A small electric dipole of dipole moment P is placed at a distance r from one infinitely long uniformly charged, straight conductor. If `lambda` is the linear charge density of straight conductor and axis of the dipole is perpendicular to the straight conductor then the force acting on the dipole is

Options

A. `(lambdaP)/(2piepsilon_0r^2)`
B. `(lambdaP)/(2piepsilon_0r)`
C. `(lambdaP)/(4piepsilon_0r^2)`
D. `(lambdaP)/(piepsilon_0r^2)`

Explanation

To solve the problem of finding the force acting on a small electric dipole placed near an infinitely long uniformly charged straight conductor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have an infinitely long straight conductor with a linear charge density \( \lambda \).
- A small electric dipole with dipole moment \( \mathbf{P} \) is placed at a distance \( r \) from the conductor.
- The axis of the dipole is perpendicular to the straight conductor.

2. **Electric Field due to the Conductor**:
- The electric field \( \mathbf{E} \) due to an infinitely long charged wire at a distance \( r \) is given by the formula:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 r} \]
- This electric field points away from the conductor if the charge is positive.

3. **Force on the Dipole**:
- The force \( \mathbf{F} \) on an electric dipole in an electric field is given by:
\[ \mathbf{F} = \mathbf{p} \cdot \nabla \mathbf{E} \]
- However, since the dipole moment \( \mathbf{P} \) is perpendicular to the electric field, we can also express the force as:
\[ F = \frac{dE}{dr} \cdot P \]

4. **Calculate the Gradient of the Electric Field**:
- We need to find the derivative of the electric field \( E \) with respect to \( r \):
\[ \frac{dE}{dr} = \frac{d}{dr} \left( \frac{\lambda}{2 \pi \epsilon_0 r} \right) = -\frac{\lambda}{2 \pi \epsilon_0 r^2} \]

5. **Substituting into the Force Equation**:
- Now, substituting \( \frac{dE}{dr} \) into the force equation:
\[ F = P \left( -\frac{\lambda}{2 \pi \epsilon_0 r^2} \right) = -\frac{\lambda P}{2 \pi \epsilon_0 r^2} \]

6. **Final Expression for the Force**:
- The magnitude of the force acting on the dipole is:
\[ F = \frac{\lambda P}{2 \pi \epsilon_0 r^2} \]
- The negative sign indicates that the force is attractive if the dipole is aligned in the direction of the electric field.

### Conclusion:
The force acting on the dipole is given by:
\[ F = \frac{\lambda P}{2 \pi \epsilon_0 r^2} \]