In the circuit shown in the figure,four capacitors are connected to a battery. The maximum energy is stored in the capacitor of.....$\mu F$

$A$ parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A$,the energy stored in the capacitor is (where $\epsilon_{0}$ is the permittivity of free space).

$A$ parallel plate capacitor has a uniform electric field $E \ (V/m)$ in the space between the plates. If the distance between the plates is $d \ (m)$ and the area of each plate is $A \ (m^2)$,then the energy (joules) stored in the capacitor is:

The amount of work done in increasing the voltage across the plates of a capacitor from $5 \ V$ to $10 \ V$ is $W$. The work done in increasing it from $10 \ V$ to $15 \ V$ will be (nearly):

Given $C_1 = 2\ \mu F$,$C_2 = 6\ \mu F$,and $C_3 = 4\ \mu F$. What is the total energy of the system?

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.