Question

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

• A. $30$C
• B. $40$C
• C. $10$C
• D. $20$C

Answer 1

The correct option is D $20$C

Let initially, the charge on the plates of capacitor be $Q$ and energy stored be $E$.

$\therefore$ $E=\frac{Q^2}{2c}$ .........(1) where $c$ is the capacitance

Now new charge on the plates $Q^{\prime }=Q+2$

New energy stored $E^{\prime }=1.21E$

$\therefore$ $1.21E=\frac{(Q+2{)}^2}{2c}$ ..........(2)

Dividing (2) by (1) we get, $1.21=\frac{(Q+2{)}^2}{Q^2}$

Or $\frac{Q+2}{Q}=1.1$

$⟹$ $Q=20$ C