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 $k=2, V=4\, V , L=80\, cm$, $d=0.1\, mm$ )

Answer 1

$m g =F_{\text {slab }}=\frac{d U}{d x}=\frac{1}{2} V^{2} \frac{d C}{d x}$
$=\frac{1}{2} V^{2} \frac{d}{d x}\left(\frac{\varepsilon_{0}}{d}\left(L^{2}+(k-1) L x\right)\right)$
$=\frac{1}{2} \frac{V^{2}(k-1) L}{d}$
After the slab falls off,
$2 m g-m g=2 m a$
$a=\frac{g}{2}$
$\frac{L}{2}=\frac{1}{2} \times \frac{g}{2} t^{2}$
$t=\sqrt{\frac{2 L}{g}}=\sqrt{\frac{2 \times 0.8}{10}}$
$=\sqrt{0.16}=0.4\, s$