The area A of a parallel plate capacitor is divided into two equal, halves and filled with two media of dielectric constants `K_1 and K_2` respectively. The capacitance of the capacitor will be

Options

A. `(epsi_0 A (K_1 + K_2))/d`
B. `(epsi_0A)/d ((K_1 + K_2)/2)`
C. `(epsi_0A)/d. (K_1K_2)/((K_1 + K_2))`
D. `(epsi_0A)/(2d). (K_1 K_2)/((K_1 +K_2))`

Answer

B

Explanation

The given arrangement is a combination of two capacitors in parallel whose capacitances are
`C_1 = (K_1 epsi_0 (A/2))/d andC_2 = (K_2 epsi_0(A/2))/drArrC = C_1 + C_2 = (epsi_0A)/d ((K_1 +K_2)/2)`