Answer to Question #146206 in Electric Circuits for Wisdom

Question

Derive the expression for the energy stored by a capacitor of charge Q potential difference between the plate of V and capacitance.

Answer 1

Let there be a capacitor with charge $Q$ , potential difference across it $V$ and it's capacitance $C.$

We have $Q=CV$

Also work done $W=VQ$

If the source delivers a small amount of charge $dQ$ at a constant potential $V$ , the work done $dW=dQ\times V =\frac{Q}{C}dQ$

Now the total work to deliver a charge $Q$ is given by

$W=\displaystyle\int_{0}^{Q} \frac{Q}{C}dQ=\frac{Q^2}{2C}$

Work done =energy stored in the capacitor.

Substituting $Q=CV$ in above equation we have,

Energy stored in capacitor $= E=\frac{Q^2}{2C}=\frac{1}{2} CV^2$