Question

A parallel plate air capacitor has a capacitance of 25 uF. The charge on each plate is Q 50 uF. There are no batteries connected to the capacitor. How much work is required to double the separation distance between the plates?

Answer 1

Initial capacitance, C1 = 25 uF = 25 x 10^-6 F

Charge on each plate, Q = 50 uC = 50 x 10^-6 C

So, energy stored in the capacitor, E1 = (1/2)*C*V^2 = (1/2)*(Q^2/C) [Since, Q = C*V]

Therefore, initial energy of the capacitor, Ei = (1/2)*(Q^2/C1)

= 0.5*[(50 x 10^-6)^2 / (25 x 10^-6)]

= 50 x 10^-6 J

When the separation between the plates is doubled, the new capacitance of the capacitor -

C2 = C1/2 = 12.5 uF = 12.5 x 10^-6 F

So, the final energy of the capacitor, Ef = (1/2)*(Q^2/C2)

= 0.5*[(50 x 10^-6)^2 / (12.5 x 10^-6)]

= 100 x 10^-6 J

Therefore the work required = Ef - Ei = 100 x 10^-6 J - 50 x 10^-6 J = 50 x 10^-6 J (Answer)