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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
Explanation
Common potential V/2 halves the stored energy.
Detailed Solution
Charge on capacitor $q = CV$
When it is connected with another uncharged capacitor: $V_c = \frac{q_1 + q_2}{C_1 + C_2} = \frac{q_0}{C + C} = \frac{V}{2}$
Initial energy $U_i = \frac{1}{2}CV^2$
Final energy $U_f = \frac{1}{2}C\left(\frac{V}{2}\right)^2 + \frac{1}{2}C\left(\frac{V}{2}\right)^2 = \frac{CV^2}{4}$
Loss of energy $= U_i - U_f = \frac{CV^2}{4}$
i.e. the energy decreases by a factor of 2.