Two short electric dipoles having dipole moment `p_(1)" and "p_(2)` are placed co-axially and uni-directionally, at a distance r apart. Calculate nature and magnitude of force between them :

Answer

To solve the problem of finding the nature and magnitude of the force between two short electric dipoles placed coaxially and unidirectionally at a distance \( r \) apart, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two dipoles, \( \mathbf{p_1} \) and \( \mathbf{p_2} \), placed along the same axis (coaxially) and pointing in the same direction (unidirectionally).
- The distance between the two dipoles is \( r \).

2. **Potential Energy of the Dipoles**:
- The potential energy \( U \) between two dipoles can be expressed as:
\[ U = -\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \mathbf{p_1} \cdot \mathbf{p_2}}{r^3} \]
- Since both dipoles are aligned in the same direction, the angle \( \theta \) between \( \mathbf{p} \) and the electric field \( \mathbf{E} \) is \( 0^\circ \), and thus \( \cos(0^\circ) = 1 \).

3. **Electric Field Due to a Dipole**:
- The electric field \( \mathbf{E} \) at a distance \( r \) due to dipole \( \mathbf{p_2} \) acting on dipole \( \mathbf{p_1} \) is given by:
\[ \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \mathbf{p_2}}{r^3} \]

4. **Substituting Electric Field into Potential Energy**:
- Substitute the expression for \( \mathbf{E} \) into the potential energy formula:
\[ U = -\mathbf{p_1} \cdot \mathbf{E} = -\mathbf{p_1} \cdot \left( \frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \mathbf{p_2}}{r^3} \right) \]
- Thus, we have:
\[ U = -\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \mathbf{p_1} \cdot \mathbf{p_2}}{r^3} \]

5. **Calculating the Force**:
- The force \( \mathbf{F} \) between the dipoles can be found by taking the derivative of the potential energy with respect to the distance \( r \):
\[ F = -\frac{dU}{dr} \]
- Differentiating \( U \):
\[ F = -\frac{d}{dr}\left(-\frac{1}{4 \pi \epsilon_0} \cdot \frac{2 \mathbf{p_1} \cdot \mathbf{p_2}}{r^3}\right) \]
- This results in:
\[ F = \frac{6 \mathbf{p_1} \cdot \mathbf{p_2}}{4 \pi \epsilon_0 r^4} \]

6. **Nature of the Force**:
- The force is positive, indicating that the force is attractive since the potential energy is negative. Therefore, the nature of the force between the dipoles is attractive.

### Final Result:
- The magnitude of the force between the two dipoles is:
\[ F = \frac{6 \mathbf{p_1} \cdot \mathbf{p_2}}{4 \pi \epsilon_0 r^4} \]
- The nature of the force is **attractive**.