Question

A $400pF$ capacitor is charged with a $100V$ battery. After disconnecting battery this capacitor is connected with another $400pF$ capacitor. Then find out energy loss.

• A. $1\times {10}^{-6}J$
• B. $2\times {10}^{-6}J$
• C. $3\times {10}^{-6}J$
• D. $4\times {10}^{-6}J$

Answer 1

The correct option is A $1\times {10}^{-6}J$

Given, $C_1=400pF$ and $V_1=100V$

$E_1=\frac{1}{2}{C}_1{V}^2=2\times {10}^{-6}J$

When another uncharged capacitor of $C_2=400pF$ is connected across Ist capacitor,

$V=\frac{C_1{V}_1}{C_1+C_2}$ as $V=\frac{q_1+q_2}{C_1+C_2}$

$V=\frac{400\times {10}^{-12}\times 100}{800\times {10}^{-12}}=50V$

$E_2=\frac{1}{2}(C_1+C_2)V^2=\frac{1}{2}\times 800\times {10}^{-12}\times 2500=4\times {10}^{-6}J$

$E_2-E_1=\mid \Delta E\mid =2\times {10}^{-6}J$