A-Level — physicsHard3 min read

Capacitance

C = Q/V; energy stored E = ½CV²; charging/discharging through resistors; time constant RC

Tutor AI Team
Tutor AI Team

# Capacitance — A-Level Physics

Capacitors store energy in electric fields. They're fundamental to electronics — found in cameras (flash), defibrillators, power supplies, and timing circuits.

1. Capacitance

C=QV\boxed{C = \frac{Q}{V}}

  • CC = capacitance (F, farads)
  • QQ = charge stored (C)
  • VV = potential difference across capacitor (V)

1 farad = 1 coulomb per volt (very large; typical capacitors are μF, nF, or pF).

Parallel Plate Capacitor

C=ε0εrAdC = \frac{\varepsilon_0 \varepsilon_r A}{d}

Where εr\varepsilon_r = relative permittivity (dielectric constant).

2. Combinations

Parallel: CT=C1+C2+...C_T = C_1 + C_2 + ... (stores more charge at same V)

Series: 1CT=1C1+1C2+...\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + ...

3. Energy Stored

E=12QV=12CV2=Q22C\boxed{E = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}}

This is the area under the Q-V graph (triangle).

4. Charging and Discharging

Discharging Through a Resistor

Q=Q0e−t/RCQ = Q_0 e^{-t/RC} V=V0e−t/RCV = V_0 e^{-t/RC} I=I0e−t/RCI = I_0 e^{-t/RC}

All quantities decay exponentially.

Time Constant

τ=RC\tau = RC

  • After one time constant (t=RCt = RC): value drops to 1/e≈37%1/e \approx 37\% of initial
  • After 5τ5\tau: capacitor is effectively fully discharged (<1%)

Charging Through a Resistor

Q=Q0(1−e−t/RC)Q = Q_0(1 - e^{-t/RC}) V=V0(1−e−t/RC)V = V_0(1 - e^{-t/RC})

Current still decays: I=I0e−t/RCI = I_0 e^{-t/RC}

Worked Example: Example 1

Problem

A 220 μF capacitor is charged to 12 V. Find the charge and energy stored.

Q=CV=220×10−6×12=2.64×10−3Q = CV = 220 \times 10^{-6} \times 12 = 2.64 \times 10^{-3} C = 2.64 mC E=12CV2=12×220×10−6×144=0.0158E = \frac{1}{2}CV^2 = \frac{1}{2} \times 220 \times 10^{-6} \times 144 = 0.0158 J = 15.8 mJ

Solution

Worked Example: Discharging

Problem

A 470 μF capacitor at 9 V discharges through a 10 kΩ resistor. Find the time constant and the voltage after 3 s.

τ=RC=10000×470×10−6=4.7\tau = RC = 10000 \times 470 \times 10^{-6} = 4.7 s V=9e−3/4.7=9e−0.638=9×0.528=4.75V = 9e^{-3/4.7} = 9e^{-0.638} = 9 \times 0.528 = 4.75 V

Solution

Worked Example: Half-Life of Discharge

Problem

Time for voltage to halve: V0/2=V0e−t/RCV_0/2 = V_0 e^{-t/RC} 1/2=e−t/RC1/2 = e^{-t/RC} t1/2=RCln⁡2=0.693RCt_{1/2} = RC\ln 2 = 0.693RC

Solution

6. Practice Questions

    1. A 100 μF and 300 μF capacitor are connected (a) in parallel, (b) in series. Find the total capacitance. (2 marks)
    1. A camera flash uses a 1000 μF capacitor charged to 300 V. Calculate the energy stored. (2 marks)
    1. A 47 μF capacitor discharges through a 100 kΩ resistor. Calculate the time constant and the time for the voltage to drop to 10% of its initial value. (3 marks)

    Answers

    1. (a) CT=100+300=400C_T = 100 + 300 = 400 μF. (b) 1/CT=1/100+1/300=4/3001/C_T = 1/100 + 1/300 = 4/300 → CT=75C_T = 75 μF.
    1. E=12CV2=12(0.001)(90000)=45E = \frac{1}{2}CV^2 = \frac{1}{2}(0.001)(90000) = 45 J.

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Summary

  • C=Q/VC = Q/V; energy =12CV2= \frac{1}{2}CV^2
  • Discharge: Q,V,IQ, V, I all decay as e−t/RCe^{-t/RC}
  • Time constant τ=RC\tau = RC; after 5τ5\tau effectively fully discharged
  • Charge: Q=Q0(1−e−t/RC)Q = Q_0(1 - e^{-t/RC})
Tags:
#a-level#physics#fields#capacitance#capacitor#time-constant#rc-circuits

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