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Prove that the energy stored in a parallel plate capacitor is given by \frac{1}{2} \mathrm{CV}^{2}
Physics | Previous Year Question Papers
Prove that the energy stored in a parallel plate capacitor is given by \frac{1}{2} \mathrm{CV}^{2}

Ans: Let us suppose a capacitor is connected to a battery and it supplies a small amount of change dq at constant potential V, then a small amount of work done by the battery is given by

    \[\begin{array}{l} \mathrm{d} W=\mathrm{V} \mathrm{dq} \\ \Rightarrow \mathrm{dW}=\frac{\mathrm{qc}}{\mathrm{dq}} \ldots \ldots(\text { Since, } \mathrm{q}=\mathrm{CV}) \end{array}\]

Total work done where capacitor is fully changed to q.

    \[\begin{array}{l} \int d W=W=\int_{0}^{q} \frac{q}{c} d q \\ \Rightarrow W=\int_{0}^{q} q d q \\ \Rightarrow W=\frac{q^{2}}{2 C}=\frac{C^{2} V^{2}}{2 C} \\ \Rightarrow W=\frac{1}{2} C V^{2} \end{array}\]

This work done is stored in the capacitor in the form of electrostatic potential energy.

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