A negative point charge ` -2q` and a positive charge `q` are fixed ar a distance `l` apart. Where should a postive test charge `Q` be placed on the line connecting the charge for it to be in equlibrium ? What is the nature of the equilibrium with respect to longitudinal motions ?

Explanation

To determine where to place a positive test charge \( Q \) between a negative point charge \( -2q \) and a positive charge \( q \) fixed at a distance \( l \) apart, we need to analyze the forces acting on the test charge \( Q \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two charges: \( -2q \) (negative) and \( q \) (positive) separated by a distance \( l \).
- We want to find a point \( x \) along the line connecting these charges where a positive test charge \( Q \) can be placed such that it experiences no net force (equilibrium).

2. **Identifying the Regions**:
- The test charge \( Q \) can be placed either to the left of \( -2q \), between \( -2q \) and \( q \), or to the right of \( q \).
- However, placing \( Q \) to the left of \( -2q \) would result in it being attracted to \( -2q \) and repelled by \( q \), making equilibrium impossible in that region.

3. **Analyzing the Region Between the Charges**:
- We will consider placing \( Q \) between \( -2q \) and \( q \). Let’s denote the distance from \( -2q \) to \( Q \) as \( x \). Therefore, the distance from \( Q \) to \( q \) would be \( l - x \).

4. **Setting Up the Force Equations**:
- The force \( F_1 \) on \( Q \) due to charge \( -2q \) is given by:
\[ F_1 = k \frac{Q \cdot 2q}{x^2} \]
- The force \( F_2 \) on \( Q \) due to charge \( q \) is given by:
\[ F_2 = k \frac{Q \cdot q}{(l - x)^2} \]
- For equilibrium, these forces must be equal in magnitude:
\[ F_1 = F_2 \implies k \frac{Q \cdot 2q}{x^2} = k \frac{Q \cdot q}{(l - x)^2} \]

5. **Simplifying the Equation**:
- Cancel \( k \) and \( Q \) from both sides (assuming \( Q \neq 0 \)):
\[ \frac{2q}{x^2} = \frac{q}{(l - x)^2} \]
- This simplifies to:
\[ 2(l - x)^2 = x^2 \]

6. **Expanding and Rearranging**:
- Expanding the equation:
\[ 2(l^2 - 2lx + x^2) = x^2 \]
- Rearranging gives:
\[ 2l^2 - 4lx + 2x^2 = x^2 \implies x^2 - 4lx + 2l^2 = 0 \]

7. **Using the Quadratic Formula**:
- Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{4l \pm \sqrt{(-4l)^2 - 4 \cdot 1 \cdot 2l^2}}{2 \cdot 1} \]
- This simplifies to:
\[ x = \frac{4l \pm \sqrt{16l^2 - 8l^2}}{2} = \frac{4l \pm \sqrt{8l^2}}{2} = \frac{4l \pm 2\sqrt{2}l}{2} = 2l \pm \sqrt{2}l \]

8. **Finding the Valid Solution**:
- The two possible solutions are:
\[ x = (2 + \sqrt{2})l \quad \text{and} \quad x = (2 - \sqrt{2})l \]
- Since \( x \) must be less than \( l \) (it is between the two charges), we take:
\[ x = l \frac{2 - \sqrt{2}}{1} \]

9. **Nature of Equilibrium**:
- To determine the nature of the equilibrium, we analyze the forces around the equilibrium position.
- If \( Q \) is displaced slightly to the left, the force due to \( -2q \) will be greater than the force due to \( q \), pulling \( Q \) back to the equilibrium position (restoring force).
- Conversely, if \( Q \) is displaced to the right, the force due to \( q \) will be greater, also pulling \( Q \) back to equilibrium.
- Thus, the equilibrium is stable.

### Final Answer:
The positive test charge \( Q \) should be placed at a distance \( \frac{l(2 - \sqrt{2})}{1} \) from the charge \( -2q \) for it to be in equilibrium. The nature of this equilibrium is stable.