Question

What is the capacitance of the capacitor, shown in figure?

Answer 1

There are three capacitors with capacitances $C_1,C_2,C_3$ and they are connected as shown in figure.

Here, $C_1=\frac{\left(A/2\right)K_1{ϵ}_0}{2d}=\frac{A{K}_1{ϵ}_0}{4d}$,

$C_2=\frac{\left(A/2\right)K_2{ϵ}_0}{d}=\frac{A{K}_2{ϵ}_0}{2d}=\frac{2C_1{K}_2}{K_1}$,

$C_3=\frac{\left(A/2\right)K_3{ϵ}_0}{d}=\frac{A{K}_3{ϵ}_0}{2d}=\frac{2C_1{K}_3}{K_1}$,

Now $C_2$ and $C_3$ are in series so equivalent capacitance for this,

$C_{23}=\frac{C_2{C}_3}{C_2+C_3}=\frac{\left(2C_1{K}_2/K_1\right)\left(2C_1{K}_3/K_1\right)}{\left(2C_1{K}_2/K_1\right)+\left(2C_1{K}_3/K_1\right)}=\left(2C_1/K_1\right)(K_2{K}_3/K_2+K_3)$

As $C_{23},C_1$ are in parallel so net capacitance is

$C_{net}=C_{23}+C_1=\left(2C_1/K_1\right)(K_2{K}_3/K_2+K_3)+C_1=C_1[\frac{2K_2{K}_3}{K_1(K_2+K_3)}+1]$