Two identical positive charges are placed at `x=-a` and `x=a`. The correct variation of potential V along the x-axis is given by

Options

A.
B.
C.
D.

Explanation

To solve the problem of finding the variation of electric potential \( V \) along the x-axis due to two identical positive charges placed at \( x = -a \) and \( x = a \), we can follow these steps:

### Step 1: Understand the Configuration
We have two identical positive charges, \( +Q \), located at positions \( x = -a \) and \( x = a \). The electric potential \( V \) at any point \( x \) on the x-axis is the sum of the potentials due to each charge.

### Step 2: Write the Expression for Electric Potential
The electric potential \( V \) at a point \( x \) due to a point charge \( Q \) is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point where we are calculating the potential.

### Step 3: Calculate the Potential at Point \( x \)
For our configuration:
- The distance from the charge at \( x = -a \) to a point \( x \) is \( |x + a| \).
- The distance from the charge at \( x = a \) to a point \( x \) is \( |x - a| \).

Thus, the total potential \( V \) at a point \( x \) is:
\[ V(x) = \frac{kQ}{|x + a|} + \frac{kQ}{|x - a|} \]

### Step 4: Analyze the Potential at Specific Points
1. **At the origin \( x = 0 \)**:
\[ V(0) = \frac{kQ}{|-a|} + \frac{kQ}{|a|} = \frac{kQ}{a} + \frac{kQ}{a} = \frac{2kQ}{a} \]

2. **As \( x \) approaches \( -a \) or \( a \)**:
- As \( x \) approaches \( -a \), \( |x + a| \) approaches 0, leading \( V \) to approach \( +\infty \).
- As \( x \) approaches \( a \), \( |x - a| \) approaches 0, leading \( V \) to approach \( +\infty \).

3. **As \( x \) approaches \( \pm \infty \)**:
- As \( x \) approaches \( +\infty \) or \( -\infty \), both \( |x + a| \) and \( |x - a| \) become very large, thus \( V \) approaches 0.

### Step 5: Sketch the Graph
- The potential \( V \) is \( +\infty \) at \( x = -a \) and \( x = a \).
- The potential is \( \frac{2kQ}{a} \) at \( x = 0 \).
- The potential approaches 0 as \( x \) goes to \( \pm \infty \).

### Conclusion
The variation of potential \( V \) along the x-axis will have a shape that approaches \( +\infty \) at the positions of the charges and approaches 0 as \( x \) moves towards \( \pm \infty \). The correct option representing this variation is option C.