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

Answer 1

The correct option is C Decreases by a factor of 2

$Q_1+Q_2=Q_0$
By KVL,
$\frac{Q_1}{C}-\frac{Q_2}{C}=0$
or, $Q_1=Q_2$
$Q_1+Q_2=Q_0$
$Q_1=Q_2=Q_{0/2}$
$E_{final}=\frac{1}{2c}(\frac{Q}{2}{)}^2+\frac{1}{2c}(\frac{Q}{2}{)}^2$
$=\frac{1}{2c}\frac{Q^2}{4}+\frac{1}{2c}\frac{Q^2}{4}$
$\frac{Q}{4c}$
$\frac{E_f}{E_Q}=\frac{1}{2}$