The energy stored in a capacitor of capacitance $C$ having a charge $Q$ under a potential $V$ is

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Answer 1

$E = \frac{1}{2} C V^{2} \times \frac{C}{C}$
$Q = C V$
$\Rightarrow E = \frac{1}{2} \frac{Q ^{2}}{C}$

$\Rightarrow E = \frac{1}{2} Q V$

Answer 2

E = \frac{1}{2} C V^{2} \times \frac{C}{C}

Q = C V

\Rightarrow E = \frac{1}{2} \frac{Q ^{2}}{C}

\Rightarrow E = \frac{1}{2} Q V

The energy stored in a capacitor of capacitance $C$ having a charge $Q$ under a potential $V$ is

$\frac{1}{2} Q^{2} V$

$\frac{1}{2} C^{2} V$

$\frac{1}{2} \frac{Q ^{2}}{V}$

$\frac{1}{2} Q V$

$\frac{1}{2} C V$

Solution

$\frac{1}{2} Q V$

$E = \frac{1}{2} C V^{2} \times \frac{C}{C}$
$Q = C V$
$\Rightarrow E = \frac{1}{2} \frac{Q ^{2}}{C}$

$\Rightarrow E = \frac{1}{2} Q V$

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$\frac{1}{2} Q V$

$E = \frac{1}{2} C V^{2} \times \frac{C}{C}$
$Q = C V$
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