A negatively charged particle is situated on a striaght line joining two other charge particle stationary of motion of the negatively charged particle will depend on

Options

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

Answer

To solve the problem of determining the conditions under which a negatively charged particle remains stationary between two other charged particles, we can break down the solution into the following steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have three charged particles: two stationary charges (let's call them \( Q_1 \) and \( Q_2 \)) and one negatively charged particle (let's call it \( -q \)). The negatively charged particle is positioned on the straight line joining \( Q_1 \) and \( Q_2 \).

2. **Electric Field Concept**:
The negatively charged particle will experience forces due to the electric fields created by \( Q_1 \) and \( Q_2 \). The direction of the electric field produced by a positive charge is away from the charge, while for a negative charge, it is towards the charge.

3. **Condition for Equilibrium**:
For the negatively charged particle \( -q \) to be stationary, the net force acting on it must be zero. This occurs when the electric field due to \( Q_1 \) at the position of \( -q \) is equal in magnitude and opposite in direction to the electric field due to \( Q_2 \).

4. **Setting Up the Equation**:
Let’s denote the distances from \( -q \) to \( Q_1 \) as \( r_1 \) and to \( Q_2 \) as \( r_2 \). The condition for equilibrium can be expressed as:
\[ E_{Q_1} = E_{Q_2} \]
where \( E_{Q_1} \) and \( E_{Q_2} \) are the magnitudes of the electric fields due to \( Q_1 \) and \( Q_2 \) respectively.

5. **Electric Field Formula**:
The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \( k \) is Coulomb's constant.

6. **Equating the Electric Fields**:
Substituting the electric field formulas into the equilibrium condition gives:
\[ \frac{k \cdot |Q_1|}{r_1^2} = \frac{k \cdot |Q_2|}{r_2^2} \]
The \( k \) cancels out, leading to:
\[ \frac{|Q_1|}{r_1^2} = \frac{|Q_2|}{r_2^2} \]

7. **Position of the Negatively Charged Particle**:
This equation shows that the position of the negatively charged particle \( -q \) depends on the magnitudes of \( Q_1 \) and \( Q_2 \) and their respective distances \( r_1 \) and \( r_2 \). Thus, the position at which \( -q \) can be stationary is determined by the ratio of the charges and the distances.

8. **Conclusion**:
Therefore, the stationary position of the negatively charged particle depends on both the magnitudes of the charges \( Q_1 \) and \( Q_2 \) and the position at which it is situated between them.

### Final Answer:
The stationary position of the negatively charged particle depends on both the magnitudes of the charges and its position.