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 factor of 2
C. Remains the same
D. Increases by a factor of 2

Answer

B

Explanation

Charge on capacitor
when it is connected with another uncharged capacitor .

`V_(c)=(q_(1)+q_(2))/(C_(1)+C_(2))=(q+0)/(C+C)`
`V_(c)=(V)/(2)`
Initial energy
`U_(i)=(1)/(2)CV^(2)`
Final energy
`U_(f)=(1)/(2)C((V)/(2))^(2)+(1)/(2)C((V)/(2))^(2)` ltbrge `(CV^(2))/(4)`
Loss of energy `=U_(i)-U_(f)`
`=(CV^(2))/(4)`
i.e. decreases by a factor (2)