A capacitor of capacitance C is charged to a potential difference V and then connected across an uncharged capacitor of capacitance 2C. When the system reaches equilibrium, what is the total electrostatic energy stored in the system?

A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is: $\left(ext{Given }\varepsilon_0=8.834 imes 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

If the area of the plates of a parallel plate capacitor is doubled, keeping the distance between the plates constant, the capacitance:

64 identical spheres of charge $q$ and capacitance $C$ each are combined to form a large sphere. The charge and capacitance of the large sphere is

Assertion (A) : When an uncharged capacitor is charged by a battery, only $50 \%$ of the energy supplied by a battery is stored in the capacitor.

Reason (R): Rest $50 \%$ is lost.

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:

A $500\mu F$ capacitor is charged at a steady rate of $100\mu C/{\rm{second}}.$ The potential difference across the capacitor will be $10\;V$ after an interval of

A parallel plate capacitor is connected to a 12V battery. If the distance between the plates is halved while connected to the battery, what happens to the capacitance and the charge stored?

A parallel plate capacitor has a capacity $2\mu F$. A dielectric slab is inserted between the plates and capacitor is charged with 100 Volt and then isolated. The potential when dielectric is removed is

A parallel plate capacitor has capacitance C. If the distance between the plates is doubled, what will be the new capacitance?

A charged spherical conductor has a surface density of $0.7C/{m^2}$ when its charge is increased by $0.44C$ the charge density changes by $0.14\,\,C/{m^2}$. The radius of the sphere is