Question

The parallel-plate capactior shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2d.


At this new plate separataion, what is the energy stored in the capacitor, neglecting fringing effects?

• A. 2E
• B. $\sqrt{2}E$
• C. E
• D. E/2

Answer 1

The correct option is A 2E

Ans : (a)

Let, $E=E_1,Energy{E}_1=\frac{Q_1^2}{2C_1}$

Electrically isolated $\Rightarrow Q_2=Q_1$

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

$E_2=\frac{Q{}_2^2}{2C_2}=\frac{Q_1^2}{\frac{2C_1}{2}}=2\left(\frac{Q_1^2}{2C_1}\right)$

$=2E_1=2E$