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

If `C` is capacity and `V` is potential of battery, then initial energy stored
`E_(1) = (1)/(2) CV^(2) = (1)/(2) (Q^(2))/(C )`
When uncharged capacitor of capaacity `C` is connected in parallel.
`C_(p) = C + C = 2C`
Charge on each capacitor `= Q//2`
Total energy `E_(2) = ((Q//2)^(2))/(2C) + ((Q//2)^(2))/(2C)`
`(2Q^(2))/(4(2C)) = (1)/(2) ((Q^(2))/(2C))`
`E_(2) = E_(1)//2`
i.e., total electrostatic energy of resulting system decreases by a factor of `2`.