The equivalent capacitance of the arrangement shown in figure, if `A` is the area of each plate is

Options

A. `C=(epsilon_0A)/d[K_1/2+(K_2+K_3)/(K_2K_3)]`
B. `C=(epsilon_0A)/d[K_1/2+(K_2K_3)/(K_2+K_3)]`
C. `C=(epsilon_0A)/(2d)[K_1+(K_2+K_3)/(K_2+K_3)]`
D. `C=(epsilon_0A)/d[K_1+(K_2K_3)/(K_2+K_3)]`

Answer

B

Explanation

`C=C_(LHS)+C_(RHS)`
`=(K_1epsilon_0(A/2))/d+(epsilon_0(A/2))/(d-d/2-d/2+(d/2)/K_2+(d/2)/K_3`
`=(epsilon_0A)/d[K_1/2+(K_2K_3)/(K_2+K_3)]`