The coulombs force between two point charges, If magnitude of each charge is doubled and distance between charges is halved, the force between them becomes:
Correct answer: D. 16F
- A. F
- B. 8F
- C. 4F
- D. 16F
Explanation
The force is F ∝ (q₁q₂)/r². The new force F' will have a factor of (2×2) in the numerator from doubling the charges and a factor of (1/2)² = 1/4 in the denominator from halving thedistance. This gives F' ∝ 4 / (1/4) = 16. The force becomes 16 times greater.
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