Two stationary particles each of +q are placed at a distacne apart. Now a negatively charged particle is placed in a straight line joining two charge. The direction of motion of the negatively charged particle will depend on

Options

A. The position at which it is situated
B. The magnitude of its cahrge
C. The magnitude of +q charge
D. The magnitude of the both the charges

Explanation

To solve the problem, we need to analyze the forces acting on the negatively charged particle when it is placed between two stationary positively charged particles. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two stationary particles, each with a charge of +q, placed at a distance R apart.
- A negatively charged particle (let's denote it as -q) is placed on the line joining these two positive charges.

2. **Identifying Forces**:
- The negatively charged particle will experience forces due to both positively charged particles.
- Let's denote the position of the negatively charged particle as point C, with point A being one positive charge and point B being the other positive charge.

3. **Distance from Charges**:
- Let the distance from charge A to charge C be R1, and the distance from charge B to charge C be R2.
- The total distance between A and B is R, so we have R1 + R2 = R.

4. **Applying Coulomb's Law**:
- The force exerted on the negatively charged particle by charge A (F1) can be calculated using Coulomb's Law:
\[ F_1 = k \frac{q \cdot q_1}{R_1^2} \]
- The force exerted on the negatively charged particle by charge B (F2) is:
\[ F_2 = k \frac{q \cdot q_1}{R_2^2} \]
- Here, \(k\) is Coulomb's constant, and \(q_1\) is the magnitude of the negatively charged particle.

5. **Direction of Forces**:
- Since both forces are attractive (because the charges are opposite), F1 will act towards charge A and F2 will act towards charge B.

6. **Net Force Calculation**:
- The net force on the negatively charged particle will depend on the relative magnitudes of F1 and F2.
- If R1 = R2 (i.e., the negatively charged particle is at the midpoint), then F1 = F2, and the net force will be zero.

7. **Effect of Position**:
- If the negatively charged particle is closer to charge A (i.e., R1 < R2), then F1 will be greater than F2, resulting in a net force towards charge A.
- Conversely, if the negatively charged particle is closer to charge B (i.e., R1 > R2), then F2 will be greater than F1, resulting in a net force towards charge B.

8. **Conclusion**:
- The direction of motion of the negatively charged particle depends on its position relative to the two positive charges.
- If it is at the midpoint, the net force is zero, and it will not move. If it is not at the midpoint, it will move towards the charge that is closer to it.