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

Answer

A

Explanation

Charge on capacitor when it is connected with battery `q = CV`

when it is connected with another uncharged capacitor.

Common potential difference across capacitors
`V_(C) = (q_(1)+q_(2))/(C_(1)+C_(2)) = (q + 0)/(C+C) = (q)/(2C) rArr V_(C) = (V)/(2)`
Initial energy, `U_(1) = (1)/(2)CV^(2)`
Final energy, `U_(f) = (1)/(2) [2C] [(V)/(2)]^(2) = (1)/(2) U_(1)`
Hence total electrostatic energy of resulting system is decreased by a factor of `2`.