Question

If the charge on a capacitor is increased by $4$C the energy stored in it increases by $44\%$. The original charge on the capacitor is?

• A. $20$C
• B. $10$C
• C. $5$C
• D. $15$C

Answer 1

Answer: (A)

$20$C

The energy stored in the capacitor is given by
$U=\frac{q^2}{2C}$
$\Rightarrow U\propto q^2$
$\therefore \frac{U_1}{U_2}={\left(\frac{q_2}{q_1}\right)}^2$
$\frac{U_2}{U_1}-1={\left(\frac{q_2}{q_1}\right)}^2-1$
$\Rightarrow \frac{U_2-U_1}{U_1}=\frac{q_2^2}{q_1^2}-1$
$\because \frac{U_2-U_1}{U_1}\times 100=\left(\frac{q_2^2}{q_1^2}-1\right)100$
$\Rightarrow 44=\left(\frac{q_2^2}{q_1^2}-1\right)\times 100$
$\Rightarrow \frac{q_2^2}{q_1^2}=0.44+1=1.44$
$\therefore q_2=1.2q_1$
Now, $q_2-q_1=4C$
$\therefore 0.2q_1=4$
$\therefore q_1=\frac{4}{0.2}=20C$.