A point charge placed on the axis of a uniformly charged disc experiences a force f due to the disc. If the charge density on the disc `sigma` is, the electric flux through the disc, due to the point charge will be

Options

A. `(2 pi f)/(sigma)`
B. `(f)/(2 pi sigma)`
C. `(f^(2))/(sigma)`
D. `(f)/(sigma)`

Explanation

To solve the problem, we need to find the electric flux through a uniformly charged disc due to a point charge placed on its axis. Let's break down the solution step by step.

### Step 1: Understand the Configuration
We have a uniformly charged disc with charge density \( \sigma \) and a point charge \( Q \) placed at a distance \( x \) along the axis of the disc. The goal is to find the electric flux \( \Phi \) through the disc due to this point charge.

### Step 2: Electric Field Due to the Disc
The electric field \( E \) at a point along the axis of a uniformly charged disc can be expressed as:
\[ E_d = \frac{\sigma}{2 \epsilon_0} \left(1 - \frac{x}{\sqrt{R^2 + x^2}}\right) \]
where \( R \) is the radius of the disc, \( \sigma \) is the surface charge density, and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Electric Flux
The electric flux \( \Phi \) through the disc due to the electric field \( E \) can be calculated using the formula:
\[ \Phi = E \cdot A \]
where \( A \) is the area of the disc. The area \( A \) of the disc is given by:
\[ A = \pi R^2 \]
Thus, the electric flux through the disc becomes:
\[ \Phi = E_d \cdot \pi R^2 \]

### Step 4: Substitute the Electric Field Expression
Substituting the expression for \( E_d \) into the flux equation, we have:
\[ \Phi = \left(\frac{\sigma}{2 \epsilon_0} \left(1 - \frac{x}{\sqrt{R^2 + x^2}}\right)\right) \cdot \pi R^2 \]

### Step 5: Simplify the Expression
Now we can simplify the expression for electric flux:
\[ \Phi = \frac{\sigma \pi R^2}{2 \epsilon_0} \left(1 - \frac{x}{\sqrt{R^2 + x^2}}\right) \]

### Step 6: Relate Flux to Force
From the previous parts, we know that the force \( F \) experienced by the point charge due to the electric field can be expressed as:
\[ F = Q \cdot E_d \]
Using the relationship between electric flux and force, we can express the flux in terms of force:
\[ \Phi = \frac{F}{\sigma} \]

### Final Result
Thus, the electric flux through the disc due to the point charge is given by:
\[ \Phi = \frac{F}{\sigma} \]