The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2\,\mu \,F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is…….$\mu \,F$

Define dielectric constant.

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 ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?

If a slab of insulating material $4 \times {10^{ – 3}}\,m$ thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $3.5 \times {10^{ – 3}}\,m$ to restore the capacity to original value. The dielectric constant of the material will be