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)

Decreases by a factor of 2

Answer 2

Charge on capacitor
$q=C V$
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)