Question

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

Answer 1

Energy stored between the plates of capacitor is given by

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

now we know that

voltage across the plates of capacitor is doubled

$V' = 2V$

now the energy of the capacitor will be given as

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

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

so here the energy stored between the plates of capacitor will becomes 4 times more