Question

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

• A. $\frac{E}{2}$
• B. $\frac{E}{4}$
• C. $\frac{E}{8}$
• D. $\frac{3}{4}E$

Answer 1

Answer: (C)

$\frac{E}{8}$

Answer 2

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
$E_{i} - E_{f} = \frac{5E}{4} - \frac{9E}{8} = \frac{E}{8}$