Energy stored in a capacitor

Practical method

Aim

Investigate how the energy stored in a capacitor depends on the p.d. across it, testing E = \(\dfrac{1}{2}CV^{2}\).

Variables

Independent variable: charging p.d. V
Dependent variable: energy released on discharge
Control variables: the same capacitor; discharge through the same joulemeter or resistor.

Apparatus

• 1000 μF capacitor
• Variable d.c. supply
• Joulemeter (or data logger measuring V and I during discharge)
• Voltmeter
• Two-way switch

Method

1. Charge the capacitor to a measured p.d. V.
2. Discharge it through the joulemeter (or a resistor with a data logger) and record the energy transferred.
3. Repeat for p.d.s from 2 V to 8 V.
4. Plot energy against V²: a straight line through the origin with gradient \(\dfrac{C}{2}\) confirms E = \(\dfrac{1}{2}CV^{2}\).

Sample results

C = 1000 μF (nominal).

Analysis

E is proportional to V²: gradient = \(\dfrac{31.5 \times 10^{-3}}{64.0}\) = 4.9 × 10⁻⁴ J V⁻², so C = 2 × gradient = 980 μF, within the capacitor's tolerance of 1000 μF.

Uncertainty

Joulemeters have a resolution of about 0.1 mJ, so the lowest reading (1.9 mJ) is uncertain by about 5%. Higher p.d.s give smaller percentage uncertainties.

Errors and improvements

Risk assessment

Common mistakes

• Plotting E against V and expecting a straight line.
• Forgetting the factor of ½.
• Using μF without converting to F.

Worked analysis

At 8.0 V, E = \(\dfrac{1}{2}CV^{2}\) = 0.5 × 1000 × 10⁻⁶ × 8.0² = 0.032 J (32 mJ); measured 31.5 mJ.

Common misconception

The battery supplies QV, but only ½QV is stored in the capacitor; the rest is dissipated in the resistance while charging.