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The energy stored in a capacitor of capacitance $C$ charged to voltage $V$ (holding charge $Q$) is
A$CV^2$
B$\dfrac{1}{2}QV^2$
C$\dfrac{1}{2}CV^2$
D$\dfrac{Q}{2C}$
Answer & Solution
Correct answer: C. $\dfrac{1}{2}CV^2$
1. Charging transfers charge against a rising potential; the work stored is $U = \dfrac{1}{2}QV$.
2. Using $Q = CV$: $U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$.
3. Option (a) is the standard form; note $\dfrac{1}{2}QV^2$ and $\dfrac{Q}{2C}$ are dimensionally wrong for energy.
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', Long Answer Q.19_
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