Question

A certain variable parallel plate capacitor has a capacitance of 33 μF. How much energy is stored in this capacitor if put at a potential difference of 7.5 V? The area of the plates is then doubled, and the plate separation is cut to 1/3 of its original value. It is then put at 7.5 V again. Now how much energy is stored in the capacitor?

Answer 1

Part A.

Energy Stored in capacitor is given by:

U = (1/2)*C*V^2

Given that C = 33 uF = 33*10^-6 F

V = 7.5 V

So,

U = (1/2)*33*10^-6*7.5^2

U = 9.3*10^-4 J = Energy Stored in capacitor initially

Part B.

Capacitance of a capacitor is given by:

C = e0*A/d

When Area of plates is doubled and plate separation is cut ro 1.3 of original then

C1 = e0*(2*A)/(d/3) = 6*e0*A/d

C1 = 6*C = 6*33*10^-6

So Now energy stored will be:

U1 = (1/2)*C1*V^2

U1 = (1/2)*(6*33*10^-6)*7.5^2

U1 = 5.57*10^-3 J

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