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

Options

A. `(sqrt(mgd))/(pi epsi_(0)r^(2))`
B. `sqrt((mgd)/(pi epsi_(0)r))`
C. `sqrt((m gd^(2))/(pi epsi_(0)r^(2)))`
D. `sqrt((mgd)/(pi epsi_(0)r^(2)))`

Explanation

To find the minimum voltage required to lift the thin conducting disc placed at the center of the bottom plate of a capacitor, we can follow these steps:

### Step 1: Understand the System
We have a capacitor made of two circular plates with radius \( R \) and separated by a distance \( d \) (where \( d \ll R \)). A thin conducting disc of radius \( r \) (where \( r \ll R \)) and thickness \( t \) (where \( t \ll r \)) is placed at the center of the bottom plate.

### Step 2: Electric Field in the Capacitor
The electric field \( E \) between the plates of the capacitor is given by:
\[ E = \frac{V}{d} \]
where \( V \) is the voltage across the plates.

### Step 3: Charge on the Disc
The charge \( Q' \) on the disc can be calculated using the capacitance \( C \) of the capacitor. The capacitance of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
where \( A \) is the area of the plates. For our circular plates, the area \( A \) is:
\[ A = \pi R^2 \]
Thus, the capacitance becomes:
\[ C = \frac{\epsilon_0 \pi R^2}{d} \]
The charge \( Q' \) on the capacitor is:
\[ Q' = C \cdot V = \frac{\epsilon_0 \pi R^2}{d} \cdot V \]

### Step 4: Force on the Disc
The force \( F \) acting on the disc due to the electric field is given by:
\[ F = Q' \cdot E \]
Substituting the expressions for \( Q' \) and \( E \):
\[ F = \left( \frac{\epsilon_0 \pi R^2}{d} \cdot V \right) \cdot \left( \frac{V}{d} \right) = \frac{\epsilon_0 \pi R^2 V^2}{d^2} \]

### Step 5: Condition for Lifting the Disc
For the disc to lift off, the force \( F \) must be equal to or greater than the weight of the disc \( mg \):
\[ \frac{\epsilon_0 \pi R^2 V^2}{d^2} = mg \]

### Step 6: Solve for Voltage \( V \)
Rearranging the equation to solve for \( V \):
\[ V^2 = \frac{mg d^2}{\epsilon_0 \pi R^2} \]
Taking the square root gives:
\[ V = \sqrt{\frac{mg d^2}{\epsilon_0 \pi R^2}} \]

### Final Expression for Minimum Voltage
Thus, the minimum voltage required to lift the disc is:
\[ V = \sqrt{\frac{mg d^2}{\epsilon_0 \pi r^2}} \]