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

A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by

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.........$\mu F$

Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.

Reason : The surface density of charge onthe plate remains constant or unchanged.

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)

Polar molecules are the molecules: