A parallel plate capacitor having capacity $\mathrm{C}_0$ is charged to $\mathrm{V}_0$. With battery…
A parallel plate capacitor having capacity $\mathrm{C}_0$ is charged to $\mathrm{V}_0$. With battery disconnected, if the separation between the plates is doubled then the energy stored in it is $E_1$. Instead if the separation between the plates is doubled, with battery in connection, the energy stored in it is $\mathrm{E}_2$. Then the value of $\frac{E_2}{E_1}$ is
$0.5$
$1.5$
$2$
$0.25$
Solution
When battery disconnected, charge is constant
$\begin{aligned}
& \mathrm{Q}=\mathrm{Q}_0=\mathrm{C}_0 \mathrm{~V}_0 \\
& \mathrm{~d}^{\prime}=2 \mathrm{~d} \quad\left(\text { Here } \mathrm{C}_0=\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right) \\
& \mathrm{C}^{\prime}=\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}^{\prime}}=\frac{\epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=\frac{\mathrm{C}_0}{2} \\
& \mathrm{E}_1=\frac{\mathrm{Q}^2}{2 \mathrm{c}^{\prime}} \\
& \mathrm{E}_1=\frac{\mathrm{Q}_0^2}{2 \times \frac{\mathrm{c}_0}{2}}=\mathrm{C}_0 \mathrm{~V}_0^2
\end{aligned}$
When battery in connection then potential will same.
$\begin{aligned}
& \mathrm{V}=\mathrm{V}_0 \\
& \mathrm{C}=\frac{\mathrm{C}_0}{2} \\
& \mathrm{E}_2=\frac{1}{2} C \mathrm{~V}^2=\frac{1}{2} \times\left(\frac{\mathrm{C}_0}{2}\right) \times \mathrm{V}_0^2=\frac{1}{4} \mathrm{C}_0 \mathrm{~V}_0^2 \\
& \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{\frac{1}{4} \mathrm{C}_0 \mathrm{~V}_0^2}{\mathrm{C}_0 \mathrm{~V}^2}=0.25
\end{aligned}$