A spherical conductor has a capacitance of 8 µF and it is charged to a potential of 2000V. The conductor is connected with a long thin wire to an another uncharged spherical conductor of capacity of 32 µF.

Determine how much heat is released.

###### Solution. Posted on Wednesday, April 29, 2015.

We should start with an electrostatic potential energy of a first spherical conductor.

U1 = C1*V^2/2

The total capacitance of Parallel Connected Capacitors is

C = C1 + C2.

According the Law Of Conservation of Charge, the total charge will not change and the electrostatic potential energy of a system of two conductors

U2 = q^2/(2*C), where q = C1*V.

Thereby, the released energy (heat) will be equal to

U2 − U1 = C1*V^2/2 − q^2/(2*C).

Finally,

ΔU = [(C1*C2)/(C1 + C2)] * (V^2/2) = [((8*10^-6)*(32*10^-6))/(8*10^-6 + 32*10^-6)] * (2000^2/2) = 12.8 J

The Answer is ΔU = 12.8 J