If `q_(1) q_(2)` is more significant than zero, then the nature of force is?

Options

A. Attractive
B. Repulsive
C. No Change
D. None of the above

Explanation

To solve the question, we need to analyze the nature of the force between two charges \( q_1 \) and \( q_2 \) when both charges are greater than zero.

### Step-by-Step Solution:

1. **Understanding the Charges**:
Given that \( q_1 q_2 > 0 \), this means that the product of the two charges is positive. This can occur in two scenarios:
- Both charges are positive: \( q_1 > 0 \) and \( q_2 > 0 \)
- Both charges are negative: \( q_1 < 0 \) and \( q_2 < 0 \)

2. **Applying Coulomb's Law**:
According to Coulomb's Law, the force \( F \) between two point charges is given by the formula:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k \) is Coulomb's constant and \( r \) is the distance between the charges.

3. **Determining the Nature of the Force**:
- If both charges are positive (\( q_1 > 0 \) and \( q_2 > 0 \)), they will repel each other because like charges repel.
- If both charges are negative (\( q_1 < 0 \) and \( q_2 < 0 \)), they will also repel each other for the same reason.

4. **Conclusion**:
In both scenarios where \( q_1 q_2 > 0 \), the force between the two charges is repulsive. Therefore, the nature of the force when \( q_1 q_2 > 0 \) is **repulsive**.

### Final Answer:
The nature of the force is **repulsive**.

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