Energy stored is the area under the Q-V graph

Capacitance & Charge/Discharge — AQA A-Level Physics

When charging, the powerThe rate of energy transfer. Measured in watts (W). supply does work pushing electrons from one plate to the other.
The p.d. across the capacitor increases as chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). builds up. Q is directly proportional to V (straight-line graph through the origin).
The energyThe capacity to do work. Measured in joules (J). stored is the area under the Q-V graph, which is a triangle: $E = 1/2 x base x height$.

E = \frac{1}{2} QV

$E$: energyThe capacity to do work. Measured in joules (J). stored (J)
$Q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). stored (C)
$V$: potential differenceThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). (V)

E = \frac{1}{2} CV^2

$E$: energy stored (J)
$C$: capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F). (F)
$V$: potential differenceThe energy transferred per unit charge between two points. Measured in volts (V). (V)

E = \frac{Q^2}{2C}

$E$: energy stored (J)
$Q$: charge (C)
$C$: capacitanceThe charge stored per unit potential difference across a capacitor. Measured in farads (F). (F)

Worked Example

Calculate the change in energy stored in a 1500 uF capacitor when the potential difference changes from 10 V to 30 V.

Show Solution

Use E = 1/2 CV^2 for each state

$$\Delta E = \frac{1}{2}C(V_2^2 - V_1^2)$$

Substitute values

$$\Delta E = \frac{1}{2} \times (1500 \times 10^{-6}) \times (30^2 - 10^2)$$ $$= \frac{1}{2} \times 1.5 \times 10^{-3} \times 800$$

Evaluate

$$\Delta E = 0.6 \text{ J}$$

Answer

$\Delta E = 0.6$ J

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