theta=\cos^{-1}(0.290)=73.1^\circ$ .	3
03.1	Flux linkage of A $5.76\times10^{-2}\,\text{Wb turns}$ ; ratio $N\Phi_B/N\Phi_A=1.00$ .	For A, $N\Phi=BAN\cos\theta=(0.320)(1.50\times10^{-3})(240)\cos60.0^\circ=5.76\times10^{-2}\,\text{Wb turns}$ . Coil B has half as many turns but twice the area, with the same $B$ and $\theta$ , so its flux linkage is equal to A's and the ratio is $1.00$ .	3

Tier 3 · Hard

Mark scheme for 3.7.5.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Initial $+4.51\times10^{-2}\,\text{Wb turns}$ ; final $-1.64\times10^{-2}\,\text{Wb turns}$ ; change $-6.15\times10^{-2}\,\text{Wb turns}$ .	Use $N\Phi=BAN\cos\theta$ with $BAN=(0.120)(8.0\times10^{-4})(500)=0.0480\,\text{Wb turns}$ . Initially, $N\Phi_i=0.0480\cos20.0^\circ=+4.51\times10^{-2}\,\text{Wb turns}$ . Finally, $N\Phi_f=0.0480\cos110^\circ=-1.64\times10^{-2}\,\text{Wb turns}$ . Therefore $\Delta(N\Phi)=N\Phi_f-N\Phi_i=-6.15\times10^{-2}\,\text{Wb turns}$ ; the negative sign records reversal relative to the chosen normal.	5
02.1	Initial flux linkage $0.176\,\text{Wb turns}$ ; increase $0.144\,\text{Wb turns}$ .	The full area is $A=(60.0\times10^{-3})(40.0\times10^{-3})=2.40\times10^{-3}\,\text{m}^2$ . Initially the linked area is $0.55A$ , so $N\Phi_i=BN(0.55A)\cos25.0^\circ=(0.420)(350)(0.55)(2.40\times10^{-3})\cos25.0^\circ=0.1759\,\text{Wb turns}$ . Fully inside, $N\Phi_f=(0.420)(350)(2.40\times10^{-3})\cos25.0^\circ=0.3197\,\text{Wb turns}$ . The increase is $0.3197-0.1759=0.1439\,\text{Wb turns}$ .	5
03.1	Magnetic flux density $0.200\,\text{T}$ ; percentage uncertainty $2.5\%$ for $180^\circ$ and $5.0\%$ for $90^\circ$ ; the $180^\circ$ rotation is better.	A $180^\circ$ rotation changes flux linkage from $+BAN$ to $-BAN$ , so $\|\Delta(N\Phi)\|=2BAN$ . Hence $B=0.240/[2(250)(2.40\times10^{-3})]=0.200\,\text{T}$ . Adding the absolute uncertainties of the two readings gives $\pm0.006\,\text{Wb turns}$ , so the percentage uncertainty for the $180^\circ$ change is $(0.006/0.240)\times100=2.5\%$ . A $90^\circ$ rotation gives half the change, $0.120\,\text{Wb turns}$ , with the same $0.006\,\text{Wb turns}$ uncertainty, giving $5.0\%$ . The $180^\circ$ rotation is better because its larger change gives the smaller percentage uncertainty.	5
04.1	Magnetic flux density $0.240\,\text{T}$ ; flux linkage $+9.76\times10^{-2}\,\text{Wb turns}$ ; angle turned $45.0^\circ$ .	The amplitude is $BAN$ , so $B=0.138/[(1.80\times10^{-3})(320)]=0.2396\,\text{T}=0.240\,\text{T}$ . In $5.00\,\text{ms}$ the coil turns through $(5.00/40.0)(360^\circ)=45.0^\circ$ . Taking the initial maximum as positive, $N\Phi=0.138\cos45.0^\circ=+9.76\times10^{-2}\,\text{Wb turns}$ .	5
05.1	Initial flux linkage $+0.281\,\text{Wb turns}$ ; final linkage $-0.281\,\text{Wb turns}$ ; signed change $-0.562\,\text{Wb turns}$ .	Add signed fluxes through the two areas. Per turn, $\Phi=A[0.600(0.350)+0.400(-0.200)]=(1.20\times10^{-2})(0.130)=1.56\times10^{-3}\,\text{Wb}$ . Thus $N\Phi_i=(180)(1.56\times10^{-3})=+0.2808\,\text{Wb turns}$ . A $180^\circ$ turn reverses the chosen normal relative to both fields, so $N\Phi_f=-0.2808\,\text{Wb turns}$ . Therefore $\Delta(N\Phi)=N\Phi_f-N\Phi_i=-0.5616\,\text{Wb turns}$ .	5

3.7.5.4 · Electromagnetic induction

Tier 1 · Easy

Mark scheme for 3.7.5.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Faraday's law: induced emf equals the rate of change of flux linkage. Lenz's law: the induced current acts to oppose the change in flux producing it.	For Faraday's law, state the rate of change of flux linkage, not merely flux. For Lenz's law, identify opposition to the change that causes the induction; this determines the polarity or current direction.	2
02.1	The flux linkage is not changing, so no emf is induced. Moving the magnet or coil, or changing the magnetic field, would produce a reading.	Award one mark for no change of flux linkage and one mark for a valid way of changing flux linkage, such as relative motion between magnet and coil.	2

Tier 2 · Standard

Mark scheme for 3.7.5.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$15\,\text{V}$ ; the polarity drives a current whose magnetic field opposes the decrease in flux.	Faraday's law gives $\|\varepsilon\|=N\|\Delta\Phi\|/\Delta t=250(3.2-0.80)\times10^{-3}/(40\times10^{-3})=15\,\text{V}$ . By Lenz's law, the induced polarity is such that any induced current produces flux in the original direction, opposing the stated decrease.	3
02.1	$0.945\,\text{V}$	In time $\Delta t$ , the conductor sweeps area $l v\Delta t$ , so the flux change is $Blv\Delta t$ . Faraday's law gives $\varepsilon=Blv=(0.750)(0.280)(4.50)=0.945\,\text{V}$ .	3
03.1	$0.840\,\text{V}$	The coil takes $t=l/v=0.0800/0.250=0.320\,\text{s}$ to leave. Its flux linkage changes by $NBA=(120)(0.350)(0.0800)^2=0.2688\,\text{Wb turns}$ . Faraday's law gives $\|\varepsilon\|=\Delta(N\Phi)/\Delta t=0.2688/0.320=0.840\,\text{V}$ .	3

Tier 3 · Hard

Mark scheme for 3.7.5.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Peak emf $88.2\,\text{V}$ ; emf magnitude at $2.50\,\text{ms}$ is $62.4\,\text{V}$ .	The angular speed is $\omega=2\pi f=314\,\text{rad s}^{-1}$ . The flux linkage is $BAN\cos\omega t$ , so Faraday's law gives $\varepsilon=BAN\omega\sin\omega t$ in magnitude. Thus $\varepsilon_0=(0.240)(6.50\times10^{-3})(180)(314)=88.2\,\text{V}$ . At $t=2.50\times10^{-3}\,\text{s}$ , $\omega t=0.785\,\text{rad}$ and $\|\varepsilon\|=88.2\sin0.785=62.4\,\text{V}$ . The emf exists because rotation changes flux linkage; its polarity, from Lenz's law, drives a current whose magnetic effect opposes that change.	5
02.1	Charge $3.84\times10^{-2}\,\text{C}$ ; thermal energy $2.75\times10^{-2}\,\text{J}$ .	The induced emf is $\varepsilon=Blv=(0.640)(0.350)(3.20)=0.7168\,\text{V}$ , so $I=\varepsilon/R=0.7168/2.80=0.256\,\text{A}$ . The travel time is $t=x/v=0.480/3.20=0.150\,\text{s}$ . Hence $Q=It=(0.256)(0.150)=3.84\times10^{-2}\,\text{C}$ . The thermal energy is $E=I^2Rt=(0.256)^2(2.80)(0.150)=2.75\times10^{-2}\,\text{J}$ to three significant figures.	4
03.1	Emf $2.00\,\text{V}$ ; current $0.500\,\text{A}$ ; charge $0.0400\,\text{C}$ ; the induced field is in the same direction as the original field, opposing the decrease.	The flux-linkage change is $\Delta(N\Phi)=NA\Delta B=(200)(4.00\times10^{-3})(-0.0500-0.150)=-0.160\,\text{Wb turns}$ . Hence $\|\varepsilon\|=0.160/(80.0\times10^{-3})=2.00\,\text{V}$ and $I=\varepsilon/R=2.00/4.00=0.500\,\text{A}$ . The charge is $Q=It=(0.500)(80.0\times10^{-3})=0.0400\,\text{C}$ . Lenz's law gives an induced field in the same direction as the original $0.150\,\text{T}$ field, opposing its decrease.	5
04.1	Initial magnetic flux density $4.80\times10^{-2}\,\text{T}$ ; the polarity drives an induced effect that opposes the loss of the original flux.	The time integral of emf magnitude equals the change in flux linkage. The triangular pulse area is $\tfrac12(30.0\times10^{-3})(4.80)=7.20\times10^{-2}\,\text{V s}=7.20\times10^{-2}\,\text{Wb turns}$ . Pulling the coil fully out changes linkage from $BAN$ to zero, so $B=(7.20\times10^{-2})/[(600)(2.50\times10^{-3})]=4.80\times10^{-2}\,\text{T}$ . Lenz's law sets the polarity so any induced current produces flux in the original direction, opposing its decrease.	5
05.1	Emf $0.264\,\text{V}$ ; current $0.147\,\text{A}$ ; magnetic force $1.94\times10^{-2}\,\text{N}$ opposing motion; mechanical and thermal powers both $3.87\times10^{-2}\,\text{W}$ .	At $0.800\,\text{s}$ the speed is $v=at=(2.50)(0.800)=2.00\,\text{m s}^{-1}$ . The motional emf is $\varepsilon=Blv=(0.550)(0.240)(2.00)=0.264\,\text{V}$ , so $I=\varepsilon/R=0.264/1.80=0.1467\,\text{A}$ . The magnetic force is $F=BIl=(0.550)(0.1467)(0.240)=1.94\times10^{-2}\,\text{N}$ . By Lenz's law it opposes the motion. The required mechanical power is $Fv=(1.936\times10^{-2})(2.00)=3.87\times10^{-2}\,\text{W}$ . The thermal power is $I^2R=(0.1467)^2(1.80)=3.87\times10^{-2}\,\text{W}$ , equal within rounding.	6

3.7.5.5 · Alternating currents

Tier 1 · Easy

Mark scheme for 3.7.5.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$8.49\,\text{V}$	For a sinusoid, $V_{\mathrm{rms}}=V_0/\sqrt2=12.0/\sqrt2=8.49\,\text{V}$ .	1
02.1	It is the direct current that would produce the same mean power, or heating effect, in a resistor.	Award one mark for comparison with a direct current and one mark for the same mean power or heating effect in a resistor.	2

Tier 2 · Standard

Mark scheme for 3.7.5.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Frequency $3.1\times10^2\,\text{Hz}$ ; peak voltage $5.6\,\text{V}$ ; rms voltage $4.0\,\text{V}$ .	The period is $T=(6.4)(0.50\,\text{ms})=3.2\times10^{-3}\,\text{s}$ , so $f=1/T=3.125\times10^2\,\text{Hz}$ . The peak-to-peak voltage is $(5.6)(2.0)=11.2\,\text{V}$ , hence $V_0=11.2/2=5.6\,\text{V}$ . For the sinusoid, $V_{\mathrm{rms}}=5.6/\sqrt2=3.96\,\text{V}$ . Values to two significant figures are $3.1\times10^2\,\text{Hz}$ and $4.0\,\text{V}$ .	4
02.1	Rms current $1.41\,\text{A}$ ; mean power $18.0\,\text{W}$ , so $P_{\text{mean}}=P_0/2$ .	The rms current is $I_{\mathrm{rms}}=V_0/(\sqrt2R)=18.0/[\sqrt2(9.00)]=1.41\,\text{A}$ . Mean power is $P_{\mathrm{mean}}=V_{\mathrm{rms}}^2/R=(18.0/\sqrt2)^2/9.00=18.0\,\text{W}$ . The peak instantaneous power is $P_0=V_0^2/R=(18.0)^2/9.00=36.0\,\text{W}$ , so $P_{\mathrm{mean}}=P_0/2$ .	3
03.1	Mean power $61.4\,\text{W}$ ; energy per cycle $1.23\,\text{J}$ .	The peak current is $I_0=6.40/2=3.20\,\text{A}$ and $I_{\text{rms}}=I_0/\sqrt2=2.26\,\text{A}$ . Mean power is $P=I_{\text{rms}}^2R=(3.20/\sqrt2)^2(12.0)=61.4\,\text{W}$ . One cycle lasts $T=1/f=0.0200\,\text{s}$ , so the energy is $PT=(61.44)(0.0200)=1.23\,\text{J}$ .	4

Tier 3 · Hard

Mark scheme for 3.7.5.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$V_0=325\,\text{V}$ ; $V_{\mathrm{pp}}=651\,\text{V}$ ; $I_{\mathrm{rms}}=7.83\,\text{A}$ ; $I_0=11.1\,\text{A}$ ; energy $1.30\,\text{MJ}$ .	For a sinusoid, $V_0=\sqrt2V_{\mathrm{rms}}=\sqrt2(230)=325\,\text{V}$ and $V_{\mathrm{pp}}=2V_0=651\,\text{V}$ . For a resistive heater, $P=V_{\mathrm{rms}}I_{\mathrm{rms}}$ , so $I_{\mathrm{rms}}=1800/230=7.83\,\text{A}$ and $I_0=\sqrt2I_{\mathrm{rms}}=11.1\,\text{A}$ . Convert $12.0\,\text{min}=720\,\text{s}$ ; then energy $E=Pt=(1800)(720)=1.296\times10^6\,\text{J}=1.30\,\text{MJ}$ .	6
02.1	Peak-to-peak height $6.22$ divisions; one cycle $10.0$ divisions wide; trace centre displaced $0.300$ division upwards.	The peak-to-peak voltage is $V_{\mathrm{pp}}=2\sqrt2V_{\mathrm{rms}}=2\sqrt2(110)=311\,\text{V}$ , giving $311/50.0=6.22$ vertical divisions. The period is $T=1/f=1/50.0=20.0\,\text{ms}$ , giving $20.0/2.00=10.0$ horizontal divisions. A positive dc offset raises the centre by $15.0/50.0=0.300$ vertical division.	5
03.1	Peak voltage $33.9\,\text{V}$ ; peak-to-peak voltage $67.9\,\text{V}$ ; use $10.0\,\text{V div}^{-1}$ , giving a peak-to-peak height of $6.79$ divisions.	Equal mean heating in the same resistor means $V_{\text{rms}}=V_{\text{dc}}=24.0\,\text{V}$ . For a sinusoid, $V_0=\sqrt2V_{\text{rms}}=33.9\,\text{V}$ and $V_{\text{pp}}=2V_0=67.9\,\text{V}$ . At $5.0\,\text{V div}^{-1}$ the trace would need $13.6$ divisions and would be clipped. The $10.0\,\text{V div}^{-1}$ setting is the most sensitive one that fits, and the height is $67.9/10.0=6.79$ divisions.	5
04.1	Period $5.80\,\text{ms}$ ; frequency $172\,\text{Hz}$ ; negative peak $-14.8\,\text{V}$ ; peak-to-peak voltage $29.6\,\text{V}$ ; rms voltage $10.5\,\text{V}$ ; next downward zero crossing at $5.05\,\text{ms}$ .	An upward zero crossing to the next positive maximum is one quarter-cycle, so $T=4(3.60-2.15)=5.80\,\text{ms}$ and $f=1/(5.80\times10^{-3})=172\,\text{Hz}$ . The centred sinusoid has negative peak $-14.8\,\text{V}$ , $V_{\text{pp}}=2(14.8)=29.6\,\text{V}$ and $V_{\text{rms}}=14.8/\sqrt2=10.5\,\text{V}$ . The next downward zero crossing is half a period after the upward crossing, at $2.15+5.80/2=5.05\,\text{ms}$ .	6
05.1	Frequency $156\,\text{Hz}$ ; Y lags X by $1.60\,\text{ms}$ , or $90.0^\circ$ ; rms voltages $6.22\,\text{V}$ for X and $3.11\,\text{V}$ for Y.	For X, the interval between corresponding upward zero crossings is the period: $T=7.85-1.45=6.40\,\text{ms}$ , so $f=1/(6.40\times10^{-3})=156\,\text{Hz}$ . The corresponding crossing of Y is delayed by $3.05-1.45=1.60\,\text{ms}$ . Hence the phase lag is $(1.60/6.40)(360^\circ)=90.0^\circ$ . The peaks are half the peak-to-peak values, so $V_{X,\text{rms}}=(17.6/2)/\sqrt2=6.22\,\text{V}$ and $V_{Y,\text{rms}}=(8.80/2)/\sqrt2=3.11\,\text{V}$ .	6

3.7.5.6 · The operation of a transformer

Tier 1 · Easy

Mark scheme for 3.7.5.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$96\,\text{V}$	Use $V_s/V_p=N_s/N_p$ . Hence $V_s=24(1200/300)=96\,\text{V}$ .	2
02.1	$30$ turns	Use $N_s/N_p=V_s/V_p$ . Therefore $N_s=600(12.0/240)=30$ turns.	2

Tier 2 · Standard

Mark scheme for 3.7.5.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Secondary current $14.5\,\text{A}$ ; dissipated power $66.2\,\text{W}$ .	Input power is $P_{\mathrm{in}}=I_pV_p=(1.80)(230)=414\,\text{W}$ . Output power is $P_{\mathrm{out}}=\eta P_{\mathrm{in}}=(0.840)(414)=347.76\,\text{W}$ . Thus $I_s=P_{\mathrm{out}}/V_s=347.76/24.0=14.49\,\text{A}$ . The dissipated power is $P_{\mathrm{in}}-P_{\mathrm{out}}=414-347.76=66.24\,\text{W}$ .	4
02.1	For example: eddy-current heating and hysteresis loss; insulated core laminations reduce eddy currents or a soft magnetic core reduces hysteresis.	Award one mark each for two named mechanisms, such as eddy currents, hysteresis, winding resistance or flux leakage. Award one mark for a matched remedy, such as mutually insulated laminations for eddy currents or a soft magnetic core for hysteresis.	3
03.1	Primary turns $300$ ; efficiency $75.0\%$ .	The transformer equation gives $V_s/N_s=V_p/N_p$ , so the graph gradient is $V_p/N_p$ . Hence $N_p=12.0/0.0400=300$ . Input power is $P_p=V_pI_p=(12.0)(0.800)=9.60\,\text{W}$ and output power is $P_s=V_sI_s=(6.00)(1.20)=7.20\,\text{W}$ . Therefore efficiency is $P_s/P_p=(7.20/9.60)\times100=75.0\%$ .	4

Tier 3 · Hard

Mark scheme for 3.7.5.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Loss at $25.0\,\text{kV}$ is $32.0\,\text{kW}$ ; loss at $250\,\text{kV}$ is $320\,\text{W}$ , one hundredth as large. A step-up transformer reduces current, and insulated laminations restrict eddy currents.	Assuming the transmitted power is the stated $2.50\,\text{MW}$ , at $25.0\,\text{kV}$ the current is $I=P/V=(2.50\times10^6)/(25.0\times10^3)=100\,\text{A}$ , so $P_{\mathrm{loss}}=I^2R=(100)^2(3.20)=3.20\times10^4\,\text{W}=32.0\,\text{kW}$ . At $250\,\text{kV}$ , $I=10.0\,\text{A}$ and loss is $(10.0)^2(3.20)=320\,\text{W}$ , a factor of $100$ smaller. A step-up transformer uses $N_s/N_p=V_s/V_p$ to raise voltage and, approximately conserving power, reduce current. Alternating flux can induce circulating eddy currents in a solid core; thin mutually insulated laminations break the current paths, raising their resistance and reducing $I^2R$ heating.	6
02.1	Alternating current produces changing magnetic flux and hence an induced secondary emf; steady dc produces no changing flux after switching. Doubling the secondary turns doubles the secondary rms voltage but does not change its frequency.	The ac primary current creates continuously changing flux linkage in the secondary, so Faraday's law gives an induced emf. A steady dc current produces constant flux after the switching transient and therefore no sustained secondary emf. For an ideal transformer, $V_s/V_p=N_s/N_p$ , so doubling $N_s$ doubles $V_s$ . The secondary emf is produced by the same changing core flux, so its frequency remains equal to the primary frequency.	5
03.1	$3017$ turns accept $3017$ to $3019$ turns where rounded intermediate values are carried	The maximum cable loss is $0.0150(600\times10^3)=9.00\times10^3\,\text{W}$ . From $P_{\text{loss}}=I^2R$ , the maximum current is $I=\sqrt{(9.00\times10^3)/5.00}=42.426\,\text{A}$ ( $42.4\,\text{A}$ to three significant figures). To transmit $600\,\text{kW}$ , the secondary voltage must be at least $V_s=P/I=(600\times10^3)/42.426=1.4142\times10^4\,\text{V}$ . Thus $N_s/N_p=V_s/V_p$ , giving $N_s\ge320(1.4142\times10^4)/(1.50\times10^3)=3016.99$ . The minimum whole number that keeps the loss within the limit is therefore $3017$ turns.	5
04.1	Step-up secondary turns $5000$ ; cable current $2.40\,\text{A}$ ; cable loss $34.6\,\text{W}$ ; delivered power $3.61\times10^5\,\text{W}$ ; overall efficiency $90.2\%$ .	The turns ratio gives $N_s=N_pV_s/V_p=250(160/8.00)=5000$ . The step-up output power is $0.960(400\times10^3)=3.84\times10^5\,\text{W}$ , so cable current is $I=P/V=(3.84\times10^5)/(160\times10^3)=2.40\,\text{A}$ . Cable loss is $I^2R=(2.40)^2(6.00)=34.56\,\text{W}$ . The second-transformer input is $384000-34.56=383965\,\text{W}$ , so delivered power is $0.940(383965)=3.61\times10^5\,\text{W}$ . Overall efficiency is $360927/(400000)\times100=90.2\%$ .	6
05.1	Secondary turns $145$ ; loaded efficiency $85.2\%$ .	The turns ratio gives $N_s=N_pV_s/V_p=870(37.5/225)=145$ turns. The loaded input power is $I_pV_p=(2.68)(225)=603\,\text{W}$ and the loaded output power is $I_sV_s=(13.7)(37.5)=513.75\,\text{W}$ . Therefore $\eta=I_sV_s/(I_pV_p)=(513.75/603)\times100=85.2\%$ .	5