Show that, for a given dipole, V & E cannot have the same magnitude at distance less than 2m from the dipole. Suppose that the distance is `sqrt(5)m`, determine the directions along which V & E are equal in magnitude.

Explanation

To solve the problem, we will break it down into two parts as outlined in the question.

### Part 1: Show that V and E cannot have the same magnitude at a distance less than 2m from the dipole.

1. **Understanding the formulas**:
- The electric potential \( V \) due to a dipole is given by:
\[ V = \frac{kp \cos \theta}{r^2} \]
- The electric field \( E \) due to a dipole is given by:
\[ E = \frac{kp \sqrt{1 + 3 \cos^2 \theta}}{r^3} \]

2. **Setting the magnitudes equal**:
- We want to find when \( |V| = |E| \):
\[ \frac{kp \cos \theta}{r^2} = \frac{kp \sqrt{1 + 3 \cos^2 \theta}}{r^3} \]

3. **Canceling common terms**:
- Cancel \( kp \) from both sides (assuming \( kp \neq 0 \)):
\[ \frac{\cos \theta}{r^2} = \frac{\sqrt{1 + 3 \cos^2 \theta}}{r^3} \]

4. **Rearranging the equation**:
- Multiply both sides by \( r^3 \):
\[ r \cos \theta = \sqrt{1 + 3 \cos^2 \theta} \]

5. **Squaring both sides**:
- To eliminate the square root, square both sides:
\[ r^2 \cos^2 \theta = 1 + 3 \cos^2 \theta \]

6. **Rearranging the equation**:
- Rearranging gives:
\[ r^2 \cos^2 \theta - 3 \cos^2 \theta = 1 \]
\[ \cos^2 \theta (r^2 - 3) = 1 \]

7. **Finding the condition for \( r \)**:
- This implies:
\[ \cos^2 \theta = \frac{1}{r^2 - 3} \]
- For \( \cos^2 \theta \) to be valid, \( r^2 - 3 \) must be positive, which means:
\[ r^2 > 3 \quad \Rightarrow \quad r > \sqrt{3} \approx 1.732 \]
- Therefore, the minimum distance \( r \) must be at least \( 2 \) meters.

### Conclusion for Part 1:
Thus, we have shown that for a given dipole, \( V \) and \( E \) cannot have the same magnitude at a distance less than \( 2 \) meters from the dipole.

---

### Part 2: Determine the directions along which \( V \) and \( E \) are equal in magnitude at \( r = \sqrt{5} \) m.

1. **Using the derived equation**:
- We will use the equation:
\[ \cos^2 \theta = \frac{1}{r^2 - 3} \]
- Substitute \( r = \sqrt{5} \):
\[ \cos^2 \theta = \frac{1}{(\sqrt{5})^2 - 3} = \frac{1}{5 - 3} = \frac{1}{2} \]

2. **Finding \( \cos \theta \)**:
- Taking the square root gives:
\[ \cos \theta = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

3. **Finding the angles**:
- The angles corresponding to \( \cos \theta = \frac{\sqrt{2}}{2} \) are:
\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{7\pi}{4} \]

### Conclusion for Part 2:
Thus, the directions along which the magnitudes of \( V \) and \( E \) are equal at a distance of \( \sqrt{5} \) meters are \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{7\pi}{4} \).

---