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2. Electric Potential and Capacitance
medium
A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be
A
$2$
B
$\frac{1}{4}$
C
$\frac{1}{2}$
D
$1$
(AIEEE-2007)
Solution
Required ratio
$=\frac{\text { Energy stored in capacitor }}{\text { Workdone by the battery }}=\frac{\frac{1}{2} C V^{2}}{C e^{2}}$
where $C=$ Capacitance of capacitor
$V=$ Potential difference
$e=\mathrm{emf}$ of battery
$=\frac{\frac{1}{2} C e^{2}}{C e^{2}}=\frac{1}{2} \quad(\because V=e)$
Standard 12
Physics
