Q130 / 180 · KCET 2013 · Physics

When an additional charge of $2 \mathrm{C}$ is given to a capacitor, energy stored in it is increased by $21 \%$. The original charge of the capacitor is

A
$30 \mathrm{C}$
B
$40 \mathrm{C}$
C
$10 \mathrm{C}$
Correct answer
D
$20 \mathrm{C}$

Solution

Energy stored by a capacitor, $U=\frac{1}{2} \frac{q^{2}}{C}$ So, here $U_{1}=\frac{1}{2} \frac{q_{1}^{2}}{C} ; \quad U_{2}=\frac{1}{2} \frac{q_{2}^{2}}{C}$ $\therefore \quad \frac{U_{1}}{U_{2}}=\left(\frac{q_{1}}{q_{2}}\right)^{2}=\left(\frac{q}{q+2}\right)^{2}$ Given, $\frac{U_{2}-U_{1}}{U_{1}}=0.21$ or $\frac{U_{2}}{U_{1}}=1.21 \Rightarrow\left(\frac{q}{q+2}\right)^{2}=1.21$ On solving we, get $q=20 \mathrm{C}$

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

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