3.7 Fields and their consequences — AQA 7408 revision notes

3.7.1

Year 2

Fields (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

State one difference between a gravitational force field and an electric force field.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.2.1

Year 2

Newton's law (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.2.2

Year 2

Gravitational field strength (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

Define gravitational field strength at a point.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.2.3

Year 2

Gravitational potential (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.2.4

Year 2

Orbits of planets and satellites (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.3.1

Year 2

Coulomb's law (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.3.2

Year 2

Electric field strength (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Electric field strength at a point is the force per unit charge on a small positive test charge placed there: $E=\dfrac{F}{Q}$ , a vector in $\text{N C}^{-1}$ (equivalently $\text{V m}^{-1}$ ), directed along the force on a positive charge. Field lines point from positive to negative and are closer where the field is stronger. In a uniform field between parallel plates $E=\dfrac{V}{d}$ (from $Fd=Q\Delta V$ ); around a point charge $E=\dfrac{Q}{4\pi\varepsilon_0r^2}$ . A charge entering a uniform field at right angles follows a parabola.

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

A small sphere carries a charge of

+6.0\,\text{nC}

. Calculate the electric field strength at a point

0.15\,\text{m}

from the centre of the sphere. Two parallel plates

40\,\text{mm}

apart produce a uniform field of the same magnitude. Calculate the potential difference between the plates.

[4 marks]

Total for this question: 4

1.Radial field: $E=\dfrac{Q}{4\pi\varepsilon_0r^2}=\dfrac{6.0\times10^{-9}}{4\pi\times8.85\times10^{-12}\times0.15^2}$ .
2. $E=2.4\times10^{3}\,\text{N C}^{-1}$ (or $\text{V m}^{-1}$ ), directed away from the positive sphere.
3.Uniform field: rearrange $E=\dfrac{V}{d}$ with $d=0.040\,\text{m}$ : $V=Ed$ .
4. $V=2.4\times10^{3}\times0.040=96\,\text{V}$ .

Answer: $E=2.4\times10^{3}\,\text{V m}^{-1}$ (radially outwards); $V=96\,\text{V}$

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.

Mark-scheme wording to use

force per unit charge; 1 mark small positive/test charge — paraphrase of AQA 7408, June 2018 Paper 2, Q02.1
“(Lines are closer together so) the field is stronger” — AQA 7408, June 2017 Paper 2, Q06.2
electric field strength is the potential gradient or that its greatest value occurs on a straight section — paraphrase of AQA 7408, June 2024 Paper 2, Q06.2
“N C⁻¹ or V m⁻¹” — AQA 7408, June 2022 Paper 2, Q03.2
forces balance or qE=Bqv … 1.8 x 10⁴ V m⁻¹ — paraphrase of AQA 7408, November 2020 Paper 2, Q03.2
excess electrons are on the outer surface; they are equally spaced/spread out because the electrons mutually repel — paraphrase of AQA 7408, June 2022 Paper 2, Q03.1

What the examiner wants

A definition needs both halves — force per unit charge AND on a small positive (test) charge — and a unit with every field value. Convert millimetres to metres before $E=V/d$ . Read a field from a potential–distance graph as the gradient: use a tangent for the local value and a wide triangle for precision, and note the field is largest where the graph is steepest. For a negative charge the force is opposite to the field, and a particle crossing a uniform field follows a parabola, not a circle.

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

ORIGINAL

Define electric field strength at a point.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.3.3

Year 2

Electric potential (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.4.1

Year 2

Capacitance (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

Capacitance is the charge stored per unit potential difference: $C=\dfrac{Q}{V}$ , in farads ( $1\,\text{F}=1\,\text{C V}^{-1}$ ). $Q$ is the magnitude of the charge on either plate — the plates carry $+Q$ and $-Q$ — and for a given capacitor $Q\propto V$ , so a graph of $Q$ against $V$ is a straight line through the origin with gradient $C$ . A parallel-plate capacitor has $C=\dfrac{A\varepsilon_0\varepsilon_r}{d}$ : a dielectric of relative permittivity $\varepsilon_r$ raises the capacitance by that factor.

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

State what is meant by a capacitance of

470\,\mu\text{F}

. An isolated conducting sphere of radius

0.15\,\text{m}

can be treated as a capacitor with

C=4\pi\varepsilon_0r

. The sphere is charged to a potential of

2.4\times10^{4}\,\text{V}

. Calculate the capacitance of the sphere and the charge on it.

[4 marks]

Total for this question: 4

1. $470\,\mu\text{F}$ means the capacitor stores $470\,\mu\text{C}$ of charge for each volt of potential difference across it.
2.Capacitance of the sphere: $C=4\pi\varepsilon_0r=4\pi\times8.85\times10^{-12}\times0.15=1.67\times10^{-11}\,\text{F}$ (about $17\,\text{pF}$ ).
3.Charge from $C=\dfrac{Q}{V}$ : $Q=CV=1.67\times10^{-11}\times2.4\times10^{4}$ .
4. $Q=4.0\times10^{-7}\,\text{C}$ .

Answer: $470\,\mu\text{C}$ per volt; $C=1.7\times10^{-11}\,\text{F}$ ; $Q=4.0\times10^{-7}\,\text{C}$

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.

Mark-scheme wording to use

per volt/change in pd — paraphrase of AQA 7408, June 2018 Paper 2, Q03.1
Use C=Q/V together with V=Q/(4pi epsilon_0 r) — paraphrase of AQA 7408, June 2022 Paper 2, Q03.3
Do not use the parallel-plate equation — paraphrase of AQA 7408, June 2022 Paper 2, Q03.3
capacitance or stored charge with dielectric is seven times that with vacuum/air/free space for the same geometry and pd — paraphrase of AQA 7408, June 2025 Paper 2, Q05.1
polar alignment … counter-field/reduced field … fixed Q and reduced V increase Q/V — paraphrase of AQA 7408, November 2021 Paper 2, Q01.2

What the examiner wants

Define capacitance with both halves — the charge stored AND per volt of potential difference — and convert microfarads and millicoulombs to SI before calculating. A dielectric constant means the capacitance is that many times the capacitance with air or vacuum for the same plates and pd; explain its action as a chain: polar molecules align, their field opposes the applied field, so for a fixed charge the pd falls and $Q/V$ rises. For a charged sphere use $C=Q/V$ with $V=Q/(4\pi\varepsilon_0r)$ , never the parallel-plate equation.

Exam tip

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

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.4.2

Year 2

Parallel plate capacitor (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.4.3

Year 2

Energy stored by a capacitor (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.4.4

Year 2

Capacitor charge and discharge (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

A capacitor discharging through a resistor loses charge exponentially: $Q=Q_0e^{-t/RC}$ , with the same form for $V$ and $I$ . The time constant $RC$ (in seconds) is the time for the quantity to fall to $1/e\approx37\%$ of its initial value; the time to halve is $T_{1/2}=0.69RC$ . When charging through a resistor, $Q=Q_0(1-e^{-t/RC})$ — charge and pd rise towards their final values while the current falls exponentially from $V_0/R$ . A graph of $\ln V$ (or $\ln I$ ) against $t$ is a straight line of gradient $-\dfrac{1}{RC}$ .

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

470\,\mu\text{F}

capacitor is charged to a potential difference of

9.0\,\text{V}

and then discharged through a

22\,\text{k}\Omega

resistor. Calculate the time constant of the circuit, the time taken for the potential difference across the capacitor to fall to

3.0\,\text{V}

, and the current in the resistor at that instant.

[5 marks]

Total for this question: 5

1.Time constant: $RC=22\times10^{3}\times470\times10^{-6}=10.3\,\text{s}$ .
2.Discharge: $V=V_0e^{-t/RC}$ , so $t=-RC\ln\left(\dfrac{V}{V_0}\right)=-10.34\times\ln\left(\dfrac{3.0}{9.0}\right)$ .
3. $t=10.34\times1.099=11.4\,\text{s}$ (about $1.1$ time constants; check: $1.58$ half-lives of $0.69RC=7.1\,\text{s}$ ).
4.Current at that instant: $I=\dfrac{V}{R}=\dfrac{3.0}{22\times10^{3}}$ .
5. $I=1.4\times10^{-4}\,\text{A}$ ( $0.14\,\text{mA}$ ).

Answer: $RC=10.3\,\text{s}$ ; $t=11\,\text{s}$ ; $I=1.4\times10^{-4}\,\text{A}$

Required practical 9

Investigation of the charge and discharge of capacitors

Apparatus

Capacitor (e.g. $470\,\mu\text{F}$ to $2200\,\mu\text{F}$ ) and a resistor chosen so that $RC$ is tens of seconds
Low-voltage dc supply and a two-way switch (charge / discharge)
Digital voltmeter or data logger across the capacitor, microammeter in series, and a stopwatch

Method

Connect the capacitor, resistor and supply so the switch charges the capacitor in one position and discharges it through the resistor in the other; note the polarity of an electrolytic capacitor.
Charge the capacitor fully, then switch to discharge and start the stopwatch; record the voltmeter (and ammeter) reading every $5$ or $10\,\text{s}$ until the pd has fallen to a few per cent of its starting value.
Repeat the discharge twice more and take mean values at each time.
Plot $V$ against $t$ : read $RC$ as the time for $V$ to fall to $0.37V_0$ , or read $T_{1/2}$ and use $RC=T_{1/2}/0.69$ .
Plot $\ln V$ against $t$ : the straight line has gradient $-1/RC$ and intercept $\ln V_0$ ; compare $RC$ with the product of the measured $R$ and the nominal $C$ . Repeat for charging, recording the rise of $V$ .

Variables

Independent:: time since the switch was thrown
Dependent:: potential difference across the capacitor (and the current)
Control:: resistance $R$ and capacitance $C$ ; supply pd $V_0$ ; starting each discharge from full charge

Hazards and precautions

An electrolytic capacitor connected the wrong way round or above its rated pd can rupture: check polarity and keep the supply at or below the rated value.
A charged capacitor can give a shock or a spark: discharge it through the resistor before touching the circuit.

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 $-1/RC$ .

Mark-scheme wording to use

“reject idea that V falls to zero in 5RC” — AQA 7408, June 2017 Paper 3, Q02.3
“take repeated readings … calculate an average value … check/correct/compensate for any zero error on the voltmeter” — AQA 7408, November 2021 Paper 3, Q03.5
determine graph gradient and use its magnitude as 1/(RC) — paraphrase of AQA 7408, November 2021 Paper 3, Q03.7
“connecting resistor in parallel with R halves [reduces by 50%] circuit [total] resistance [time constant]” — AQA 7408, June 2017 Paper 3, Q02.5
Do not credit a discharge equation — paraphrase of AQA 7408, June 2023 Paper 2, Q05.1

What the examiner wants

The June 2017 Paper 3 report found weak oscilloscope vocabulary and scale reading, misuse of 'five time constants', and vague uncertainty claims; AQA rejects the claim that discharge is complete after $5RC$ . Use the charging equation with the $(1-e^{-t/RC})$ bracket for charging — a discharge equation there scores nothing. To find $RC$ from data use the half-life ( $T_{1/2}/\ln2$ ), the $37\%$ point, or the gradient of $\ln V$ against $t$ with its minus sign, and read a graph with a large triangle.

Exam tip

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

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.5.1

Year 2

Magnetic flux density (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

A current-carrying wire in a magnetic field experiences a force. When the field is perpendicular to the current, $F=BIl$ , where $l$ is the length of wire in the field. Magnetic flux density $B$ is the force per unit current per unit length on a wire at right angles to the field; its unit, the tesla, is defined by $1\,\text{T}$ giving a force of $1\,\text{N}$ on each metre of a wire carrying $1\,\text{A}$ at right angles to the field. Fleming's left-hand rule gives the directions: first finger field, second finger conventional current, thumb force.

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.

Required practical 10: a U-magnet stands on a top-pan balance and a rigid wire is clamped horizontally through its gap. Switching on the current puts an upward force on the wire and, by Newton's third law, an equal downward force on the magnet — read as a change in the balance reading.

Worked example

A U-shaped magnet is placed on a top-pan balance and a rigid horizontal wire is clamped so that a length of

0.050\,\text{m}

of the wire lies at right angles to the field between the poles. When a current of

2.5\,\text{A}

is switched on, the balance reading increases by

1.30\,\text{g}

. Calculate the magnetic flux density between the poles, and explain why the balance reading increases.

[4 marks]

Total for this question: 4

1.Convert the change in balance reading to a force: $F=\Delta m\,g=1.30\times10^{-3}\times9.81=1.28\times10^{-2}\,\text{N}$ .
2.The field is perpendicular to the current, so $F=BIl$ and $B=\dfrac{F}{Il}=\dfrac{1.28\times10^{-2}}{2.5\times0.050}$ .
3. $B=0.10\,\text{T}$ (2 significant figures).
4.The balance reading rises because the field exerts an upward force on the wire (Fleming's left-hand rule), so by Newton's third law the wire exerts an equal and opposite, downward, force on the magnet — which the balance measures.

Answer: $B=0.10\,\text{T}$ ; the magnet feels the equal and opposite (downward) partner of the upward force on the wire

Required practical 10

Investigate how the force on a wire varies with flux density, current and length of wire using a top pan balance

Apparatus

Top-pan balance reading to $0.01\,\text{g}$ , with a U-shaped magnet (or magnets on a steel yoke) standing on the pan
Rigid wire or thick copper rod clamped horizontally through the field, connected to a dc supply, ammeter and variable resistor
Metre rule and set square to check the rod is horizontal and to measure the length of wire in the field

Method

Clamp the rod so it passes horizontally through the centre of the gap at right angles to the field; with no current, zero the balance.
Switch on the current, record the ammeter reading and the change in balance reading, then switch off between readings so the rod does not heat up.
Repeat for at least six currents up to about $5\,\text{A}$ , repeating each and averaging; convert each mass change to a force with $F=\Delta m\,g$ .
Plot $F$ against $I$ : a straight line through the origin whose gradient is $Bl$ , so $B=\text{gradient}/l$ .
Change the length of wire in the field by adding or removing magnets on the yoke (or using rods bent to different lengths) at a fixed current, and confirm $F\propto l$ ; change $B$ by using different magnets.

Variables

Independent:: current $I$ (then length $l$ , then flux density $B$ — one at a time)
Dependent:: force on the wire, from the change in balance reading
Control:: the two variables not being tested; rod horizontal and perpendicular to the field, at the same position in the gap; temperature of the rod (switch off between readings)

Hazards and precautions

A current of several amps heats the rod and leads: switch off between readings and do not touch the rod while current flows.
Strong magnets pinch fingers and can damage the balance and phones: keep them apart, and keep steel tools away from the gap.

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.

Mark-scheme wording to use

“the tesla is the (strength) of the magnetic field/flux density that produces a force of 1 newton in a wire of length 1m with 1 ampere (flowing perpendicular to the field)” — AQA 7408, June 2017 Paper 2, Q02.3
“Use of (B = F / I l) = m g / I l … Lack of use of ‘g’ is a PE and scores zero.” — AQA 7408, June 2017 Paper 2, Q02.4
“force on rod is down(wards) … the current (in rod) is from left / to right / rightwards … predicts direction of field based on their force and their current using valid (left-hand) rule” — AQA 7408, June 2024 Paper 3, Q03.2
upward force; rightward current P to Q; left-hand-rule deduction that B is into page — paraphrase of AQA 7408, June 2018 Paper 3, Q01.2
measure or check the rod height at two separated positions — paraphrase of AQA 7408, June 2024 Paper 3, Q03.1
X and Y explicitly in series, neither shorted — paraphrase of AQA 7408, June 2018 Paper 3, Q01.5

What the examiner wants

The tesla definition needs every part — $1\,\text{N}$ , per metre, $1\,\text{A}$ , AND at right angles to the field. In the balance practical, convert the change in mass reading to a force with $F=\Delta m\,g$ before $F=BIl$ , use only the length of wire inside the field, and argue the direction through Newton's third law: the balance measures the force on the MAGNET, which is equal and opposite to the force on the wire. To check the rod is horizontal, measure its height at both ends with a rule and set square and adjust until they are equal.

Exam tip

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

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.5.2

Year 2

Moving charges in a magnetic field (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.5.3

Year 2

Magnetic flux and flux linkage (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.5.4

Year 2

Electromagnetic induction (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

State Faraday's law and Lenz's law for electromagnetic induction.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.5.5

Year 2

Alternating currents (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

A sinusoidal voltage has peak value

12.0\,\text{V}

. Calculate its rms value.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.7.5.6

Year 2

The operation of a transformer (A-level only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

Fields (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Newton's law (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Gravitational field strength (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Gravitational potential (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Orbits of planets and satellites (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Coulomb's law (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Electric field strength (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Electric potential (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Capacitance (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Parallel plate capacitor (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Energy stored by a capacitor (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Capacitor charge and discharge (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Magnetic flux density (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Moving charges in a magnetic field (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Magnetic flux and flux linkage (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Electromagnetic induction (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Alternating currents (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The operation of a transformer (A-level only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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