A parallel plate capacitor of capacity $100 \mu \mathrm{F}$ is charged by a battery of 50 volts. The battery…
A parallel plate capacitor of capacity $100 \mu \mathrm{F}$ is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is
$\frac{125}{2} \times 10^{-3}$
$125 \times 10^{-3}$
$1.25 \times 10^{-3}$
$0.0125 \times 10^{-3}$
Solution
$
C=100 \mu \mathrm{F}, V=50 \text { volt }
$
Capacitance of parallel plate capacitor
$
C=\frac{\varepsilon_0 A}{d}
$
$d=$ separation between the plates
Initial energy
$
\begin{aligned}
E_1 & =\frac{1}{2} C V^2 \\
& =\frac{1}{2} \times\left(100 \times 10^{-6}\right) \times(50)^2 \\
& =50 \times 10^{-6} \times 2500 \\
& =125 \times 10^{-3} \mathrm{~J}
\end{aligned}
$
When distance between the plates becomes double, then capacitance $C^{\prime}=\frac{C}{2}=\frac{100}{2}=50 \mu \mathrm{F}$
Final energy $E_2=\frac{1}{2} C^{\prime} V^2$
$\begin{aligned} & =\frac{1}{2} \times 50 \times 10^{-6} \times(50)^2 \\ & =25 \times 10^{-6} \times 2500 \\ & =625 \times 10^{-4} \\ & =625 \times 10^{-3} \mathrm{~J}\end{aligned}$
$\begin{aligned} & \text { Additional energy }=E_1-E_2 \\ & =125 \times 10^{-3}-625 \times 10^{-3} \\ & =62.5 \times 10^{-3} \\ & =\frac{125}{2} \times 10^{-3} \mathrm{~J} \\ & \end{aligned}$