Coulomb's Law (numerical)

Question

Two charges 2×10⁻⁶ C and 3×10⁻⁶ C are 0.20 m apart. Force magnitude? (k = 9×10⁺⁹ N·m²/C²)

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1. **Identify Given Data**:
   - $q_1 = 2 	imes 10^{-6} 	ext{ C}$
   - $q_2 = 3 	imes 10^{-6} 	ext{ C}$
   - $r = 0.20 	ext{ m}$
   - $k = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$
2. **Substitute in Coulomb's Law Formula**:
   - $F = k \frac{q_1 q_2}{r^2}$
   - $F = 9 	imes 10^9 	imes \frac{(2 	imes 10^{-6}) 	imes (3 	imes 10^{-6})}{(0.2)^2}$
3. **Calculation**:
   - $F = 9 	imes 10^9 	imes \frac{6 	imes 10^{-12}}{0.04}$
   - $F = \frac{54 	imes 10^{-3}}{0.04} = \frac{0.054}{0.04}$
   - $F = 1.35$ N.
4. **Result**: The magnitude of the force is **1.35 N**.

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Coulomb's Law (numerical)

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