Question

If the area of the plates of a parallel plate capacitor is halved, and the separation between the plates tripled, all while the charge on the capacitor remains constant, then by what factor does the electric field potential energy stored in the capacitor change?

Answer 1

Answer:

Explanation:

The capacitance of a parallel plate capacitor is given by

$C=\frac{\epsilon _{0}A}{d}$

where A is the area of plates and d is the separation between the plates.

Now the area is halved and the separation is tripled.

The new capacitance is

C' = C / 6

Initial potential energy is given by

$U=\frac{q^{2}}{2C}$

here the charge is constant

The new energy is given by

$U'=\frac{6q^{2}}{2C}$

U' = 6 U

The energy becomes six times the initial value.