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 a circular path of radius \( r \) with charge \( q_2 \) at the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charge \( q_2 \) located at the center of a circular path with radius \( r \).
- We are moving another charge \( q_1 \) around this circular path.

2. **Identifying the Nature of the Path**:
- The circular path around the charge \( q_2 \) is an equipotential surface. This means that the electric potential at every point on this path is the same.

3. **Work Done in Electric Fields**:
- The work done \( W \) in moving a charge in an electric field is given by the formula:
\[ W = q \cdot V \]
where \( V \) is the potential difference.

4. **Potential Difference on the Equipotential Surface**:
- Since the circular path is an equipotential surface, the potential difference \( V \) when moving charge \( q_1 \) around the path is zero:
\[ V = 0 \]

5. **Calculating the Work Done**:
- Substituting the potential difference into the work done equation:
\[ W = q_1 \cdot 0 = 0 \]

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 the equipotential path is \( 0 \).

---