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).
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