Question

A capacitor of $200pF$ is charged by a $300V$ battery. The battery is then disconnected and the charged capacitor is connected to another uncharged capacitor of $100pF$. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor.

Answer 1

$q_o=300\times 200\times {10}^{-12}$

$=6\times {10}^{-8}C$

$E_o=\frac{1}{2}C{V}^2$

$=\frac{1}{2}\times 200\times {10}^{-12}\times (300{)}^2$

$=9\times {10}^{-6}J$

Capacitor 1 will charge capacitor 2 until $V_{AB}=V_{CD}$

$\frac{q_1}{200\times {10}^{-12}}=\frac{q_2}{100\times {10}^{-12}}$

or, $q_1=2q_2$

But, $q_1+q_2=q_o=6\times {10}^{-8}$

$\therefore 3q_2=6\times {10}^{-8}$

$q_2=2\times {10}^{-8}$

$q_1=4\times {10}^{-8}$

Energy stored in capacitor 1

$=\frac{(q_1{)}^2}{2C_1}=\frac{(4\times {10}^{-8}{)}^2}{2\times 200\times {10}^{-12}}$

$=4\times {10}^{-6}J$

Energy stored in capacitor 2

$=\frac{(q_2{)}^2}{2C_2}=\frac{(2\times {10}^{-8}{)}^2}{2\times 200\times {10}^{-12}}$

$=2\times {10}^{-6}J$

Difference in energy stored

$(9\times {10}^{-6}J)-[4\times {10}^{-6}+2\times {10}^{-6}]J$

$=3\times {10}^{-6}J$