Two identical capacitors are charged individually to voltages $V_1$ and $V_2$. They are then connected in parallel with plates of like polarity joined together. After the system comes to equilibrium, what is the electrostatic energy stored in the system?

A capacitor stores $10 \mu \mathrm{C}$ charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of $20 \mu \mathrm{C}$ flows through the battery. Find the dielectric constant of the dielectric.

A capacitor of capacity ${C_1}$ is charged upto $V$ volt and then connected to an uncharged capacitor of capacity ${C_2}$. Then final potential difference across each will be

A capacitor of capacitance $9 \mathrm{~nF}$ having dielectric slab of ${\varepsilon _r}=2.4,$ dielectric strength $20 \mathrm{~MV} / \mathrm{m}$ and $P.D. = 20 \mathrm{~V}$ calculate area of plates.

If the circumference of a sphere is 2 m, then capacitance of sphere in water could be

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be

A capacitor of capacity $\mathrm{C}_1=1 \mu \mathrm{F}$ can with stand the maximum voltage $\mathrm{V}_1=16 \mathrm{KV}$ while another capacitor $\mathrm{C}_2=2 \mu \mathrm{F}$ can with stand the maximum voltage $\mathrm{V}_2=4 \mathrm{KV}$. The maximum voltage of capacitance connected in series will be-

A parallel plate capacitor connected to a battery has a charge Q. If the distance between the plates is reduced to one-third of its original value, while remaining connected to the battery, what is the new charge on the capacitor?

Two capacitors of capacitance C each are connected in series. The equivalent capacitance is:

The potential to which a conductor is raised, depends on

The energy stored in a capacitor of capacitance $C$ charged to a potential $V$ is $U$. If the voltage is doubled, the new stored energy will be: