A-Level Physics

Capacitors

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A capacitor stores separated charge and energy in an electric field. Capacitance is defined by C = \(\dfrac{Q}{V}\), so a capacitor of capacitance C stores charge magnitude Q on each plate when the potential difference is V. One farad is one coulomb per volt. For an ideal capacitor the energy stored can be written E = \(\dfrac{1}{2}QV\) = \(\dfrac{1}{2}CV^{2}\) = \(\dfrac{Q^{2}}{2C}\), depending on which quantities are known.

When an initially uncharged capacitor is connected to a d.c. supply through a resistor, the charging current is initially largest. As charge accumulates, the capacitor potential difference rises and the potential difference across the resistor falls, so current decreases. In the ideal long-time limit the capacitor potential difference approaches the supply emf and current approaches zero. During discharge through a resistor, charge, capacitor p.d. and current magnitude fall exponentially.

The characteristic time constant is τ = RC. For discharge, quantities such as charge and p.d. follow forms like Q = Q0 \(e^{-t/RC}\) and V = V0 \(e^{-t/RC}\). After one time constant a discharging quantity has fallen to e⁻¹ ≈ 0.368 of its initial value. Log-linear plots can be used to determine RC because ln V = ln V0 - \(\dfrac{t}{RC}\) for an ideal discharge.

Capacitance and stored charge

Charge resides as equal and opposite quantities on the two plates; the capacitor as a whole can remain electrically neutral.

Energy stored

Charging a capacitor requires work because later charge is moved against an increasing potential difference. The stored energy is E = \(\dfrac{1}{2}QV\) = \(\dfrac{1}{2}CV^{2}\) = \(\dfrac{Q^{2}}{2C}\). The factor of one half comes from the p.d. rising from zero to its final value during ideal charging.

Energy is stored in the electric field. When the capacitor discharges through a resistor, this energy is transferred mainly to thermal energy in the resistor.

Charging through a resistor

For a d.c. supply through a resistor, charge and capacitor p.d. rise exponentially towards final values while current falls towards zero. Initially the uncharged capacitor has zero p.d. and current is maximum. At long times the capacitor p.d. approaches the supply p.d. and ideal current approaches zero.

Discharging

During ideal discharge, Q, V and current magnitude fall exponentially. The time constant is τ = RC. After one time constant, a discharging quantity has fallen to e⁻¹ ≈ 0.368 of its initial value; after several time constants it is close to zero.

Linearising exponential data

Taking natural logarithms gives a straight-line form. For discharge, ln V = ln V0 - \(\dfrac{t}{RC}\), so a graph of ln V against t has gradient -\(\dfrac{1}{RC}\). Similar treatment applies to charge or current.

This method can be more reliable than trying to judge a curved exponential graph by eye.

Practical RC measurements

Use a voltmeter or data logger with sufficiently high input resistance so the measurement device does not significantly alter the circuit. Choose R and C to give a measurable time constant. Repeating the discharge from the same initial condition improves comparability.

Exam technique

Distinguish capacitor charge from circuit current. Current is the rate at which charge changes, not the amount stored. In exponential calculations, identify whether the quantity is charging towards a final value or discharging towards zero before choosing the form of the equation.

Charging equation and graph areas

The area under a current–time graph equals the charge that has flowed, and the area under a p.d.–charge graph equals the energy stored.

Worked example

Worked example 1: A 220 μF capacitor is charged to 12 V. Q = CV = 220×10⁻⁶ × 12 = 2.64×10⁻³ C. Stored energy = \(\dfrac{1}{2}CV^{2}\) ≈ 1.58×10⁻² J.

Worked example 2: A 220 μF capacitor charged to 12 V stores Q = CV = 220 × 10⁻⁶ × 12 = 2.64 × 10⁻³ C and energy \(\dfrac{1}{2}CV^{2}\) = 1.584 × 10⁻² J.

A 220 μF capacitor at 12 V stores Q = 2.64 × 10⁻³ C and E = \(\dfrac{1}{2}CV^{2}\) ≈ 1.58 × 10⁻² J.

Worked example 3: A 150 μF capacitor charged to 10 V stores Q = 1.50 × 10⁻³ C.

Misconception check

When current has fallen to approximately zero after charging from a d.c. supply, the capacitor is not uncharged. It stores charge while its p.d. is close to the supply voltage.

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State the definition of capacitance.

C = \(\dfrac{Q}{V}\).

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