The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it, one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the condenser is

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 $q_{0}$ )

The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$

The plates of a parallel plate capacitor are charged up to $100 \,volt$ . A $2 \,mm$ thick plate is  inserted between the plates, then to maintain the same potential difference, the distance  between the capacitor plates is increased by $1.6\, mm$. The dielectric constant of the plate  is :-

Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.

Reason : Capacity of the capacitor does not depend upon the nature of the material.

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants ${k_1},{k_2}$ and ${k_3}$ as shown. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by