$A$ capacitor of capacitance $C$ has stored energy $W$ and charge $Q$. If the charge is increased to $2Q$,what will be the new stored energy?

The energy stored in the capacitor is

$A$ parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:

Two capacitors $C_1$ and $C_2$ are connected in parallel to a battery. The charge-time graph is shown below for the two capacitors. The energy stored in them is $U_1$ and $U_2$,respectively. Which of the given statements is true?

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.

$A$ parallel plate capacitor has a plate separation $d$ and plate area $A$. If it is charged to a potential $V$ and then disconnected from the battery,calculate the work done in increasing the separation between the plates to $2d$.