(D) Case $1$: Battery is disconnected. The charge $Q$ remains constant.$Q = Q_0 = C_0 V_0$New separation $d' = 2d$. The new capacitance is $C' = \frac

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(D) Case $1$: Battery is disconnected. The charge $Q$ remains constant.$Q = Q_0 = C_0 V_0$New separation $d' = 2d$. The new capacitance is $C' = \frac{\epsilon_0 A}{d'} = \frac{\epsilon_0 A}{2d} = \frac{C_0}{2}$.The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0/2)} = C_0 V_0^2$.Case $2$: Battery remains connected. The potential $V$ remains constant.$V = V_0$New capacitance $C' = \frac{C_0}{2}$.The energy stored is $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} \times (\frac{C_0}{2}) \times V_0^2 = \frac{1}{4} C_0 V_0^2$.Ratio: $\frac{E_2}{E_1} = \frac{\frac{1}{4} C_0 V_0^2}{C_0 V_0^2} = 0.25$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

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$A$ parallel plate capacitor having capacity $C_0$ is charged to $V_0$. With the 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 the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

AP EAMCET 2023 All AP EAMCET

A
$0.5$
B
$1.5$
C
$2$
D
$0.25$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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