Q144 / 160 · AP EAMCET 2015 · Physics

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 the electrostatic energy becomes

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

Solution

$\begin{aligned} & \text { } E=\frac{1}{2} C V^2 \Rightarrow C \propto \frac{1}{\mathrm{~d}} \\ & \Rightarrow \frac{C_1}{C_2}=\frac{d_1}{d_2}=\frac{2 d}{d}=2\left[\because C=\frac{A \in_0}{d}\right] \\ & \text { Now, } \frac{E_1}{E_2}=\frac{C_1}{C_2}=\frac{C}{C / 2}=\frac{2}{1}\end{aligned}$ $\Rightarrow \frac{E}{E_2}=\frac{2}{1} \Rightarrow E_2=\frac{E}{2}$

Hence, the correct answer is option: 'C'.

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