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) $\dfrac{1}{2}$
(B) $1$
(C) $2$
(D) $\dfrac{1}{4}$

Hint:A battery consists of two plates of opposite charges. Here we consider the capacitance for calculating the energy as well as the charge.

Complete answer:
Let us take the electromotive force of the battery as $E$. Now the charge can be calculated as the product of the capacitance and the electromotive force. That is, we can write it mathematically as, $q = CE$. Now as we have the charge, we can define the work done as the product of charge and the electromotive force. That is, it can be expressed mathematically as, $W = qE$. As we know, the value of charge, $q = CE$, we can substitute this value for charge in the expression for work.
$
W = qE\\
\Rightarrow W = CE \times E\\
\Rightarrow W = C{E^2}
$
Now we can calculate the energy stored in the capacitor. As the mathematical expression for the energy stored in the capacitor is, $U = \dfrac{1}{2}C{E^2}$. So taking the ratio of the energy stored in the battery to the work done we have,
$\therefore\dfrac{U}{W} = \dfrac{{\dfrac{1}{2}C{E^2}}}{{C{E^2}}} = \dfrac{1}{2}$

Therefore we can conclude that option (A ) is true.

Note: battery can be made up of one or more cells.Since there is an electric field inside the capacitor, there is also energy stored in the capacitor.So obviously, a capacitor can be used to store energy or as battery. When connected with a DC battery, plates get charged in a capacitor. When this happens, there is also an electric potential difference across the plates. This means that there will be less current coming out of the battery and less charge going to the plates. Technically, the capacitor will take forever to become fully charged.