A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is

In one design of capacitor thin sheets ot metal of area $80\  mm \times 80\  mm$ sandwich between them a piece of paper whose thickness is $40\ μm$. The relative permittivity of the paper is $4.0$ and its dielectric strength is $20\ MVm^{-1}$. Calculate the maximum charge that can be put on the capacitor

[permittivity of free space $ = 9 \times 10^{-12}\  Fm^{-1}$]

A parallel - plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be

(If initial charge is $q_{0}$ )

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air cored parallel capacitor charged to a potential $V$. The two capacitors share charges and the common potential is $V$. The dielectric constant $K$ is

An insulator plate is passed between the plates of a capacitor. Then the displacement current