A charge Q is kept at the centre of a circle of radius `'r'` If permittivity of the free space is e, then work done in carrying a charge q along the diameter of the circle from one end to the other will be

Options

A. A) `(qQ )/( 4 pi in_0 in_r r)`
B. B) `(qQ)/( 8 pi in_0 r)`
C. C) `(qQ)/(2 pi in_0)`
D. D) zero

Answer

To solve the problem, we need to determine the work done in moving a charge \( q \) from one end of the diameter of a circle to the other end, with a charge \( Q \) located at the center of the circle.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charge \( Q \) at the center of a circle with radius \( r \).
- We want to move a charge \( q \) from point A to point B along the diameter of the circle.

2. **Electric Potential Due to Charge \( Q \)**:
- The electric potential \( V \) at a distance \( r \) from a point charge \( Q \) is given by the formula:
\[ V = \frac{Q}{4\pi \epsilon r} \]
- Since charge \( q \) is moved along the diameter, the distance from the charge \( Q \) to any point on the diameter remains constant (equal to \( r \)).

3. **Equipotential Surface**:
- The charge \( Q \) creates an equipotential surface around it. This means that the potential is the same at all points equidistant from the charge.
- Therefore, the potential at point A (one end of the diameter) is equal to the potential at point B (the other end of the diameter):
\[ V_A = V_B \]

4. **Work Done Calculation**:
- The work done \( W \) in moving a charge in an electric field is given by the formula:
\[ W = q(V_B - V_A) \]
- Since \( V_A = V_B \), we have:
\[ W = q(0) = 0 \]

5. **Conclusion**:
- The work done in carrying the charge \( q \) along the diameter of the circle from one end to the other is:
\[ W = 0 \]

### Final Answer:
The work done in carrying a charge \( q \) along the diameter of the circle from one end to the other is \( 0 \).