Question

A $10\mu F$ capacitor is charged by a battery of emf $200V$. The energy drawn from the battery, and the energy stored in the capacitor, are respectively

• A. 0.05 J and 0 m J
• B. 0.05 mJ and 0.05 m J
• C. 0.4 J and 0.2 J
• D. 1.0 m J and 0.5 m J

Answer 1

Answer: (C)

Energy drawn from battery
$=$ potential difference $(V)\times$ charge $(q)$
Also, charge $(q)=$ capacitance $(C)$
$\times$ potential difference $(V)$
$\therefore E=V\times CV=C{V}^2$
Given, $C=10\mu F=10\times 10^{-6}F$,
$V=200$ volt
$\therefore E=\left(10\times 10^{-6}\right)(200{)}^2=0.4J$
Work done in charging the conductor is stored as potential energy in the electric field, given by
$U=\frac{1}{2}C{V}^2$
Given, $C=10\mu F=10\times 10^{-6}F$,
$V=200$ volt
$U=\frac{1}{2}\times 10\times 10^{-6}\times (200{)}^2$
$=0.2J$