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.
$$\frac{1}{2}$$
B.
$$1$$
C.
$$2$$
D.
$$\frac{1}{4}$$
Answer :
$$\frac{1}{2}$$
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,
$$e =$$ emf of battery
$$\frac{{\frac{1}{2}C{e^2}}}{{C{e^2}}} = \frac{1}{2}\,\,\,\,\,\,\,\,\left( {\because V = e} \right)$$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Current
Releted Question 1
The temperature coefficient of resistance of a wire is 0.00125 per $$^ \circ C$$ At $$300\,K,$$ its resistance is $$1\,ohm.$$ This resistance of the wire will be $$2\,ohm$$ at.
The electrostatic field due to a point charge depends on the distance $$r$$ as $$\frac{1}{{{r^2}}}.$$ Indicate which of the following quantities shows same dependence on $$r.$$
A.
Intensity of light from a point source.
B.
Electrostatic potential due to a point charge.
C.
Electrostatic potential at a distance r from the centre of a charged metallic sphere. Given $$r$$ < radius of the sphere.