Question

The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection, if the plates of the capacitor are separated so that the distance between them is twice the original distance, then electrostatic energy becomes :

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

Answer 1

Answer: (C)

$\frac{E}{2}$

$E=\frac{1}{2}C{v}^2$

$\Rightarrow C\propto \frac{1}{d}$

$\Rightarrow \frac{C_1}{C_2}=\frac{d_2}{d_1}=\frac{2d}{d}=2$

$\Rightarrow \frac{E_1}{E_2}=\frac{C_1}{C_2}=\frac{2}{1}$

$\Rightarrow \frac{E}{E_2}=\frac{2}{1}$

$\Rightarrow E_2=\frac{E}{2}$