Question

A $900pF$ capacitor is charged by $100V$ 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 $900pF$ 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
$q=CV=900\times 10^{-12}F\times 100V=9\times 10^{-8}C$
(ii) The energy stored by the capacitor is
$=\frac{1}{2}C{V}^2=\frac{1}{2}QV$
$=(1/2)\times 9\times 10^{-8}C\times 100V$
$=4.5\times 10^{-6}J$
(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
$=2\times \frac{1}{2}{Q}^{\prime }{V}^{\prime }=\frac{1}{2}QV$
$=2.25\times 10^{-6}J$