Question

A condenser of capacitance $12 \mu F$ was initially charged to $20 V$. Now potential difference is made $40 V$. The increase in potential energy is

• A. $7.2 \times 10^{-3} J$
• B. $4 \times 10^{-3} J$
• C. $3 \times 10^{-4} J$
• D. $5 \times 10^{-6} J$

Answer 1

Answer: (A)

$\Delta E =\frac{1}{2} CV _{2}^{2}-\frac{1}{2} CV _{1}^{2}$
$ \therefore \Delta E =\frac{1}{2} C \left( V _{2}^{2}- V _{1}^{2}\right) $
$\therefore \Delta E =\frac{1}{2} \times 12 \times 10^{-6}\left((40)^{2}-(20)^{2}\right) $
$\Delta E =6 \times 10^{-6}(1600-400)$
$\Delta E =6 \times 10^{-6}(1200)$
$ \Delta E =72 \times 10^{-4} J$
$\simeq 7.2 \times 10^{-3} J$