A cylindrical layer of a homogeneous dielectric with the dielectric constant $\epsilon$ is introduced into a cylindrical capacitor so that the layer fills the gap of width $d$ between the plates. The mean radius of the plates is $R$ such that $R\gg d.$ The capacitor is connected to a source of a permanent voltage $V.$ Find the force pulling the dielectric inside the capacitor.

Using the formula $W=\raisebox{1ex}{$q^2$}\!\left/ \!\raisebox{-1ex}{$2C$}\right.$ for the energy of a capacitor, we find that, in accordance with W, the required force is:

$F_x={\overline{)\frac{\partial W}{\partial x}}}_q=\frac{q^2}{2}\frac{{\displaystyle \raisebox{1ex}{$\partial C$}\!\left/ \!\raisebox{-1ex}{$\partial x$}\right.}}{C^2}=\frac{V^2}{2}.\frac{\partial C}{\partial x} .....\left(1\right)$

Since $d\ll R$ the capacitance of the given capacitor can be calculated by the formula for a parallel-plate capacitor. Therefore, if the dielectric is introduced to a depth $x$ and the capacitor length is $\mathcal{l}$, we have:

$C=\frac{\epsilon {\epsilon }_{0}x2\pi R}{d}+\frac{{\epsilon }_0(\mathcal{l}-x)2\pi R}{d}=\frac{{\epsilon }_{0}2\pi R}{d}(\epsilon x+\mathcal{l}-x) ......\left(2\right)$

Substituting (2) into (1), we obtain

$F_x=\raisebox{1ex}{${\epsilon }_0(\epsilon -1)\pi R{V}^2$}\!\left/ \!\raisebox{-1ex}{$d$}\right.$