Question

A parallel plate capacitor carries a charge q the distance between the plate is doubled by application of a force the work done by the force is

Answer 1

Answer:

The energy required to move the plates is q²/2C.

Explanation:

Let, us consider the present Capacitance of the capacitor to be C.

Also, let us consider the distance between the plates of the capacitor to be d.

Hence, C = εA/d, Where  ε = permittivity or dielectric constant and A = area of the capacitor plates.

Given, that the Charge stored on the capacitor is q.

Hence, q = CV, where V = voltage in the capacitor

Given, the distance between the plates of the capacitor be doubled.

So, new distance = 2d

Hence, new Capacitance, Cnew = εA/2d = C/2

Initial Energy stored in the capacitor = q²/2C

Final Energy stored in the capacitor = q²/2Cnew = q²/C

Energy required for the movement of plates

= Final Energy – Initial energy

= q²/C – q²/2C

= q²/2(1-1/2)

= q²/2c

Hence the energy required to move the plates is q²/2C.