Question

You have a parallel-plate 6.33 × 10-6 F capacitor that is charged to 0.00861 C. While the capacitor is isolated, you change the plate separation so that the capacitance becomes 2.41 × 10-6 F. How much work do you perform in this process?

Answer 1

Concept:

i. Energy stored on a capacitor

------------------(1)

Q= Charge stored on the Capacitor

C= Capacitance

U= Energy stored on the capacitor

V= Voltage difference between the capacitor plates.

ii. Charge conservation: If the capacitor is isolated i.e. no battery or charge source is connected to it then the charge stored in the capacitor remain conserved.

Idea: Idea behind to solve this problem is to find the energy stored in two of the situation i.e. before moving and after moving, The change in energy is equal to the work done.

Solution: Energy stored in first configuration (Using equation 1)

Q= 0.00861C

C= 6.33 x10^-6 F

U₁ = Q²​​​​​/2C = (0.00861)²/(6.33x10^-6) J

= 11.711 J

Energy stored in second configuration (Using equation 1)

Q= 0.00861C (Charge remains same)

C= 2.41 x10^-6 F

U₂ = Q²​​​​​/2C = (0.00861)²/(2.41x10^-6) J

= 30.760 J

NET WORK = Change in Energy = U₂ - U₁

= (30.760-11.711) J

= 19.049 J

19.049 J of work is performed to achieve configuration 2 from configuration 1.