Question

A parallel plate capacitor s charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles, then ​

Answer 1

Explanation:

Q = C * V

Q = charge on the capacitor plates (each)

V = voltage across the plates

C = capacitance of the capacitor

C = ε A / d , Where A = area of each of the plates

d = distance between the plates.

ε = permittivity of the medium in between the plates.

If the distance d increases, then the capacitance C decreases. As a consequence, the charge on the capacitance is not affected. So Voltage difference between the plates increases.

Electrostatic energy = 1/2 C V² = 1/2 Q V = Q² / 2 C

So as the capacitance decreases, the electrostatic energy increases. The work done in moving the capacitor plates against electrostatic attraction, is stored in the capacitor as electrostatic potential energy.