AQA 7408 · Section 3.7.4 · Paper 2
Capacitors: charge, discharge and energy
Connect capacitance and dielectrics to stored energy, exponential RC curves and the graph methods AQA assesses. Diagnose the time-constant misconception, test predictions in the rebuilt simulator, then complete the exam assessment.
Content diagnostic
What does one time constant mean?
Active revision
Use a three-pass RC routine
Identify the state
Is the capacitor charging towards CV₀ or discharging from Q₀? This chooses 1 − e⁻ˣ or e⁻ˣ.
Find RC first
Convert R to ohms and C to farads. RC has units of seconds and sets the horizontal scale.
Check the limit
At t = 0 and as t becomes large, verify that Q, V and I approach physically sensible values.
Capacitance and dielectrics
Link stored charge to geometry and polarisation
Capacitance
One farad stores one coulomb per volt. At fixed capacitance, Q against V is a straight line through the origin with gradient C.
Parallel plates
Increasing plate area A or relative permittivity εr increases C. Increasing plate separation d reduces C. Use ε₀ = 8.85 × 10⁻¹² F m⁻¹.
Dielectric action
In a polar dielectric, molecular dipoles rotate towards alignment with the applied electric field. A non-polar molecule can also become polarised when its charge distribution is distorted. Bound surface charges produce an opposing field, reducing the resultant field and p.d. for a given free charge.
Energy stored
Read energy as an area, then choose the useful equation
On a graph of V against Q, energy is the triangular area under the line from 0 to Q.
Use this when capacitance and potential difference are known. At fixed C, doubling V quadruples E.
Use this when charge is fixed. Increasing C then reduces the stored energy.
During discharge, stored electrical energy is transferred mainly to thermal energy in the resistor.
Useful simulation
Measure the exponential, do not just watch it
The rebuilt white-and-green model puts the controls, signed current and three aligned graphs in one view. Drag time to exact RC multiples or play the model; every result is also reported numerically.
Time-constant check
Select charging and set t = 1.00RC. Record Q/Qfinal, V/V₀ and I/I₀. Switch to discharging and explain why all three remaining magnitudes are 36.8%.
Change one component
Double R while C and V₀ stay fixed. Compare RC, initial current and the fraction reached after one RC. Then restore R and double C.
Find the half-life
In discharge mode, move the scrubber until Q is 50.0% of Q₀. Check that t/RC is about 0.69, then connect this to T½ = 0.69RC.
Track energy
Halve the capacitor p.d. during discharge and observe that E is one quarter of its initial value, not one half.
Charge and discharge graphs
Use the same exponential structure for Q, V and I
Discharging
If current during charging is chosen positive, discharge current is negative. Many questions plot its magnitude. State the convention before interpreting signs.
Charging
Q₀ = CV₀ is the final charge in the charging equation. Charging current starts at V₀/R and decays to zero as capacitor p.d. approaches supply p.d.
Graph methods AQA can assess
- One time constant: discharge leaves e⁻¹ = 0.368 of Q₀, V₀ and |I₀|; charging reaches 1 − e⁻¹ = 0.632 of the final Q and V.
- Half-life: set Q/Q₀ = ½ to obtain T½ = RC ln 2 = 0.69RC.
- Gradient: I = dQ/dt for signed charge. The magnitude of the slope of a falling Q–t curve is current magnitude.
- Area: area under an I–t graph is charge transferred. Area under a V–Q graph is stored energy.
- Straight-line form: ln Q = ln Q₀ − t/RC and ln V = ln V₀ − t/RC, so the gradient is −1/RC.
Practical and data analysis
Determine a time constant from discharge data
Method
- Charge the capacitor fully to a measured V₀ using a suitable d.c. supply.
- Disconnect the supply and discharge through a known resistor while recording capacitor p.d. against time.
- Choose R and C so RC is long enough for the logger or timer to resolve.
- Plot V against t to inspect the exponential, then plot ln V against t.
- Use RC = −1/gradient and compare with the component-value product.
Quality and safety
- Use a high-input-resistance voltmeter or voltage sensor so measurement does not create a significant parallel discharge path.
- Fully discharge safely before changing the circuit; observe voltage and polarity ratings for electrolytic capacitors.
- Repeat from the same V₀ and keep resistor temperature steady.
- Account for resistor tolerance, capacitor tolerance, lead resistance and the sensor input resistance.
Misconception repair
Seven statements to replace
- “A capacitor charges linearly.” In an RC circuit, current falls as capacitor p.d. rises, producing an exponential approach.
- “RC is the time for complete charge or discharge.” The mathematical curve approaches its limit asymptotically; one RC means 63.2% changed or 36.8% remaining.
- “The capacitor stores current.” It stores separated charge and energy; current is the rate at which charge moves.
- “Doubling voltage doubles energy.” At fixed C, E = ½CV², so doubling V quadruples energy.
- “A dielectric adds free charge.” Its bound charges polarise and reduce the internal field; the effect on free charge depends on whether the supply remains connected.
- “Discharge current has to be positive.” Its sign depends on the chosen reference direction; its magnitude decays exponentially.
- “Area under Q against V is always energy.” Energy is ∫V dQ: area under V plotted vertically against Q horizontally.
Original exam-style assessment
33 marks with worked feedback
Attempt every part before opening its answer. Convert μF to F and kΩ to Ω, show the exponent as a dimensionless ratio and state current sign conventions when discussing graph gradients.
1. Define capacitance and state its SI unit.[2 marks]
Capacitance is charge stored per unit potential difference: C = Q/V. Its SI unit is the farad, F (equivalent to C V⁻¹).
2. A parallel-plate capacitor has plate area 3.6 × 10⁻² m² and separation 0.80 mm. The dielectric has relative permittivity 2.4. Calculate the capacitance. Use ε₀ = 8.85 × 10⁻¹² F m⁻¹.[3 marks]
C = Aε₀εᵣ/d = (3.6 × 10⁻² × 8.85 × 10⁻¹² × 2.4)/(0.80 × 10⁻³) = 9.56 × 10⁻¹⁰ F, or 0.956 nF.
3. A dielectric is inserted fully between the plates of an isolated charged capacitor. Explain the microscopic action of the dielectric and state what happens to capacitance and potential difference.[4 marks]
Polar molecules rotate so that their dipoles tend to align with the field; in a non-polar dielectric, charge distributions become polarised. Bound surface charges create a field opposing the original field, reducing the resultant field and p.d. for the same free charge. Since C = Q/V and Q is fixed for an isolated capacitor, capacitance increases and potential difference decreases.
4. A charge–potential-difference graph is a straight line from the origin to Q = 6.0 mC at V = 10 V. Determine the capacitance and the energy stored.[3 marks]
C = Q/V = 6.0 × 10⁻³/10 = 6.0 × 10⁻⁴ F. Energy is the area under the V-against-Q relationship: E = ½QV = ½ × 6.0 × 10⁻³ × 10 = 3.0 × 10⁻² J.
5. A 220 μF capacitor is charged to 12 V. Calculate its stored energy. The potential difference is then halved without changing C. State the new energy as a fraction of the original.[3 marks]
E = ½CV² = ½ × 220 × 10⁻⁶ × 12² = 1.58 × 10⁻² J. Because E ∝ V² at fixed C, halving V leaves one quarter of the original energy.
6. A 470 μF capacitor initially at 9.0 V discharges through a 2.2 kΩ resistor. Calculate the charge and potential difference after 1.50 s.[4 marks]
RC = 2200 × 470 × 10⁻⁶ = 1.034 s and Q₀ = CV₀ = 4.23 × 10⁻³ C. Q = Q₀e^(−t/RC) = 4.23 × 10⁻³e^(−1.50/1.034) = 9.91 × 10⁻⁴ C. V = Q/C = 2.11 V (equivalently V = V₀e^(−t/RC)).
7. During discharge, charge falls from 4.8 mC to 2.4 mC in 0.76 s. The resistance is 1.5 kΩ. Use the time to halve to determine the capacitance.[3 marks]
T½ = 0.69RC, so C = T½/(0.69R) = 0.76/(0.69 × 1500) = 7.34 × 10⁻⁴ F, or 734 μF.
8. For a discharging capacitor, explain what the gradient of a Q–t graph and the area under an I–t graph represent. Describe how the magnitude of the initial gradient changes if R is doubled.[4 marks]
The gradient dQ/dt is the signed current; because Q decreases, it is negative for the chosen positive plate. Its magnitude is the current. The area under an I–t graph is the charge transferred, with the sign depending on the current convention. Initially |I₀| = V₀/R, so doubling R halves the magnitude of the initial Q–t gradient.
9. Describe how measurements of capacitor potential difference during discharge can be used to determine RC using a straight-line graph. Include two experimental precautions or improvements.[5 marks]
Measure V at known times while the capacitor discharges through a known resistor. From V = V₀e^(−t/RC), take natural logarithms: ln V = ln V₀ − t/RC. Plot ln V against t; the gradient is −1/RC, so RC = −1/gradient. Suitable precautions include using a high-resistance voltage sensor so it does not provide a significant extra discharge path, fully charging before each run, using data logging for short time constants, keeping component temperature steady, repeating measurements, or selecting R and C to give a measurable time constant.
10. A charging circuit has its resistance doubled while C and V₀ stay constant. State the effect on the time constant and on the fraction of final charge reached after one new time constant.[2 marks]
The time constant doubles because τ = RC. After one new time constant the fraction reached is unchanged at 1 − e⁻¹ = 0.632, or 63.2%.
Specification coverage
AQA 7408 section 3.7.4
- 3.7.4.1(a) Define capacitance using C = Q/V.
- 3.7.4.2(a) Describe dielectric action in a capacitor.
- 3.7.4.2(b) Use C = Aε0εr/d for a parallel-plate capacitor with dielectric.
- 3.7.4.2(c) Define relative permittivity and dielectric constant.
- 3.7.4.2(d) Describe how a simple polar molecule rotates in an electric field.
- 3.7.4.3(a) Interpret area under a charge-potential difference graph as energy stored.
- 3.7.4.3(b) Use capacitor energy equations E = ½QV = ½CV² = ½Q²/C.
- 3.7.4.4(a) Interpret charging and discharging graphs for Q, V and I against time.
- 3.7.4.4(b) Use gradients and areas under capacitor graphs where appropriate.
- 3.7.4.4(c) Define and calculate time constant RC.
- 3.7.4.4(d) Determine RC from graphical data.
- 3.7.4.4(e) Use time to halve T½ = 0.69RC.
- 3.7.4.4(f) Use Q = Q0e^(−t/RC) for capacitor discharge.
- 3.7.4.4(g) Use corresponding exponential equations for V and I during discharge.
- 3.7.4.4(h) Use Q = Q0(1 − e^(−t/RC)) for capacitor charge.
Build the AQA Paper 2 connections
Return to the AQA 3.7 fields module hub. Revisit current, charge and electrical energy, resistance and current–voltage characteristics, and circuits, e.m.f. and internal resistance. Connect stored energy to mechanics and materials, then continue through the AQA Paper 2 revision hub, problem-solving practice and MCQ practice.
Written against all 15 AQA Physics 7408 points in sections 3.7.4.1–3.7.4.4. Questions are original. The static circuit is an original generated PNG with no assessed text; the interactive model is the rebuilt PhysicsUK capacitor simulator.