JEE MAIN - Physics (2007 - No. 16)
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
$$1/2$$
$$1$$
$$2$$
$$1/4$$
Explication
Required ratio
$$ = {{Energy\,\,\,stored\,\,in\,\,capacitor} \over {Workdone\,\,by\,\,the\,\,battery}} = {{{1 \over 2}C{V^2}} \over {C{e^2}}}$$
where $$C=$$ Capacitance of capacitor
$$V=$$ Potential difference,
$$e=$$ $$emf$$ of battery
$$ = {{{1 \over 2}C{e^2}} \over {C{e^2}}} = {1 \over 2}$$
$$\left( \, \right.$$ as $$V=e$$ $$\left. \, \right)$$
$$ = {{Energy\,\,\,stored\,\,in\,\,capacitor} \over {Workdone\,\,by\,\,the\,\,battery}} = {{{1 \over 2}C{V^2}} \over {C{e^2}}}$$
where $$C=$$ Capacitance of capacitor
$$V=$$ Potential difference,
$$e=$$ $$emf$$ of battery
$$ = {{{1 \over 2}C{e^2}} \over {C{e^2}}} = {1 \over 2}$$
$$\left( \, \right.$$ as $$V=e$$ $$\left. \, \right)$$