Question

A dielectric slab of dielectric constant $k$, mass $m$, thickness $d$ and area $L\times L$ is hanging vertically in equilibrium under the influence of gravity and electrostatic pull of a capacitor connected to a battery of voltage $V$. The capacitor has plates of area $L\times L$ and distance between the plates is $d$. The capacitor is half filled by the dielectric. Suddenly a mass of $m$ is attached to dielectric without any impulse on the system. The slab falls off in time $t$. Evaluate $t$ (in $s$).
(Take $k=2,V=4V,L=80cm$, $d=0.1mm$ )

Answer 1

Answer: 0.40

$mg=F_{\text{slab }}=\frac{dU}{dx}=\frac{1}{2}{V}^2\frac{dC}{dx}$
$=\frac{1}{2}{V}^2\frac{d}{dx}\left(\frac{{\epsilon }_0}{d}\left(L^2+(k-1)Lx\right)\right)$
$=\frac{1}{2}\frac{V^2(k-1)L}{d}$
After the slab falls off,
$2mg-mg=2ma$
$a=\frac{g}{2}$
$\frac{L}{2}=\frac{1}{2}\times \frac{g}{2}{t}^2$
$t=\sqrt{\frac{2L}{g}}=\sqrt{\frac{2\times 0.8}{10}}$
$=\sqrt{0.16}=0.4s$