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A capacitor is made of two circular plates of radius $\mathrm{R}$ each, separated by a distance $\mathrm{d} \ll \mathrm{R}$. The capacitor is connected to a constant voltage. A thin conducting disc of radius $r$ $\ll \mathrm{R}$ and thickness $\mathrm{t} \ll \mathrm{r}$ is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is $\mathrm{m}$.
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Verified Answer
Let initially the disc is in touch with the bottom plate, so the entire plate is a equipotential.
The Magnitude of electric field $\mathrm{E}$ on the disc (when potential difference V) is applied between plates of capacitor is :
$$
\mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}} \quad\left[\therefore \mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dr}}\right]
$$
Let charge $q$ ' is transferred to the disc during the process. Therefore by Gauss' theorem,
$$
\begin{aligned}
& \oint \mathrm{E} . \mathrm{dS}=\frac{\mathrm{q}^{\prime}}{\epsilon_0} \\
\therefore \quad & \mathrm{q}^{\prime}=+\varepsilon_0 \frac{\mathrm{V}}{\mathrm{d}} \pi \mathrm{r}^2
\end{aligned}
$$
Electrostatic repulsive force acting on disc in upward direction.
So, $\mathrm{F}=\mathrm{q}^{\prime} \mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}} \pi \mathrm{r}^2 \varepsilon_0 \cdot \frac{\mathrm{V}}{\mathrm{d}}=\frac{\mathrm{V}^2}{\mathrm{~d}^2} \pi \mathrm{r}^2 \in_0$
The repulsive force acting on the disc will be balanced by weight $(\mathrm{mg})$.
$$
\varepsilon_0 \frac{\mathrm{V}^2}{\mathrm{~d}^2} \pi \mathrm{r}^2=\mathrm{mg}
$$
If the disc is to be lifted, then the minimum voltage
$$
\mathrm{V}=\sqrt{\frac{\mathrm{md}^2}{\pi \varepsilon_0 \mathrm{r}^2}}
$$
This is the required expression.
The Magnitude of electric field $\mathrm{E}$ on the disc (when potential difference V) is applied between plates of capacitor is :
$$
\mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}} \quad\left[\therefore \mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dr}}\right]
$$
Let charge $q$ ' is transferred to the disc during the process. Therefore by Gauss' theorem,
$$
\begin{aligned}
& \oint \mathrm{E} . \mathrm{dS}=\frac{\mathrm{q}^{\prime}}{\epsilon_0} \\
\therefore \quad & \mathrm{q}^{\prime}=+\varepsilon_0 \frac{\mathrm{V}}{\mathrm{d}} \pi \mathrm{r}^2
\end{aligned}
$$
Electrostatic repulsive force acting on disc in upward direction.
So, $\mathrm{F}=\mathrm{q}^{\prime} \mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}} \pi \mathrm{r}^2 \varepsilon_0 \cdot \frac{\mathrm{V}}{\mathrm{d}}=\frac{\mathrm{V}^2}{\mathrm{~d}^2} \pi \mathrm{r}^2 \in_0$
The repulsive force acting on the disc will be balanced by weight $(\mathrm{mg})$.
$$
\varepsilon_0 \frac{\mathrm{V}^2}{\mathrm{~d}^2} \pi \mathrm{r}^2=\mathrm{mg}
$$
If the disc is to be lifted, then the minimum voltage
$$
\mathrm{V}=\sqrt{\frac{\mathrm{md}^2}{\pi \varepsilon_0 \mathrm{r}^2}}
$$
This is the required expression.
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