A capacitor is made of two concentric cylinders of radius $r_1$ and $r_2(r_1<r_2)$ and length $L>>r_2.$ The region between $r_1$ and $r_2$ is filled with a circular cylinder of length $L$ , inner radius $r_1 , outer radius r_2$ and dielectric constant $K$(the remaining volume is an air gap). How much mechanical work must be done to remove the dielectric cylinder while maintaining constant potential difference of V between $r_1$ and $r_2?$

From potential difference between the two cylinders,  we find the capacitance $C$

$C=\frac{\lambda L}{V}=\frac{2\pi {\epsilon }_{0}L}{{\displaystyle \frac{1}{K}}\mathcal{ln}{\displaystyle \frac{r_3}{r_1}}+\mathcal{ln}\frac{r_2}{r_3}} ........\left(1\right)$

When the potential difference between the two cylinders is kept constant, the system can no longer be isolates. There has to be some source of charge(battery) to supply energy. Let $C\text{'}$ be the capacitance of the system without the dielectric material. From (1) we find

$C\text{'}=\frac{2\pi {\epsilon }_{0}L}{{\displaystyle \mathcal{ln}\frac{r_2}{r_1}}}$

The worl needed is

work needed $=\frac{1}{2}C\text{'}{V}^2-\frac{1}{2}C{V}^2-(Q\text{'}-Q)V$

where $Q\text{'}$ is the final total charge on one cylinder. Using $Q\text{'}=C\text{'}V,$ we get

Work needed $=\frac{V^2}{2}(C-C\text{'})$

                      $=\frac{V^2}{2}2\pi {\epsilon }_{0}L\left(\frac{1}{\frac{1}{K}\mathcal{ln}\frac{r_3}{r_1}+\mathcal{ln}\frac{r_2}{r_3}}-\frac{1}{\mathcal{ln}\frac{r_2}{r_1}}\right)$