Question

When an additional charge of $2C$ 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

Answer: (D)

Energy stored by a capacitor, $U=\frac{1}{2}\frac{q^2}{C}$
So, here $U_1=\frac{1}{2}\frac{q_1^2}{C};U_2=\frac{1}{2}\frac{q_2^2}{C}$
$\therefore \frac{U_1}{U_2}={\left(\frac{q_1}{q_2}\right)}^2={\left(\frac{q}{q+2}\right)}^2$
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
$q=20C$