Find the flux of point charge q through the square surface ABCD as shown

Options

A. (A)`q/6 epsilon_0`
(B)`q/epsilon_0`
(C)`q/4 epsilon_0`
(D)`q/2 epsilon_0`
B. Find the flux of point charge q through the square surface ABCD as shown `(##JM_21_M2_20210224_PHY_09_Q01##)`
C. not varying with `r`
D. varying as `r^2`

Explanation

To find the flux of a point charge \( q \) through the square surface ABCD, we can use Gauss's law, which states that the electric flux \( \Phi \) through a closed surface is proportional to the charge enclosed by that surface. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand Gauss's Law**:
Gauss's law states that the electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the total charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Configuration**:
In this problem, we have a point charge \( q \) and a square surface ABCD. The charge \( q \) is not enclosed by the square surface ABCD.

3. **Create an Enclosing Surface**:
To apply Gauss's law, we can imagine a cube that encloses the charge \( q \). Let’s assume the cube has a side length \( A \) and the charge \( q \) is located at one of its corners.

4. **Calculate Total Flux Through the Cube**:
Since the charge \( q \) is at one corner of the cube, the total electric flux through the entire cube is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

5. **Determine the Flux Through One Face**:
By symmetry, since the charge is at a corner, the flux will be equally distributed among the three faces that meet at that corner. Therefore, the flux through one face (like the square surface ABCD) will be:
\[ \Phi_{\text{face}} = \frac{1}{3} \cdot \Phi_{\text{total}} = \frac{1}{3} \cdot \frac{q}{\epsilon_0} \]

6. **Calculate the Flux Through the Square Surface ABCD**:
Since the square surface ABCD is one of the three faces sharing the corner with the charge, the flux through the surface ABCD is:
\[ \Phi_{ABCD} = \frac{1}{3} \cdot \frac{q}{\epsilon_0} \]

### Final Answer:
The electric flux \( \Phi \) through the square surface ABCD is:
\[ \Phi_{ABCD} = \frac{q}{3\epsilon_0} \]