Question

A smooth dielectric slab $A$ of mass $m$ and dielectric constant $k$ is placed between the plates of a parallel plate capacitor and connected to another block $B$ of equal mass $m$ through a string and pulley arrangement as shown in the figure.
The capacitor plates with separation $\delta$ and width $b$ are connected to a battery of emf $E$ as shown in the figure. The pulley and string are massless and $E=\sqrt{\frac{mg\delta }{b(k-1){\in }_0}}$

Answer 1

Let the length of the plate be ${\ell }_1$ width be $b$ and plate separation be $\delta$. The capacitance, $C$ as a function of $x$ is given by
$C\left(x\right)=\frac{{\in }_{0}bx}{\delta }+\frac{k{\in }_{0}b(\ell -x)}{\delta }$
The force on the dielectric is,
$F_x=\frac{1}{2}{E}^2\frac{dC}{dx}=-\frac{{\in }_0}{2}(k-1)E^2\frac{b}{\delta }$
For the two blocks, we can write,
$-T+mg=ma$ ...(1)
$T\frac{{\in }_0}{2}(k-1)E^2\frac{b}{\delta }=ma$ ...(2)
$\therefore \frac{d^{2}x}{d{t}^2}=a=\frac{1}{2m}[mg-\frac{{\in }_0}{2}(k-1\left)\frac{E^{2}b}{\delta }\right]=\frac{g}{4}$
or, $x=\frac{1}{2}\left(\frac{g}{4}\right)t^2$