Question

A capacitor of capacitance C is charged to potential V. If it carries a charge Q, then the energy stored in it is

• A. $\frac{1}{2}CV$
• B. QV
• C. $\frac{1}{2}Q{V}^2$
• D. $C{V}^2$
• E. $\frac{1}{2}QV$

Answer 1

Answer: (E)

By energy stored in a charged conductor Suppose, a conductor of capacity C is charged to a potential $V$ and $Q$ the charge on conductor at a instant. The potential of the conductor
$<br/>V=\frac{Q}{C}<br/>$
Now, work done in bringing a small charge $dQ$ at this potential is,
$<br/>dW=vdQ=\frac{Q}{C}dQ<br/>$
$\therefore$ Total work done in charging it from o to $Q$ is,
$<br/>\begin{array}{l}<br/>W={\int }_0^{Q}dW={\int }_0^Q\frac{Q}{C}dQ\\ <br/>=\frac{1}{2}\frac{Q^2}{C}<br/>\end{array}<br/>$
This work is stored as the potential energy.
$<br/>U=\frac{1}{2}\frac{Q^2}{C}(\because Q=VC)<br/>$
$<br/>U=\frac{1}{2}C{V}^2=\frac{1}{2}QV<br/>$