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:

Three plates $A, B, C$ each of area $50 \, cm^2$ have a separation of $3 \, mm$ between $A$ and $B$ and $3 \, mm$ between $B$ and $C$. The energy stored when the plates are fully charged is (as shown in the figure):

In a charged capacitor,the energy resides in:

If the distance between the plates of a capacitor with capacitance $C$ and charge $Q$ is doubled,the work done is:

$A$ parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates,the time rate of change of electrostatic energy of the capacitor is proportional to

The plates of a parallel plate capacitor have an area of $90 \,cm^{2}$ each and are separated by $2.5 \,mm$. The capacitor is charged by connecting it to a $400 \,V$ supply.
$(a)$ How much electrostatic energy is stored by the capacitor?
$(b)$ View this energy as stored in the electrostatic field between the plates,and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.