Question

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 of 2
• B. Remains the same
• C. Increases by a factor of 2
• D. Increases by a factor of 4

Answer 1

Answer: (A)

Charge on capacitor
$q=CV$
when it is connected with another uncharged capacitor.

$V_c=\frac{q_1+q_2}{C_1+C_2}=\frac{q+0}{C+C}$
$V_c=\frac{V}{2}$
Initial energy
$U_i=\frac{1}{2}C{V}^2$
Final energy
$U_f=\frac{1}{2}C{\left(\frac{V}{2}\right)}^2+\frac{1}{2}C{\left(\frac{V}{2}\right)}^2$
$=\frac{C{V}^2}{4}$
Loss of energy $=U_i-U_f$
$=\frac{C{V}^2}{4}$
i.e. decreases by a factor (2)