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

When the capacitor is charged by a battery of potential V, then energy stored in the capacitor,
`U_(i) = (1)/(2)CV^(2)` …(i)
When the battery is removed and another identical uncharged capacitor is connected in parallel

Common potential,
`V. = (CV)/(C +C) = (V)/(2)`
`:.` Then the enregy stored in the capacitor,
`U_(f) = (1)/(2)(2C) ((V)/(2))^(2) = (1)/(4)CV^(2)` ...(ii)
`:.` From eqns. (i) and (ii)
`U_(f) = (U_(i))/(2)`
that means the total electrostatic energy of resulting system will decreases by a factor of 2.