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 resulting system
- decreases by a factor of 2
- remains the same
- increases by a factor of 2
- increases by a factor of 4
Answer
(1) decreases by a factor of 2
Solution
Initial energy: Ui = Q22C
The battery is removed, so charge Q is conserved and is shared; the combination has capacitance 2C.
Uf = Q22(2C) = Q24C
UfUi = 12, i.e., the energy decreases by a factor of 2 (the rest is lost as heat during charge flow).
Answer: (1)