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:

Options

A. Increases by a factor of 4
B. Decreases by a factor of 2
C. Remains the same
D. Increases by a factor of 2

Answer

B

Explanation

Initial condition –

`U_i=1/2 CV^2`
`V_f=(CV)/(C+C)=V/2`
Final energy `=2xx1/2 Cxx(V/2)^2 =(CV^2)/4`
Clearly, energy decreases by a factor of 2.