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. The electrostatic energy is lost in the process is:

Answer 1

Capacitance of the capacitor $C=600pF$

Potential difference $V=200V$

The electrostatic energy stored in the capacitor $E=\frac{1}{2}C{V}^2$

$\therefore E=\frac{1}{2}\times (600\times {10}^{-12})(200{)}^2=12\times {10}^{-6}J$

Now if the charged capacitor $C$ is disconnected and connected to an uncharged capacitor $C^{\prime }=600pF$ the equivanlent capacitance of the combination is $\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C^{\prime }}$

$\therefore C_{eq}=300pF$

Now the Electrostatic energy is $E^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2$

$\therefore E^{\prime }=\frac{1}{2}\times (300\times {10}^{-12})\times \left({200}^2\right)=6\times {10}^{-6}J$

Energy loss $(E-E^{\prime })=6\times {10}^{-6}J$

Hence the electrostatic energy lost in the process is $6\times {10}^{-6}J$