An electron having charge e located at A, in the presence of a point charge +q located at O, is moved to the point B such that OAB forms an equilateral triangle the work done in the process is equal to

Options

A. `(q)/(AB)`
B. `(eq)/(AB)`
C. `-eq(AB)`
D. zero

Explanation

To solve the problem, we need to calculate the work done when an electron is moved from point A to point B in the electric field created by a point charge +q located at point O. The points A, B, and O form an equilateral triangle.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a point charge +q at point O.
- An electron (which has a charge of -e) is initially at point A.
- The electron is moved to point B such that OAB forms an equilateral triangle.

2. **Identifying the Potential**:
- The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by the formula:
\[ V = \frac{kq}{r} \]
where \( k \) is Coulomb's constant.

3. **Distance Between Points**:
- Since OAB forms an equilateral triangle, the distance OA (from O to A) and OB (from O to B) are equal. Let's denote this distance as \( r \).

4. **Calculating the Potential at Points A and B**:
- The potential at point A due to charge +q is:
\[ V_A = \frac{kq}{r} \]
- The potential at point B due to charge +q is also:
\[ V_B = \frac{kq}{r} \]

5. **Finding the Potential Difference**:
- The potential difference \( V_B - V_A \) is:
\[ V_B - V_A = \frac{kq}{r} - \frac{kq}{r} = 0 \]

6. **Calculating the Work Done**:
- The work done \( W \) in moving a charge \( Q \) through a potential difference \( V \) is given by:
\[ W = Q \cdot (V_B - V_A) \]
- Here, \( Q \) is the charge of the electron, which is -e:
\[ W = -e \cdot 0 = 0 \]

### Conclusion:
The work done in moving the electron from point A to point B is **0**.

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