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Question:

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

Increase by a factor of 2

Increase by a factor of 4

Decreases by a factor of 2

Remains the same

Solution:

Q1+Q2=Q0
By KVL, Q1/C - Q2/C = 0
or, Q1 = Q2
Q1 + Q2 = Q0
Q1 = Q2 = Q0/2
Efinal = (1/2)c(Q2)^2 + (1/2)c(Q2)^2 = (1/2)c(Q0/2)^2 + (1/2)c(Q0/2)^2
Q0/4c
Ef/EQ = 1/2