Two point charges are located on the x - axis : `q_1 = -e` at `x = 0` and `q_2 = + e` at `x = a`.
The work done by an external force to bring a third point charge `q_3 = +e` from infinity to `x = 2a` is.

Options

A. `(e^2)/(4 pi epsilon_0 a)`
B. `(e^2)/(8 pi epsilon_0 a)`
C. `(-e^2)/(8 pi epsilon_0 a)`
D. `(-e^2)/(4 pi epsilon_0 a)`

Explanation

To solve the problem of finding the work done by an external force to bring a third point charge \( q_3 = +e \) from infinity to \( x = 2a \), we will follow these steps:

### Step 1: Understand the System
We have two point charges:
- \( q_1 = -e \) located at \( x = 0 \)
- \( q_2 = +e \) located at \( x = a \)

We need to bring a third charge \( q_3 = +e \) from infinity to the position \( x = 2a \).

### Step 2: Calculate Initial Potential Energy
The initial potential energy \( U_i \) of the system, which consists of \( q_1 \) and \( q_2 \), is given by the formula:
\[ U_i = k \frac{q_1 q_2}{r_{12}} \]
where \( k \) is Coulomb's constant, \( r_{12} \) is the distance between \( q_1 \) and \( q_2 \), which is \( a \).

Substituting the values:
\[ U_i = k \frac{(-e)(e)}{a} = -\frac{k e^2}{a} \]

### Step 3: Calculate Final Potential Energy
Now, we need to calculate the final potential energy \( U_f \) of the system when \( q_3 \) is added at \( x = 2a \). The final potential energy includes contributions from all three charges:
\[ U_f = k \frac{q_1 q_2}{a} + k \frac{q_2 q_3}{a} + k \frac{q_1 q_3}{2a} \]

Calculating each term:
1. \( k \frac{q_1 q_2}{a} = -\frac{k e^2}{a} \)
2. \( k \frac{q_2 q_3}{a} = k \frac{(e)(e)}{a} = \frac{k e^2}{a} \)
3. \( k \frac{q_1 q_3}{2a} = k \frac{(-e)(e)}{2a} = -\frac{k e^2}{2a} \)

Adding these contributions together:
\[ U_f = -\frac{k e^2}{a} + \frac{k e^2}{a} - \frac{k e^2}{2a} \]
The first two terms cancel each other out:
\[ U_f = -\frac{k e^2}{2a} \]

### Step 4: Calculate Work Done
The work done \( W \) by the external force is equal to the change in potential energy:
\[ W = U_f - U_i \]
Substituting the values we found:
\[ W = \left(-\frac{k e^2}{2a}\right) - \left(-\frac{k e^2}{a}\right) \]
This simplifies to:
\[ W = -\frac{k e^2}{2a} + \frac{k e^2}{a} = \frac{k e^2}{2a} \]

### Step 5: Substitute the Value of \( k \)
Using \( k = \frac{1}{4 \pi \epsilon_0} \):
\[ W = \frac{1}{2} \cdot \frac{e^2}{4 \pi \epsilon_0 a} = \frac{e^2}{8 \pi \epsilon_0 a} \]

### Final Answer
The work done by the external force to bring the charge \( q_3 \) from infinity to \( x = 2a \) is:
\[ W = \frac{e^2}{8 \pi \epsilon_0 a} \]