Q75 / 90 · JEE Main 2022 (24 Jun Shift 2) · Physics

If the charge on a capacitor is increased by $2 C$ , the energy stored in it increases by $44 %$ . The original charge on the capacitor is (in $C$ )

Solution

Energy stored in a capacitor is given by $E = \frac{Q^{2}}{2 C}$ .

Let initial charge be $Q_{0}$ .

Then, $1.44 (\frac{Q_{0}^{2}}{2 C}) = \frac{{(Q_{0} + 2)}^{2}}{2 C}$

$\Rightarrow 1.2 Q_{0} = Q_{0} + 2$

$Q_{0} = \frac{2}{0.2} = 10 C$

Hence, the correct answer is option: 'A'.