Two point charges `q` and `-q` are separated by a distance `2l`. Find the flux strength vector across the circle of radius R placed with its centre coinciding with the of line joining the two charges in the perpendicular plane.

Explanation

To solve the problem of finding the electric flux strength vector across a circle of radius \( R \) placed with its center coinciding with the midpoint of the line joining two point charges \( q \) and \( -q \) separated by a distance \( 2l \), we can follow these steps:

### Step 1: Understand the Configuration
We have two charges, \( q \) and \( -q \), separated by a distance of \( 2l \). The midpoint between these charges is at a distance \( l \) from each charge. The circle of radius \( R \) is placed in a plane that is perpendicular to the line joining the two charges.

**Hint:** Visualize the setup by sketching the two charges and the circle in the perpendicular plane.

### Step 2: Determine the Electric Field
The electric field \( \vec{E} \) at a point in the plane due to a point charge is given by:
\[ \vec{E} = \frac{k \cdot |q|}{r^2} \hat{r} \]
where \( k \) is Coulomb's constant, \( r \) is the distance from the charge to the point of interest, and \( \hat{r} \) is the unit vector pointing away from the charge.

**Hint:** Remember that the electric field due to a positive charge points away from the charge, while the field due to a negative charge points towards it.

### Step 3: Calculate the Electric Field at the Center of the Circle
At the center of the circle, the contributions to the electric field from both charges need to be considered. The distance from each charge to the center of the circle is \( l \).

The electric field due to charge \( q \) at the center is:
\[ \vec{E}_q = \frac{kq}{l^2} \hat{r}_q \]
The electric field due to charge \( -q \) at the center is:
\[ \vec{E}_{-q} = \frac{k(-q)}{l^2} \hat{r}_{-q} \]

Since the circle is centered at the midpoint, the directions of these fields will be opposite.

**Hint:** Use vector addition to find the net electric field at the center of the circle.

### Step 4: Calculate the Total Electric Field
The total electric field \( \vec{E}_{\text{total}} \) at the center of the circle can be calculated as:
\[ \vec{E}_{\text{total}} = \vec{E}_q + \vec{E}_{-q} \]
Since \( \vec{E}_q \) and \( \vec{E}_{-q} \) are equal in magnitude but opposite in direction, they will add up to zero at the center.

**Hint:** Consider symmetry when calculating the net electric field.

### Step 5: Determine the Electric Flux
The electric flux \( \Phi \) through the circle can be calculated using Gauss's law:
\[ \Phi = \int \vec{E} \cdot d\vec{A} \]
Since the electric field at the center is zero, the total electric flux through the circle is also zero.

**Hint:** Remember that the flux is related to the net electric field passing through the area.

### Final Answer
The electric flux strength vector across the circle of radius \( R \) is:
\[ \Phi = 0 \]