Question

The energy stored in a capacitor of capacitance C having a charge Q under a potential V is:
(a) ${\displaystyle \frac{1}{2}}{Q}^{2}V$
(b) ${\displaystyle \frac{1}{2}}{C}^{2}V$
(c) ${\displaystyle \frac{1}{2}}{\displaystyle \frac{Q^2}{V}}$
(d) ${\displaystyle \frac{1}{2}}QV$
(e) ${\displaystyle \frac{1}{2}}CV$

Answer 1

Hint: To derive the energy stored in the capacitor, bring a small test charge element, and then integrate to get total work. The total energy stored by the capacitor is network done in assembling charge on plates of the capacitor.
1. The potential difference between the plates of a capacitor with a charge $q$ and Capacitance $C$:
$V={\displaystyle \frac{q}{C}}$ …… (1)
where,
$Q$ is the charge on the capacitor,
$C$ is the capacitance of the capacitor.
2. Work done = change in potential energy of two configurations:
$W=V(Q_{final}-Q_{in})$
Writing above expression for infinitesimally small charge $\Rightarrow dW=Vdq$ …… (A)
Where,
$dq$ is the extra charge added (initially $q$ finally $q+dq$)
$dW$ is the Work done in adding extra charge $dq$.
$V$ is the instantaneous potential due to the initial charge of $q$.

Complete step by step solution:
Given:
1. The capacitance of capacitor = $C$
2. The potential across the capacitor = $V$
3. The charge on a capacitor = $Q$

To find: The expression for energy stored in the capacitor.

Step 1 of 3:
Let’s say at any random time $t$, the charge is $q$. Bring extra charge $dq$ into the system.
From equation (A), we can say work done in bringing extra charge given by,
$dW=VdQ$ where, $V$ is potential due to already present charge $q$.

Step 2 of 3:
Substitute the value of V from equation (1) we get
$dW={\displaystyle \frac{q}{C}}dq$
Integrate both sides to get total work done to store charge up to $Q$ units:
$$\underset{0}{\overset{W}{\int }}dW=\underset{0}{\overset{Q}{\int }}{\displaystyle \frac{q}{C}}dq$$
$W={\displaystyle \frac{Q^2}{2C}}$ …… (2)

Step 3 of 3:
Substitute the value of C by rearranging equation (1) and putting in equation (2) with q=Q (as finally stored charge) we get,
$$\begin{gathered} W={\displaystyle \frac{Q^2}{2({\displaystyle \frac{Q}{V}})}} \\ W={\displaystyle \frac{QV}{2}} \end{gathered}$$
This work done is stored as potential energy in the capacitor.

The energy stored in the capacitor is ${\displaystyle \frac{1}{2}}QV$. Hence option (D) is the correct answer.

Note: Work done in bringing infinitesimal charge $dq$ (the very first element) is $0$. As there is no repulsion present for the first element. But after this, every next element would be repelled by already present charge would give rise to dW work. Then, we can do the integration to get total work and total charge stored with this work.