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 find 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: Identify the charges and their positions
- We have two point charges:
- \( q_1 = -e \) located at \( x = 0 \)
- \( q_2 = +e \) located at \( x = a \)
- The third charge \( q_3 = +e \) will be brought from infinity to \( x = 2a \).

### Step 2: Calculate the potential energy when \( q_3 \) is at \( x = 2a \)
The total potential energy \( U_f \) when \( q_3 \) is at \( x = 2a \) can be calculated using the potential energy contributions from the interactions between \( q_3 \) and the other two charges.

1. **Potential energy between \( q_1 \) and \( q_3 \)**:
\[ U_{13} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(-e)(+e)}{2a} = -\frac{e^2}{8 \pi \epsilon_0 a} \]

2. **Potential energy between \( q_2 \) and \( q_3 \)**:
\[ U_{23} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(+e)(+e)}{a} = \frac{e^2}{4 \pi \epsilon_0 a} \]

3. **Total potential energy \( U_f \)**:
\[ U_f = U_{13} + U_{23} = -\frac{e^2}{8 \pi \epsilon_0 a} + \frac{e^2}{4 \pi \epsilon_0 a} \]
To combine these, we convert \( \frac{e^2}{4 \pi \epsilon_0 a} \) to have a common denominator:
\[ U_f = -\frac{e^2}{8 \pi \epsilon_0 a} + \frac{2e^2}{8 \pi \epsilon_0 a} = \frac{e^2}{8 \pi \epsilon_0 a} \]

### Step 3: Calculate the initial potential energy \( U_i \) when \( q_3 \) is at infinity
When \( q_3 \) is at infinity, the potential energy \( U_i \) is zero because the force between the charges is negligible at infinite distance:
\[ U_i = 0 \]

### Step 4: Calculate the work done \( W \)
The work done by the external force \( W \) is given by the change in potential energy:
\[ W = U_f - U_i = \frac{e^2}{8 \pi \epsilon_0 a} - 0 = \frac{e^2}{8 \pi \epsilon_0 a} \]

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

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