The dipole moment of a system of charge ` +q` distribyted uniformly on an arc of radius `R` subtending an angle ` pi //2` at its centre where another charge ` -q` is placed is :

Options

A. `(2sqrt(2)qR)/(pi)`
B. `(sqrt(2)qR)/(pi)`
C. `(qR)/(pi)`
D. `(2qR)/(pi)`

Answer

To find the dipole moment of a system consisting of a charge `+q` distributed uniformly on an arc of radius `R` that subtends an angle of `π/2` at its center, where another charge `-q` is placed at the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have a quarter ring (arc) of radius `R` that subtends an angle of `π/2` at the center. The charge `+q` is uniformly distributed along this arc.
- The charge `-q` is placed at the center of the arc.

2. **Find the Center of Mass (CM) of the Quarter Ring**:
- The coordinates of the center of mass of a quarter ring can be derived from symmetry and are given by:
\[ (x_{cm}, y_{cm}) = \left(\frac{2R}{\pi}, \frac{2R}{\pi}\right) \]

3. **Determine the Distance Between Charges**:
- The distance `OP` from the origin (where `-q` is located) to the center of mass of the quarter ring (where `+q` is effectively concentrated) can be calculated using the distance formula:
\[ OP = \sqrt{\left(\frac{2R}{\pi}\right)^2 + \left(\frac{2R}{\pi}\right)^2} \]
- Simplifying this gives:
\[ OP = \sqrt{2 \left(\frac{2R}{\pi}\right)^2} = \frac{2R}{\pi} \sqrt{2} \]

4. **Calculate the Dipole Moment**:
- The dipole moment \( \vec{p} \) is given by the formula:
\[ \vec{p} = q \cdot \vec{d} \]
- Here, \( q \) is the magnitude of the charge `+q`, and \( \vec{d} \) is the vector pointing from the charge `-q` at the origin to the center of mass of the quarter ring.
- Thus, the magnitude of the dipole moment is:
\[ p = q \cdot OP = q \cdot \left(\frac{2R}{\pi} \sqrt{2}\right) \]
- Therefore, we can express the dipole moment as:
\[ p = \frac{2\sqrt{2} q R}{\pi} \]

5. **Final Result**:
- The dipole moment of the system is:
\[ \vec{p} = \frac{2\sqrt{2} q R}{\pi} \]