When a parallel plate capacitor is fully charged by a battery,the potential difference between the two plates becomes equal to the electromotive force $(EMF)$ of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

$A$ $6\,\mu F$ capacitor is charged from $10\,V$ to $20\,V$. The increase in energy will be:

$A$ capacitor of capacitance $C$ is fully charged. It is discharged through a small coil of negligible resistance embedded in a thermally insulated block of mass $m$ and specific heat capacity $S$. If the temperature of the block rises by $\Delta T$,find the potential difference across the capacitor.

$A$ $12 \text{ pF}$ capacitor is connected to a $50 \text{ V}$ battery. The electrostatic energy stored in the capacitor will be . . . . . . $\text{J}$.

If $V$ denotes the potential difference across the plates of a capacitor of capacitance $C$,the dimensions of $CV^2$ are

$A$ $12 \;pF$ capacitor is connected to a $50 \;V$ battery. How much electrostatic energy is stored in the capacitor?