The space between the plates of a parallel plate capacitor of capacitance C is filled with three dielectric slabs of identical siza as shown in figure. If the dielectric constants of the three slabs are `K_1 , K_2 and K_3` find the new capacitance.

Options

A. `C_eq = (K_1 + K_2 + K_3) (C)/(3)`
B. `C_eq = (K_1 - K_2 + K_3) (C)/(3)`
C. `C_eq = (K_1 + K_2 - K_3) (C)/(3)`
D. `C_eq = (K_1 - K_2 - K_3) (C)/(3)`

Answer

A

Explanation

Consider each one third of the assembly as a separte capacitor. The three positive plates are connected together at point A and the three negative plates are connected together at point B. Thus, the three capacitors are joined in parallel. As the plate area is one third of the original for each part, the capacitances of these parts will be `
`K_1 C/3, K_2/3, K_2 C/3. The equivalent capacitance is, therfore, `
`C_eq = (K-1 + K_2 + K_3) (C)/(3).