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