Mid way between the two equal and similar charges , we place the third equal and similar charge. Which of the following statements is correct , concerned to the equilibrium along , the line joining the charges ?

Options

A. The third charge experienced a net force inclined to the line joining the charges
B. The third charge is in stable equilibrium
C. The third charge is in unstable equilibrium
D. The third charge experiences a net force perpendicular to the line joining the charges

Explanation

To solve the problem, we need to analyze the situation involving three equal and similar charges placed along a straight line. Let's denote the charges as \( +Q \), \( +Q \), and \( +Q \) for the first, second, and third charges respectively. The first two charges are placed at positions \( A \) and \( B \), and the third charge is placed at position \( C \), which is midway between \( A \) and \( B \).

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- Place two equal charges \( +Q \) at points \( A \) and \( B \).
- The distance between \( A \) and \( B \) is \( r \).
- The third charge \( +Q \) is placed at point \( C \), which is at a distance \( \frac{r}{2} \) from both \( A \) and \( B \).

2. **Calculating Forces on the Third Charge**:
- According to Coulomb's law, the force between two point charges is given by:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
- The force \( F_{AC} \) exerted on charge \( C \) by charge \( A \) is:
\[ F_{AC} = k \frac{Q \cdot Q}{\left(\frac{r}{2}\right)^2} = k \frac{Q^2}{\frac{r^2}{4}} = \frac{4kQ^2}{r^2} \]
- The direction of this force is away from charge \( A \) (to the right).

- Similarly, the force \( F_{BC} \) exerted on charge \( C \) by charge \( B \) is:
\[ F_{BC} = k \frac{Q \cdot Q}{\left(\frac{r}{2}\right)^2} = \frac{4kQ^2}{r^2} \]
- The direction of this force is away from charge \( B \) (to the left).

3. **Net Force on the Third Charge**:
- Since both forces \( F_{AC} \) and \( F_{BC} \) are equal in magnitude but opposite in direction, the net force \( F_{net} \) on charge \( C \) is:
\[ F_{net} = F_{AC} - F_{BC} = \frac{4kQ^2}{r^2} - \frac{4kQ^2}{r^2} = 0 \]
- Therefore, the net force acting on the third charge \( C \) is zero.

4. **Equilibrium Analysis**:
- Since the net force on charge \( C \) is zero, it is in equilibrium.
- To determine whether this equilibrium is stable or unstable, we consider small displacements from the midpoint \( C \):
- If charge \( C \) is slightly displaced towards \( A \) or \( B \), it will experience a net force pushing it further away from the midpoint due to the repulsive forces from the other two charges.
- Thus, the equilibrium is unstable.

5. **Conclusion**:
- The correct statement regarding the equilibrium of the third charge is that it is in unstable equilibrium.

### Final Answer:
The correct statement is: The third charge is in unstable equilibrium.