Capacitors

Question

A parallel-plate capacitor of capacitance 5 μF is charged to a potential of 200 V. After charging, the battery is disconnected. What is the energy stored in the capacitor?

RankGuru · Accepted Answer

The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$, which can also be written as $U = \frac{Q^2}{2C} = \frac{1}{2}QV$. All three forms are equivalent; $U = \frac{1}{2}CV^2$ is most convenient when capacitance and voltage are given. The energy is stored in the electric field between the plates, with energy density $u = \frac{1}{2}\varepsilon_0 E^2$. Here, $C = 5 \times 10^{-6}$ F and $V = 200$ V. Substituting: $U = \frac{1}{2} \times 5 \times 10^{-6} \times (200)^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 4 \times 10^4 = \frac{1}{2} \times 0.2 = 0.1$ J. Option 0.05 J is incorrect because it omits the factor of $(200)^2$ and uses V instead of $V^2$. Option 0.2 J is incorrect because it drops the factor of $1/2$. Option 0.025 J is incorrect because it uses $\frac{1}{4}CV^2$ instead of $\frac{1}{2}CV^2$. The disconnect after charging ensures the charge remains constant. Plausibility check: 0.1 J for a 5 μF capacitor at 200 V is reasonable — a large capacitor at high voltage stores measurable energy.

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