Coulomb's Law

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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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1. **Coulomb's Law**: The electrostatic force ($F$) between two stationary point charges is:
   - $F = k \frac{|q_1 q_2|}{r^2}$.
2. **Identify Data**:
   - $q_1 = 3 	imes 10^{-6} 	ext{ C}$.
   - $q_2 = 5 	imes 10^{-6} 	ext{ C}$.
   - $r = 0.3$ m.
   - $k = 9 	imes 10^9 	ext{ N}\cdot	ext{m}^2/	ext{C}^2$.
3. **Calculation**:
   - $F = \frac{9 	imes 10^9 	imes (3 	imes 10^{-6}) 	imes (5 	imes 10^{-6})}{(0.3)^2}$
   - $F = \frac{135 	imes 10^{-3}}{0.09}$.
4. **Result**: $F = 1.5 	ext{ N}$. Since both are positive, this is a **repulsive** force.

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