Question

A capacitor is charged by a battery and the energy stored is $U$. The battery is now removed and the separation distance between the plates is doubled. The energy stored now is

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

Answer 1

Answer: (C)

Energy stored $U = \frac{1}{2}qV$
As the distance $d$ is increased between the two plates.
Now, stored energy,
$U' = \frac{1}{2}qV'$
$ = \frac{1}{2}q [\frac{q}{C}]$
$= \frac{1}{2} \frac{q^2 \, d}{\varepsilon_0 A}$
$U \propto d$
Hence, $U' = 2U$