Two dielectric slabs of dielectric constants `K_1 and K_2` are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.

Explanation

The arrangement is equivalent to two capacitors joined in parallel where area of plates of either capacitor is `A/2`. Thus,
`C_1 =(K_1 epsi_0 (A/2))/d = (K_1 epsi_0A)/(2d)`
and `C_2 =(K_2epsi_0(A/2))/(d) =(K_2 epsi_0A)/(2d)`
`:.` Net capacitance of the capacitor
`C = C_1 + C_2 = (K_1 epsi_0 A)/(2d) + (K_2 epsi_0A)/(2d) = (epsi_0A)/d ((K_1 + K_2)/2)`.