Question

A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure (31-E9). The width of each stair is a and the height is b. Find the capacitance of the assembly.

Answer 1

The total area of the flat plate is A. The width of each stair is the same. Therefore, the area of the surface of each stair facing the flat plate is the same, that is, $\frac{A}{3}$.

From the figure, it can be observed that the capacitor assembly is made up of three capacitors. The three capacitors are connected in parallel.
For capacitor C₁, the area of the plates is $\frac{A}{3}$ and the separation between the plates is d.
For capacitor C₂, the area of the plates is $\frac{A}{3}$ and the separation between the plates is (d + b).
For capacitor C₃, the area of the plates is $\frac{A}{3}$ and the separation between the plates is (d + 2b).
Therefore,

$$\begin{gathered} C_1=\frac{{\in }_{0}A}{3d} \\ {C}_2=\frac{{\in }_{0}A}{3\left(d+b\right)} \\ {C}_3=\frac{{\in }_0\mathit{}A}{3\left(d+2b\right)} \end{gathered}$$

As the three capacitors are in parallel combination,

$$\begin{gathered} C=C_1+C_2+C_3 \\ \Rightarrow C=\frac{{\in }_{0}A}{3d}+\frac{{\in }_{0}A}{3\left(d+b\right)}+\frac{{\in }_{0}A}{3\left(d+2b\right)} \\ \Rightarrow C=\frac{{\in }_{0}A}{3}\frac{\left(3d^2+6bd+2b^2\right)}{d\left(d+b\right)\left(d+2b\right)} \end{gathered}$$