Charge `q_(2)` is at the centre of a circular path with radius r. work done in carrying charge `q_(1)` once around this equipotential path , would be

Options

A. `(1)/(4 pi epsilon_(0)) xx (q_(1)q_(2))/(r^(2))`
B. `(1)/(4 pi epsilon_(0)) xx (q_(1)q_(2))/(r)`
C. zero
D. Infinity

Answer

To solve the problem of finding the work done in carrying charge \( q_1 \) once around an equipotential path where charge \( q_2 \) is at the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Equipotential Surfaces**:
- An equipotential surface is defined as a surface where the electric potential is constant throughout. This means that no work is done when moving a charge along this surface.

2. **Identifying the Charges**:
- We have two charges: \( q_1 \) (the charge being moved) and \( q_2 \) (the charge at the center of the circular path).

3. **Analyzing the Work Done**:
- The work done \( W \) in moving a charge in an electric field is given by the formula:
\[ W = q \cdot \Delta V \]
where \( q \) is the charge being moved and \( \Delta V \) is the change in electric potential.

4. **Determining the Change in Potential**:
- Since the path is an equipotential path, the potential difference \( \Delta V \) when moving from one point to another on this path is zero:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 0 \]

5. **Calculating the Work Done**:
- Substituting \( \Delta V = 0 \) into the work done formula:
\[ W = q_1 \cdot 0 = 0 \]

6. **Conclusion**:
- Therefore, the work done in carrying charge \( q_1 \) once around the equipotential path is:
\[ W = 0 \]

### Final Answer:
The work done in carrying charge \( q_1 \) once around this equipotential path is **0**.

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