The energy stored in a capacitor of capacitance $C$ charged to a potential $V$ is:

Four capacitors each of capacity $2 \mu F$ are connected as shown in the figure. If $V_{A}-V_{B}=10 \,V$,the energy stored in the system is:

$A$ capacitor with capacitance $5\,\mu F$ is charged to $5\,\mu C.$ If the plates are pulled apart to reduce the capacitance to $2\,\mu F,$ how much work is done?

If the energy density in a parallel plate capacitor is given as $1.8 \times 10^{-9} \ J/m^3$,then the value of the electric field in the region between the plates is ....... $NC^{-1}$. (Given: $\epsilon_0 = 9 \times 10^{-12} \ C^2/N \cdot m^2$)

If $C$ is the capacitance and $V$ is the electric potential,then the dimensional formula of $C V^{2}$ is

$A$ capacitor is used to store $24 \, \text{watt-hour}$ of energy at $1200 \, \text{V}$. What should be the capacitance of the capacitor?