Coulomb's Law
Two point charges of +3 μC and +5 μC are separated by a distance of 0.3 m in air. What is the magnitude of the force between them? (k = 9 × 10⁹ N·m²/C²)
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- coulomb's law
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
1.5 N
- Coulomb's Law: The electrostatic force (F) between two stationary point charges is:
- F=kr2∣q1q2∣.
- Identify Data:
- q1=3×10−6 C.
- q2=5×10−6 C.
- r=0.3 m.
- k=9×109 N⋅m2/C2.
- Calculation:
- F=(0.3)29×109×(3×10−6)×(5×10−6)
- F=0.09135×10−3.
- Result: F=1.5 N. Since both are positive, this is a repulsive force.
This easy difficulty physics question is from the chapter electrostatics, covering the topic of coulomb's law. It appeared in the 2025 exam.
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