An uncharged capacitor $C$ is connected to a battery through a resistance $R$. Show that by the time the capacitor gets fully charged, the energy dissipated in $R$ is the same as the energy stored in $C$.

Hint:A capacitor is an electrical device which stores energy in an electrical field. It has two terminals. In the given question, we will find out the work required to charge the capacitor against a potential difference. Then we have to find the energy that will dissipate when a resistance $R$ is present.

Complete step by step answer:
Let us first find the energy stored in the capacitor, consider an uncharged capacitor of capacitance $C$. When this capacitor is connected to a cell of potential difference $V$, a total charge $Q$ appears on its positive plate such that $Q = CV$.

Let, at any instant during charging, a charge $q$ is present on the capacitor, such that $q = CV$. Now, the work done in transferring a charge $dq$ to the capacitor is given by,
$dW = Vdq$
$ \Rightarrow dW = \dfrac{q}{C}dq$
So, the total work done in supplying charge $Q$ is,
$W = \int {dW = \int\limits_0^Q {\dfrac{{qdq}}{C}} } \\
\Rightarrow W= \dfrac{1}{C}\left[ {\dfrac{{{q^2}}}{2}} \right]_0^Q \\
\Rightarrow W = \dfrac{{{Q^2}}}{{2C}}$
Thus, the energy stored in the capacitor is,
$U = \dfrac{{{Q^2}}}{{2C}} \\
\Rightarrow U= \dfrac{1}{2}C{V^2}{\text{ }}\left[ {Q = CV} \right]$

Let the work done be $W$ when the capacitor is charged using the battery. The battery transfers a charge of $Q$ across terminals having potential difference $V$.Work done by battery is,
$W = QV = C{V^2}$
Thus, the energy dissipated due to resistance $R$ is,
$W - U = C{V^2} - \dfrac{1}{2}C{V^2} \\
\therefore W= \dfrac{1}{2}CV$
Therefore, the energy dissipated due to resistance $R$ is $\dfrac{1}{2}C{V^2} = U = $ energy stored in the capacitor. It is shown that by the time the capacitor gets fully charged, the energy dissipated in $R$ is the same as the energy stored in $C$.

Note: It must be noted that the charging of a capacitor is done by using a battery. It transfers charge from lower potential to higher potential. When a capacitor is fully charged it acts as an open circuit with infinite resistance.