Question

A $600pF$ capacitor is charged by a $200V$ supply. It is then disconnected from the supply and is connected to another uncharged $600pF$ capacitor. How much electrostatic energy is lost in the process? (1 pF = ${10}^{-12}$ F)

Answer 1

Energy stored in a capacitor :-

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

$E=1.2\times {10}^{-5}J$

Another capacitor is connected after disconnecting the battery.

Equivalent capacitance (C') is;

$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{C}$

$\Rightarrow C^{\prime }=300pF$

New Energy, $E=\frac{1}{2}C{V}^2$

$\Rightarrow E^{\prime }=0.6\times {10}^{-5}J$

Loss in energy, $\Delta E=E^{\prime }-E=6\times {10}^{-5}J$