Question

The voltage applied across a given parallel-plate capacitor is doubled. How is the energy stored in the capacitor affected?

Answer 1

Answer:

The energy stored in the capacitor quadruples its original value.

Explanation:

The energy stored in a capacitor is given by the equation

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the voltage across the plates

The capacitance, C, depends only on the properties of the capacitor, so it does not change when the voltage applied is changed.

Instead, in this problem the voltage applied is doubled:

V' = 2V

So the new energy stored is

$U'=\frac{1}{2}C(2V)^2=4(\frac{1}{2}CV^2)=4U$

so, the energy stored has quadrupled.