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)

Energy of a charged capacitor, $E=\frac{1}{2}C{V}^2$
Energy of another identical capacitor charged to $\frac{V}{2},E^{\prime }=\frac{1}{2}C{\left(\frac{V}{2}\right)}^2=\frac{E}{4}$
Initial energy of the system $E_i=E+\frac{E}{4}=\frac{5E}{4}$
Common potential of the system after sharing charge
$V^{\prime }=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}=\frac{CV+C\frac{V}{2}}{C+C}=\frac{3V}{4}$
$\therefore$ Final energy of the system,
$E_f=2\times \frac{1}{2}C{V}^{\prime 2}$
$=C{\left(\frac{3V}{4}\right)}^2=\frac{9}{8}E$
$\therefore$ Loss in energy of the system
$E_i-E_f=\frac{5E}{4}-\frac{9E}{8}=\frac{E}{8}$