$A$ capacitor is made of two circular plates of radius $R$ each,separated by a distance $d \ll R$. The capacitor is connected to a constant voltage $V$. $A$ thin conducting disc of radius $r \ll R$ and thickness $t \ll r$ is placed at the center of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is $m$.

(D) Initially,the thin conducting disc is placed at the center of the bottom plate. The bottom plate is an equipotential surface. The electric field between the plates of the capacitor is $E = \frac{V}{d}$.
When the disc is placed on the bottom plate,it acquires a charge $q'$ due to the electric field. By Gauss's law,the charge on the disc is $q' = \epsilon_0 E A$,where $A = \pi r^2$ is the area of the disc.
Substituting $E = \frac{V}{d}$,we get $q' = \epsilon_0 \left( \frac{V}{d} \right) \pi r^2$.
The repulsive force $F$ acting on the disc in the upward direction is $F = q' E$.
Substituting the values,$F = \left( \epsilon_0 \frac{V}{d} \pi r^2 \right) \left( \frac{V}{d} \right) = \frac{\epsilon_0 \pi r^2 V^2}{d^2}$.
For the disc to be lifted,this repulsive force must be equal to the weight of the disc $(mg)$:
$\frac{\epsilon_0 \pi r^2 V^2}{d^2} = mg$.
Solving for $V$,we get $V^2 = \frac{mg d^2}{\pi \epsilon_0 r^2}$.
Therefore,the minimum voltage required is $V = d \sqrt{\frac{mg}{\pi \epsilon_0 r^2}}$.