The space between the plates of a parallel plate capacitor is completely filled in two ways. In the first case, it is filled with a slab of dielectric constant K . In the second case, it is filled with two slabs of equal thickness and dielectric constants K 1 and K 2 respectively as shown in the figure. The capacitance of the capacitor is same in the two cases. Obtain the relationship between K , K 1 and K 2 .

In case I : Dielectric constant is K , area of the cross-section is A and width of dielectric slab is d . Then, the capacitance will be: C I = K ε ∘ A d In case II : For first d 2 space is covered by dielectric slab K 1 with cross-sectional area A and the next d 2 space is covered by dielectric slab K 2 with cross-sectional area A . Then the total capacitance here is: 1 C II = 1 C K 1 + 1 C K 2 1 C II = d 2 K 1 ε ∘ A + d 2 K 2 ε ∘ A = d 2 ε ∘ A K 1 + K 2 K 1 K 2 ⇒ C II = 2 ε ∘ A d K 1 K 2 K 1 + K 2 Since, both the capacitance are equal. Then, C I = C II K ε ∘ A d = 2 ε ∘ A d K 1 K 2 K 1 + K 2 ⇒ K = 2 K 1 K 2 K 1 + K 2 Hence, the relation between K , K 1 and K 2 is K = 2 K 1 K 2 K 1 + K 2 .