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

A capacitor of $4 \mu \mathrm{F}$ charged to $50 \mathrm{V}$ is connected to another capacitor of $2 \mu \mathrm{F}$ , charged to $100 \mathrm{V}$. The total energy of the combination is

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

A capacitor of capacitance $C$ is charged to potential difference $V$ and then disconnected from the battery. The air dielectric of capacitor is replaced by another dielectric of dielectric constant $k$. The fractional decrease in energy of the capacitor is