Question

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

• A. $C$
• B. $(K_1+K_2+K_3)\frac{C}{3}$
• C. $(K_1+K_2+K_3)C$
  
• D. $(K_1{K}_2+K_2{K}_3+K_3{K}_1)C$

Answer 1

The correct option is B $(K_1+K_2+K_3)\frac{C}{3}$

consider each one third of the assembly as a seperate capacitor. Thus, three capacitors are joined in parallel.
Plate area is one third of original area.
$A^1=\frac{A}{3}$
$\Rightarrow C=\frac{A{ϵ}_0}{d}$
$\Rightarrow C_1=\frac{K_1\frac{A}{3}{ϵ}_0}{d}$= $\frac{K_{1}C}{3}$
$\Rightarrow C_2=\frac{K_2\frac{A}{3}{ϵ}_0}{d}$= $\frac{K_{2}C}{3}$
$\Rightarrow C_3=\frac{K_3\frac{A}{3}{ϵ}_0}{d}$= $\frac{K_{3}C}{3}$
Their equivalent $C_{eq}=C_1+C_2+C_3$
$\Rightarrow C_{eq}=(K_1+K_2+K_3)\frac{C}{3}$