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/3, 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. (2)

Answer 1

Initially when there is vacuum between the two plates, the capacitance of the two parallel plate 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 insert the dielectric slab of thickness, the electric field reduces to E.Now the gap between plates is divided in two parts, for distance t there is electric field E and for the remaining distance (d– t) the electric field is E0.If V be the potential difference between the plates of the capacitor, then V = Et + E0(d–t)