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.

Initially when there is vacuum between the two plates, the capacitance of the capacitor is
C0=ε0A/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 reduce 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+E0d/2=d/2(E+E0) (t=d/2)
V=d/2(E0/K+E0)=dE0/2K(K+1) (as E0/E=K)
now E0=σ/ε0=q/ε0A ⇒V=d/2Kq/ε0A(K+1)
we know C=q/V=2Kε0A/(K+1)d