Question

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d / 2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Answer 1

Initially when there is vacuum between the two plates, the capacitance of the two parallel plates is,

where A is the area of parallel plates. Suppose that the capacitor is connected to a battery, an electric field E₀ is Produced.

Now if we lnsert the dielectric slab of thickness t = d/2, the electric field reduces to E.

Now the gap between the plates is divided in two parts, for distance f there is electric field E and for the remaining distance (d - t), the electric field is E_0.

If V be the potential difference between the plates of the capacitor,  then