The work done in carrying a charge q once round a circle of radius r with a charge Q at the centre is

Options

A. `(qQ)/(4pi epsi_(0)a)`
B. `(qQ)/(4pi epsi_(0)a^(2))`
C. `(q)/(4pi epsi_(0)a)`
D. zero

Explanation

To solve the problem of finding the work done in carrying a charge \( q \) once around a circle of radius \( r \) with a charge \( Q \) at the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charge \( Q \) located at the center of a circle.
- We are moving a charge \( q \) along the circumference of this circle.

2. **Electric Potential Due to Charge \( Q \)**:
- The electric potential \( V \) at any point on the circumference of the circle (distance \( r \) from charge \( Q \)) is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant.

3. **Potential at Different Points**:
- Since the charge \( q \) is moved along a circular path, every point on the circumference is at the same distance \( r \) from the charge \( Q \).
- Therefore, the electric potential \( V \) is constant along the entire circular path.

4. **Calculating Potential Difference**:
- The potential difference \( \Delta V \) between any two points on the circle is:
\[ \Delta V = V_a - V_b = 0 \]
because \( V_a \) and \( V_b \) are equal (both equal to \( \frac{kQ}{r} \)).

5. **Work Done Calculation**:
- The work done \( W \) in moving the charge \( q \) around the circle is given by:
\[ W = q \cdot \Delta V \]
- Substituting \( \Delta V = 0 \):
\[ W = q \cdot 0 = 0 \]

6. **Conclusion**:
- The work done in carrying the charge \( q \) once around the circle is \( 0 \).

### Final Answer:
The work done in carrying a charge \( q \) once round a circle of radius \( r \) with a charge \( Q \) at the center is **0**.

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