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AQA GCSE Physics revision notes

Electricity

Section 4.2
12 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8463 section 4.2

Checked against AQA 8463 section 4.2. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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4.2.1.1

Standard circuit diagram symbols

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

  • Circuit diagrams use standard symbols so that the components and their connections are unambiguous. Follow the conducting lines to decide which components are in the same loop and where branches begin.
  • An ammeter measures the current through a component, so connect it in series with that component.
  • A voltmeter measures the potential difference between two points, so connect it in parallel across the component.
  • A working measuring circuit also needs a source of potential difference and a closed conducting path.
  • The examiner expects recognisable standard symbols joined by lines, not pictures of apparatus.
A measuring circuit with the ammeter in series and voltmeter in parallel across the resistor.
Worked example

Describe how to connect meters to measure the current through and potential difference across a lamp.

  1. 1.Place the ammeter in the same series path as the lamp.
  2. 2.Connect the voltmeter between the two terminals of the lamp, forming a parallel branch.

Answer: The ammeter is in series with the lamp and the voltmeter is in parallel across it.

Common mistakes

  • Don't fall into the trap of connecting the ammeter in parallel across the resistor.
  • Don't fall into the trap of putting the voltmeter in the main series loop instead of across the component.
  • Don't fall into the trap of drawing pictures of bulbs and cells rather than their standard circuit symbols.

Exam tip

For a ‘draw the circuit’ question, trace the main loop first, then add the voltmeter branch across the measured component.

Tier 1 · Easy

ORIGINAL

Draw a circuit diagram containing one cell, one open switch and one lamp, all connected in a single loop.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

Draw a circuit that can measure both the current through a fixed resistor and the potential difference across it. Include a battery and a switch.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A student draws a resistance-measuring circuit with the ammeter connected across the resistor and the voltmeter inserted in the main loop. State both corrections and describe the corrected circuit.

[4 marks]

Total for this question: 4

Your progress and exam materials
4.2.1.2

Electrical charge and current

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

  • Electric current is the rate of flow of electrical charge.
  • Charge flow, current and time are linked by Q=ItQ=It, where QQ is in coulombs, II is in amperes and tt is in seconds.
  • A source of potential difference and a closed circuit are required for charge to flow.
  • At every point in a single closed loop, the current has the same value because charge cannot accumulate at one point.
  • Current is not the amount of charge present; it tells you how many coulombs pass a point each second.
Worked example

240C240\,\text{C} passes a point in 3.03.0 minutes. Calculate the current.

  1. 1.Convert the time: 3.0min=180s3.0\,\text{min}=180\,\text{s}.
  2. 2.Rearrange Q=ItQ=It to I=QtI=\dfrac{Q}{t}.
  3. 3.I=240180=1.33AI=\dfrac{240}{180}=1.33\,\text{A}.

Answer: 1.3A1.3\,\text{A} to two significant figures

Common mistakes

  • Don't fall into the trap of using 3.03.0 directly as the time when the equation requires seconds.
  • Don't fall into the trap of treating current as a stored quantity of charge rather than charge flow per second.
  • Don't fall into the trap of claiming current is used up as charge travels around a single loop.

Exam tip

For Q=ItQ=It, convert time to seconds before substituting and give charge in coulombs.

Tier 1 · Easy

ORIGINAL

A current of 0.35A0.35\,\text{A} flows for 40s40\,\text{s}. Calculate the charge that flows.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

240C240\,\text{C} of charge passes a point in a circuit in 3.03.0 minutes. Calculate the current.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A device carries 0.75A0.75\,\text{A} for 80s80\,\text{s} and then 0.30A0.30\,\text{A} for a further 150s150\,\text{s}. Determine the total charge transferred.

[4 marks]

Total for this question: 4

4.2.1.3

Current, resistance and potential difference

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

  • The current through a component depends on its resistance and the potential difference across it. Use V=IRV=IR, where potential difference is in volts, current in amperes and resistance in ohms.
  • For a fixed potential difference, increasing resistance decreases current.
  • Measure current with an ammeter in series and potential difference with a voltmeter in parallel; corresponding readings give R=V/IR=V/I.
  • Potential difference describes energy transferred per unit charge, while resistance describes how strongly the component opposes current.
  • Use the values for the same component and the same operating condition.
Worked example

A 15Ω15\,\Omega resistor has a potential difference of 6.0V6.0\,\text{V} across it. Calculate the current.

  1. 1.Rearrange V=IRV=IR to I=VRI=\dfrac{V}{R}.
  2. 2.I=6.015=0.40AI=\dfrac{6.0}{15}=0.40\,\text{A}.

Answer: 0.40A0.40\,\text{A}

Common mistakes

  • Don't fall into the trap of calculating resistance as I/VI/V instead of V/IV/I.
  • Don't fall into the trap of using a voltmeter reading from a different component from the ammeter reading.
  • Don't fall into the trap of saying greater resistance gives greater current at fixed potential difference.

Exam tip

Write V=IRV=IR first and rearrange symbolically before substituting the values.

Tier 1 · Easy

ORIGINAL

A current of 0.50A0.50\,\text{A} passes through an 8.0Ω8.0\,\Omega resistor. Calculate the potential difference across the resistor.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A component carries 0.24A0.24\,\text{A} when the potential difference across it is 3.6V3.6\,\text{V}. Its resistance remains constant. Calculate its resistance and the current when the potential difference is 9.0V9.0\,\text{V}.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A 0.80m0.80\,\text{m} uniform wire at constant temperature has 3.0V3.0\,\text{V} across it and carries 0.25A0.25\,\text{A}. The wire is replaced by 0.50m0.50\,\text{m} of the same wire. Assume resistance is proportional to length. Calculate the new resistance and current at 3.0V3.0\,\text{V}.

[5 marks]

Total for this question: 5

4.2.1.4

Resistors

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

  • At constant temperature, an ohmic conductor has current directly proportional to potential difference, so its resistance is constant and its current–potential difference graph is a straight line through the origin.
  • A filament lamp heats as current increases, so its resistance rises and the graph becomes less steep.
  • A diode conducts readily in one direction but has very high resistance in reverse.
  • Thermistor resistance decreases as temperature increases, while LDR resistance decreases as light intensity increases.
  • Investigate a component using an ammeter in series, a voltmeter in parallel and paired readings as the supply is varied.
Characteristic current–potential difference graphs for a resistor, filament lamp and diode.
Worked example

Explain why a filament lamp's current–potential difference graph becomes less steep at larger currents.

  1. 1.A larger current heats the filament to a higher temperature.
  2. 2.The hotter filament has greater resistance.
  3. 3.Current therefore rises by less for each further increase in potential difference.

Answer: Heating increases the filament's resistance, so the graph's gradient decreases.

Common mistakes

  • Don't fall into the trap of calling a curved lamp graph an experimental error instead of linking it to heating and rising resistance.
  • Don't fall into the trap of saying thermistor resistance increases when temperature increases.
  • Don't fall into the trap of drawing substantial reverse current for a diode.

Exam tip

When explaining a characteristic graph, describe how its gradient changes and link that to resistance.

Tier 1 · Easy

ORIGINAL

An automatic garden light must switch on when it becomes dark. State how the resistance of its LDR changes as darkness increases.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A filament lamp carries 0.40A0.40\,\text{A} at 2.0V2.0\,\text{V} and 0.80A0.80\,\text{A} at 6.0V6.0\,\text{V}. Calculate its resistance at each potential difference and explain the change.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Describe an investigation of how the resistance of a thermistor changes with temperature. Include the circuit, measurements, one control and the processing of results.

[6 marks]

Total for this question: 6

4.2.2

Series and parallel circuits

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

  • In a series circuit, current is the same through every component, the supply potential difference is shared, and resistances add: Rtotal=R1+R2+R_{\text{total}}=R_1+R_2+\ldots.
  • In a parallel circuit, the potential difference is the same across every branch and the total current equals the sum of the branch currents.
  • Adding a parallel branch gives charge another path, so total resistance decreases and is less than the smallest branch resistance.
  • Use junctions to decide whether components share one path or sit on separate branches; do not apply series rules just because symbols are drawn next to each other.
Two resistors in series and two resistors on parallel branches.
Worked example

Two series resistors are 8Ω8\,\Omega and 12Ω12\,\Omega. The current is 0.30A0.30\,\text{A}. Calculate the total resistance and supply potential difference.

  1. 1.Rtotal=8+12=20ΩR_{\text{total}}=8+12=20\,\Omega.
  2. 2.V=IR=0.30×20=6.0VV=IR=0.30\times20=6.0\,\text{V}.

Answer: 20Ω20\,\Omega and 6.0V6.0\,\text{V}

Common mistakes

  • Don't fall into the trap of adding branch currents and then claiming that same total flows through each parallel branch.
  • Don't fall into the trap of adding parallel resistances as though the components were in series.
  • Don't fall into the trap of sharing supply potential difference between parallel branches.

Exam tip

Mark junctions first: current splits at a junction, while potential difference is common across branches.

Tier 1 · Easy

ORIGINAL

Two resistors of 4Ω4\,\Omega and 7Ω7\,\Omega are connected in series. Calculate their total resistance.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A 2.0Ω2.0\,\Omega resistor and a 4.0Ω4.0\,\Omega resistor are connected in series to a 12V12\,\text{V} supply. Calculate the circuit current and the potential difference across each resistor.

[5 marks]

Total for this question: 5

Tier 3 · Hard

ORIGINAL

Two branches are connected in parallel across a 12V12\,\text{V} supply. One branch contains a 6.0Ω6.0\,\Omega resistor. The other branch carries 3.0A3.0\,\text{A}. Calculate the resistance in the second branch and the current from the supply.

[5 marks]

Total for this question: 5

4.2.3.1

Direct and alternating potential difference

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 direct potential difference has one polarity, so it drives current in one direction; cells and batteries provide dc.
  • An alternating potential difference repeatedly reverses polarity, so the current also repeatedly reverses direction.
  • The UK mains supply is approximately 230V230\,\text{V} ac at 50Hz50\,\text{Hz}.
  • A frequency of 50Hz50\,\text{Hz} means 5050 complete cycles each second, with a period of 1/50=0.020s1/50=0.020\,\text{s}.
  • On a potential difference–time graph, dc stays on one side of zero, whereas ac crosses zero and alternates between positive and negative values.
Direct and alternating potential difference plotted against time.
Worked example

The mains frequency is 50Hz50\,\text{Hz}. Calculate the duration of one complete cycle.

  1. 1.Period T=1fT=\dfrac{1}{f}.
  2. 2.T=150=0.020sT=\dfrac{1}{50}=0.020\,\text{s}.

Answer: 0.020s0.020\,\text{s}

Common mistakes

  • Don't fall into the trap of saying alternating potential difference only changes size and never reverses polarity.
  • Don't fall into the trap of interpreting 50Hz50\,\text{Hz} as one cycle lasting 5050 seconds.
  • Don't fall into the trap of drawing a dc trace that alternates above and below zero.

Exam tip

A comparison must mention direction: dc keeps one polarity, while ac repeatedly reverses.

Tier 1 · Easy

ORIGINAL

State whether a battery supplies direct or alternating potential difference, and give the defining feature of that supply.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

The UK mains supply has a frequency of 50Hz50\,\text{Hz}. Calculate the number of complete cycles in 0.30s0.30\,\text{s} and describe what happens to the polarity.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Source A maintains a potential difference of +6.0V+6.0\,\text{V}. Source B varies between positive and negative values and completes 2525 cycles in 0.50s0.50\,\text{s}. Identify each supply type, calculate the frequency of B and compare the current directions they produce.

[5 marks]

Total for this question: 5

4.2.3.2

Mains electricity

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 three-core mains cable contains a brown live wire, blue neutral wire and green-and-yellow earth wire.
  • The live wire carries the alternating potential difference from the supply; the neutral completes the circuit and is close to 0V0\,\text{V}.
  • The earth is a safety wire connected to a metal case and normally carries no current.
  • If a fault makes the case live, a large current flows through the low-resistance earth path so the fuse melts or circuit breaker opens.
  • The live wire is dangerous even when a switch is open, so switches and fuses must be placed in the live wire.
The three insulated conductors inside a mains cable.
Worked example

Explain how the earth wire and fuse protect a person if a fault connects the live wire to a metal appliance case.

  1. 1.The earth wire provides a low-resistance path from the case to Earth.
  2. 2.A large fault current flows through the live wire, case, earth wire and fuse.
  3. 3.The fuse melts and disconnects the live supply.

Answer: The case is earthed and the large fault current causes the fuse to break the circuit.

Common mistakes

  • Don't fall into the trap of swapping the colours of the live and neutral wires.
  • Don't fall into the trap of saying the earth wire normally carries the appliance current.
  • Don't fall into the trap of putting a switch or fuse in the neutral wire and leaving the appliance connected to live.

Exam tip

For an electrical-safety explanation, trace the fault current through earth to the fuse or circuit breaker.

Tier 1 · Easy

ORIGINAL

State the insulation colour of the live, neutral and earth wires in a UK three-core mains cable.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

A fault makes the metal case of a mains appliance touch the live wire. Explain how the earth wire reduces the danger to a user.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A mains lamp is wired so that its switch opens the neutral wire rather than the live wire. The lamp goes out when the switch is opened. Explain why the lamp holder may still be dangerous to touch.

[4 marks]

Total for this question: 4

4.2.4.1

Power

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

  • Power is the rate at which energy is transferred, so one watt is one joule per second.
  • For an electrical component, use P=VIP=VI when potential difference and current are known.
  • Combining this with V=IRV=IR gives P=I2RP=I^2R, useful when current and resistance are known.
  • A larger power rating means the appliance transfers more energy each second, not that it necessarily transfers more energy overall; total energy also depends on operating time.
  • Use potential difference in volts, current in amperes and resistance in ohms to obtain power in watts.
Worked example

A heater has resistance 24Ω24\,\Omega and carries 5.0A5.0\,\text{A}. Calculate its power.

  1. 1.Use P=I2RP=I^2R because current and resistance are given.
  2. 2.P=5.02×24=25×24P=5.0^2\times24=25\times24.

Answer: 600W600\,\text{W}

Common mistakes

  • Don't fall into the trap of using P=IRP=IR and omitting the square on current.
  • Don't fall into the trap of treating watts as a unit of energy rather than energy transfer per second.
  • Don't fall into the trap of comparing power ratings without considering how long appliances operate.

Exam tip

Choose the power equation that uses only the given quantities, then show the squared current explicitly.

Tier 1 · Easy

ORIGINAL

A motor has a potential difference of 12V12\,\text{V} across it and a current of 3.0A3.0\,\text{A}. Calculate its power.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A heating resistor has resistance 8.0Ω8.0\,\Omega and carries 2.5A2.5\,\text{A}. Calculate the power transferred.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A mains heating element is rated at 1.8kW1.8\,\text{kW} when connected to 230V230\,\text{V}. Calculate its current and resistance at this operating point.

[5 marks]

Total for this question: 5

4.2.4.2

Energy transfers in everyday appliances

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

  • Electrical appliances transfer energy from the mains or a battery to other energy stores. A motor increases a kinetic energy store, while a heater increases a thermal energy store.
  • Calculate transferred energy using E=PtE=Pt, with power in watts and time in seconds for energy in joules.
  • The alternative equation E=QVE=QV links energy transferred to charge flow and potential difference.
  • Appliance efficiency depends on how much of the input energy reaches the intended store; unwanted transfers usually heat the surroundings.
  • A high-power appliance transfers energy quickly, but energy used also depends on its operating time.
Worked example

A 1.8kW1.8\,\text{kW} kettle operates for 150s150\,\text{s}. Calculate the energy transferred.

  1. 1.Convert power: 1.8kW=1800W1.8\,\text{kW}=1800\,\text{W}.
  2. 2.E=Pt=1800×150E=Pt=1800\times150.

Answer: 270000J270000\,\text{J} or 270kJ270\,\text{kJ}

Common mistakes

  • Don't fall into the trap of using 1.81.8 as the power without converting kilowatts to watts.
  • Don't fall into the trap of treating a power rating as the total energy transferred.
  • Don't fall into the trap of naming thermal energy as useful for every appliance, including a motor.

Exam tip

For E=PtE=Pt, convert kilowatts to watts and minutes to seconds before calculating joules.

Tier 1 · Easy

ORIGINAL

A 60W60\,\text{W} fan runs for 5.05.0 minutes. Calculate the energy transferred.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

9000C9000\,\text{C} of charge flows through a 12V12\,\text{V} motor. Calculate the energy transferred and name the main useful energy transfer.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 1.5kW1.5\,\text{kW} appliance operates for 1818 minutes from a 230V230\,\text{V} supply. Calculate the energy transferred and the charge that flows through the appliance.

[5 marks]

Total for this question: 5

4.2.4.3

The National Grid

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

  • The National Grid is the network of cables and transformers linking power stations to consumers. A step-up transformer increases potential difference before transmission.
  • For a given power, P=VIP=VI shows that a higher potential difference allows a smaller current.
  • Cable heating depends on Ploss=I2RP_{\text{loss}}=I^2R, so reducing current greatly reduces energy wasted heating the cables.
  • Near consumers, step-down transformers reduce the potential difference to safer, useful values.
  • Transformers do not create energy: they change the potential difference and current so electrical energy can be transferred efficiently over long distances.
Potential difference is stepped up for transmission and stepped down before consumers.
Worked example

A transmission line carries 2.0MW2.0\,\text{MW} at 200kV200\,\text{kV}. Calculate the current.

  1. 1.Convert prefixes: P=2.0×106WP=2.0\times10^6\,\text{W} and V=2.00×105VV=2.00\times10^5\,\text{V}.
  2. 2.I=PV=2.0×1062.00×105I=\dfrac{P}{V}=\dfrac{2.0\times10^6}{2.00\times10^5}.

Answer: 10A10\,\text{A}

Common mistakes

  • Don't fall into the trap of saying a step-up transformer increases power or creates energy.
  • Don't fall into the trap of claiming a high transmission current reduces cable heating.
  • Don't fall into the trap of forgetting that the step-down transformer is used before electricity reaches consumers.

Exam tip

A full efficiency explanation links higher potential difference to lower current and then to smaller I2RI^2R losses.

Tier 1 · Easy

ORIGINAL

Name the transformer used before long-distance transmission and the transformer used before electricity enters homes.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A transmission line transfers 6.0MW6.0\,\text{MW} at 300kV300\,\text{kV}. Calculate the current in the line and explain why this is preferable to transmitting the same power at a much lower potential difference.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A cable of resistance 0.80Ω0.80\,\Omega transfers 2.4MW2.4\,\text{MW}. Compare the power lost in the cable when transmission is at 12kV12\,\text{kV} and at 240kV240\,\text{kV}.

[6 marks]

Total for this question: 6

4.2.5.1

Static charge (physics 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

  • When certain insulating materials are rubbed together, electrons transfer from one material to the other. The material gaining electrons becomes negatively charged; the material losing electrons is left with an equal positive charge.
  • Protons do not move between the materials.
  • Like charges repel and unlike charges attract through non-contact electrostatic forces.
  • Because charge cannot easily flow away from an insulator, it may build up until a spark transfers charge through the air.
  • Static discharge can be useful, but it can also ignite flammable vapours or damage electronic components.
Worked example

A cloth loses electrons when rubbed against a plastic rod. State the charge on each material.

  1. 1.The rod gains electrons, so it has an excess of negative charge.
  2. 2.The cloth loses electrons, so it has an electron deficit and is positive.

Answer: The plastic rod becomes negative and the cloth becomes equally positive.

Common mistakes

  • Don't fall into the trap of saying positive charge particles move from the rod to the cloth.
  • Don't fall into the trap of reversing the signs so the object that gains electrons is called positive.
  • Don't fall into the trap of describing attraction between like charges.

Exam tip

Always state which material gains electrons before assigning the two charge signs.

Tier 1 · Easy

ORIGINAL

During rubbing, material X gains electrons from material Y. State the charge left on each material.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A negatively charged insulating strip repels a second charged strip without touching it. What does this show about the charge on the second strip and the type of force involved?

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A student rubs an insulating rod with a cloth and the rod becomes negatively charged. The rod is then brought close to a metal object and a spark occurs. Explain the charging and the spark in terms of electrons.

[5 marks]

Total for this question: 5

4.2.5.2

Electric fields (physics 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 charged object creates an electric field around itself. Another charged object placed in that field experiences a force; the interaction is stronger where the field is stronger.
  • Field lines show the direction of force on a positive test charge.
  • Around an isolated positive sphere they point radially outwards, while around a negative sphere they point radially inwards.
  • Lines drawn closer together represent a stronger field, so the field is strongest near the charged object.
  • Field lines never cross because the force on a positive test charge at one point has only one direction.
Electric field directions around isolated positive and negative spheres.
Worked example

Describe the electric field around an isolated negatively charged sphere.

  1. 1.A positive test charge would be attracted towards the negative sphere.
  2. 2.Draw radial field lines with arrowheads pointing inwards.
  3. 3.Place the lines closer together near the sphere to show the stronger field.

Answer: Radial field lines point towards the sphere and are most closely spaced near it.

Common mistakes

  • Don't fall into the trap of drawing arrows towards a positive sphere or away from a negative sphere.
  • Don't fall into the trap of allowing electric field lines to cross.
  • Don't fall into the trap of spacing field lines farther apart near the sphere even though the field is stronger there.

Exam tip

Field arrows show the force direction on a positive test charge: away from positive and towards negative.

Tier 1 · Easy

ORIGINAL

Draw the electric field pattern around an isolated positively charged sphere.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A small positive test charge is placed first near a positively charged sphere and then farther away. State the force direction in both positions and compare the force sizes.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A charged metal dome is brought progressively closer to an earthed metal sphere until a spark crosses the gap. Use the electric-field model to explain why there is a force before contact and why a spark becomes more likely as the gap decreases.

[5 marks]

Total for this question: 5

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