Question

A $900\, pF$ capacitor is charged by $100\, V$ battery.

• A. The charge on the capacitor is $9 \times 10^{-8} C$
• B. The energy stored by the capacitor is $4.5 \times 10^{-6} J$
• C. If capacitor is disconnected from the battery and connected to another $900\, pF$ capacitor, then energy stored by the system is $2.25 \times 10^{-6} J$
• D. None of the above

Answer 1

Answer: (A), (B), (C)

(i) The charge on the capacitor is
$q=C V=900 \times 10^{-12} F \times 100 V =9 \times 10^{-8} C$
(ii) The energy stored by the capacitor is
$=\frac{1}{2} C V^{2}=\frac{1}{2} Q V$
$=(1 / 2) \times 9 \times 10^{-8} C \times 100\, V$
$=4.5 \times 10^{-6} J$
(iii) In the steady situation, the two capacitors have their positive plates at the same potential and their negative plates at the same potential. Let the common potential difference be $V^{\prime}$.
The charge on each capacitor is then $q^{\prime}=C V^{\prime}$. By charge conservation $q^{\prime}=Q / 2$. This implies
$V^{\prime}=V / 2$. The total energy of the system is
$=2 \times \frac{1}{2} Q^{\prime} V^{\prime}=\frac{1}{2} Q V$
$=2.25 \times 10^{-6} J$