Two capacitor are joined in series shown in figure. The central rigid H-shaped part is movable. Find the equivalent capacitance of the combination and hence show that it is independent of position of the central H-shaped part. The area of each plate is $A$.

Let $d$ be the separation between the plates of first capacitor, the separation between the plates of second capacitor will be $\left[a-\left(b+d\right)\right]$.

The capacitances of these capacitors are

$C_1=\frac{{\epsilon }_{0}A}{d}$ and $C_2=\frac{{\epsilon }_{0}A}{a-\left(b+d\right)}$

Since the capacitors are in series, so the equivalent capacitance of the combination $C$ is given by

$$\begin{gathered} \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{\left({\displaystyle \frac{{\epsilon }_{0}A}{d}}\right)}+\frac{1}{\left({\displaystyle \frac{{\epsilon }_{0}A}{\left[a-\left(b+d\right)\right]}}\right)} \\ \Rightarrow \frac{1}{C}=\frac{1}{\left({\displaystyle {\epsilon }_{0}A}\right)}\left[d+a-b-d\right]=\frac{a-b}{{\epsilon }_{0}A} \\ \Rightarrow C=\frac{{\epsilon }_{0}A}{a-b} \end{gathered}$$

This is the required expression for the net capacitance and is independent of $d$ and hence of the position of the central H-shaped part.