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):

$A$ parallel plate capacitor has plate area $A$ and separation between plates is $d$. It is charged to a potential difference of $V_0$ volt. The charging battery is then disconnected and plates are pulled apart to three times the initial distance. The work done to increase the distance between the plates is $(\varepsilon_0 = \text{permittivity of free space})$

The energy stored in a capacitor is $W$. To double the charge on the plates of the capacitor,the additional work to be done is

$A$ 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

$A$ parallel plate capacitor has a uniform electric field $\overrightarrow{E}$ 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: ($\varepsilon_{0} =$ permittivity of free space)

$A$ capacitor of capacitance $10 \mu F$ is charged to $10 \text{ V}$. The energy stored in it is (in $\mu J$)