Magnitude of work done during the charging of a condenser from $q=0$ to $q=Q$ is

The energy of a charged capacitor is given by the expression $(q=$ charge on the conductor and $C=$ its capacity)

A parallel plate air condenser of capacity $10 \mu \mathrm{F}$ is charged to a potential of $1000 \mathrm{V}$. The energy of the condenser is

When a capacitor having a capacitance $8\times {10}^{-6} \mathrm{F}$ and potential difference of $100 \mathrm{V}$ is discharged, the energy released in joules is

A capacitor carries a charge of $6 \mu \mathrm{C}$ at a potential $500 \mathrm{V}$. The electrostatic energy stored in it is

A $100 \mathrm{\mu F}$ capacitor is to have an energy content of $50 \mathrm{J}$ in order to operate a flashlamp. The voltage required to charge the capacitor is

A parallel plate capacitor is charged to a potential difference of $50 \mathrm{V}$ . It is discharged through a resistance. After $1 \mathrm{s}$, the potential difference between plates becomes $40 \mathrm{V} .$ Then

If the potential difference across a capacitor is increased from $10 \mathrm{V}$ to $30 \mathrm{V}$, then the energy stored within the capacitor:

A conductor of capacity $10 \mu \mathrm{F}$ is at a potential of $10 \mathrm{V}$. If the potential increases by $1 \mathrm{V}$, the increase in energy is