$A$ charge of $40\,\mu C$ is given to a capacitor having capacitance $C = 10\,\mu F$. The stored energy in ergs is

$A$ 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 $(\varepsilon_0 = \text{permittivity of free space})$

$A$ parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

If the charge on the capacitor is increased by $3 \ C$,the energy stored in it increases by $21 \%$. The original charge on the capacitor is (in $C$)

$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the energy stored in the capacitor.

$A$ $10\, \mu F$ capacitor is charged to $500\, V$ and then its plates are joined together through a resistance of $10\, \Omega$. The heat produced in the resistance is ........ $J$.