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Q.
Energy of a charged capacitor is $E$. It is allowed to share its charge with an identical capacitor charged to half its potential. Loss in the energy of the system is
COMEDKCOMEDK 2008Electrostatic Potential and Capacitance
Solution:
Energy of a charged capacitor, $E = \frac{1}{2} CV^2$
Energy of another identical capacitor charged to $\frac{V}{2} , E' = \frac{1}{2} C \left( \frac{V}{2} \right)^2 = \frac{E}{4} $
Initial energy of the system $E_{i } = E +\frac{E}{4} = \frac{5E}{ 4}$
Common potential of the system after sharing charge
$ V' = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} = \frac{CV + C \frac{V}{2}}{C + C} = \frac{3V}{4}$
$ \therefore $ Final energy of the system,
$ E_{f} = 2 \times \frac{1}{2} CV'^{2} $
$ \, \, \, \, \, = C\left(\frac{3V}{4}\right)^{2} = \frac{9}{8}E $
$ \therefore $ Loss in energy of the system
$E_{i} - E_{f} = \frac{5E}{4} - \frac{9E}{8} = \frac{E}{8} $