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 plate is, `C_0 = (epsi_0A)/d`

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 t=d/2 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)

`V=(Ed)/2 + (E_0d)/2 =d/2 (E +E_0)      (therefore t =d/2)`

`=> V=d/2 ((E_0)/K +E_0) (dE_0)/(2K)  (K+1)     (As,(E_0)/E = K)`

Now, `E_0 =σ/epsi_0  = q/(epsiA )=>V = d/(2K) q/(epsi_0A)  (K +1)`

we know, `C = q/V = (2Kepsi_0A)/(d(K+1))`