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

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) (U/2) (B) U (C) 2U (D) 4U

Tardigrade Admin · Accepted Answer

Energy stored U = (1/2)qV As the distance d is increased between the two plates. Now, stored energy, U' = (1/2)qV' = (1/2)q [(q/C)] = (1/2) (q2 d/ε0 A) U propto d Hence, U' = 2U

Q. 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

$\frac{U}{2}$

$U$

$2 U$

$4 U$

Energy stored $U = \frac{1}{2} q V$
As the distance $d$ is increased between the two plates.
Now, stored energy,
$U^{'} = \frac{1}{2} q V^{'}$
$= \frac{1}{2} q [\frac{q}{C}]$
$= \frac{1}{2} \frac{q ^{2} d}{ε _{0} A}$
$U \propto d$
Hence, $U^{'} = 2 U$

Q. 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

2U​

U

2U

4U

Energy stored U=21​qV As the distance d is increased between the two plates. Now, stored energy, U′=21​qV′ =21​q[Cq​] =21​ε0​Aq2d​ U∝d Hence, U′=2U

Q. 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

$\frac{U}{2}$

$U$

$2 U$

$4 U$

Energy stored $U = \frac{1}{2} q V$
As the distance $d$ is increased between the two plates.
Now, stored energy,
$U^{'} = \frac{1}{2} q V^{'}$
$= \frac{1}{2} q [\frac{q}{C}]$
$= \frac{1}{2} \frac{q ^{2} d}{ε _{0} A}$
$U \propto d$
Hence, $U^{'} = 2 U$