Q: 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 :

(a) Decreases by a factor 2

(b) Remains same

(c) Increases by a factor 2

(d) Increases by a factor 4

**Click to See Answer : **

Ans: (a)

Sol: $\large U_i = \frac{1}{2}CV^2$

Common Potential $V’ = \frac{C_1V_1 + C_2 V_2}{C_1+C_2}$

$V’ = \frac{CV + 0}{C+C}$

$V’ = \frac{V}{2}$

$\large U_f = \frac{1}{2}(C+C)(\frac{V}{2})^2$

$\large U_f = \frac{1}{4}CV^2 $

$\large \frac{U_f}{U_i} = \frac{1}{2}$