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?

Answer 1

Capacitance of the capacitor $C=600pF$

Potential difference $V=200V$

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

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

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

Effective combination in second case

$C^{\prime }=\frac{600\times 600}{600+600}=300pF$

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

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

$0.6\times {10}^{-5}J$

Loss in electrostatic energy=$E-E^{\prime }$

$=\frac{1}{2}\times {10}^{-5}-0.6\times {10}^{-5}$

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