Question

Define the capacity of a condenser. Prove that the potential energy of a charged condenser is $U=\frac{1}{2}C{V}^2$, where $C$ is the capacity of the condenser and $V$ is potential difference.

Answer 1

We know that, $C=\frac{Q}{V}$ and hence $Q=CV$

Here, $C$ is the capacitance of the capacitor, $Q$ is the charge stored in it and $V$ is the potential difference between the two plates of the capacitor.

Consider that the capacitor is getting charged slowly.

Let us consider an intermediate stage in which a charge of magnitude $q$ is present on the capacitor and ${}_V$ be its potential difference.

Whenever some extra charge gets stored, the P.D. goes on increasing.

Hence, for the convenience of calculation, let us consider a very very small charge $dq$ getting deposited.

Since it's very small, the change in P.D. can be ignored and can be considered as constant as ${}_V$.

Work needs to be done in charging it. Let the small work done in depositing the charge $dq$ be $dW.$

This work done gets stored in the form of potential energy P.E. in the capacitor.

Hence the small gain in P.E. is $dW$

We know that work done is $w=q{}_V$

$\therefore dW={}_{V}dq$.

substituting from the very 1st equation, we get,

$dW=\frac{q}{C}\times dq$

Hence the total work done in the complete charging is given by the intergral.

$W=\int dW=\int \frac{qdq}{C}$

Since the min and max values of q are 0 and Q, we take them as limits$.$

$\therefore W={\int }_{{}_0}^{{}^Q}dW={\int }_{{}_0}^{{}^Q}\frac{qdq}{C}=\frac{Q^2}{2C}=\frac{(CV{)}^2}{2C}=\frac{1}{2}C{V}^2$