Question

1.30 [2pt] Imagine 2 identical capacitors with C = 1F having a charge Q =1 C. Suppose you connect their terminals, positive to positive, negative to negative. Now you have an effective capacitance 2C carrying a charge 2Q, right? Work out the total energy of the two capacitors before and after the connection and speculate on the meaning of your result.

1.30 ANSWER

Answer 1

Solution) C = 1 F

Q = 1 C

connected from positive to positive and negative to negative means they are in parallel

Equivalent capacitance , Ceq = C + C = 1 + 1 = 2 F

Total charge QT = Q + Q = 1+ 1 = 2 C

Before connecting individual energies are

U1 = (Q^2)/(2C) = (1^2)/(2×1) = (1/2) J

U2 = (Q^2)/(2C) = (1^2)/(2×1) = (1/2) J

So total energy before connecting U = U1 + U2

U = (1/2) + (1/2) = 1 J

U = 1 J

After connecting , U = (QT^2)/(2Ceq)

U = ((2Q)^2)/(2×2C)

U = (4×1)/(4×1)

U = 1 J

So from above calculations total energy is same in both cases .