Practice question
Question
Three charges \( +5 \, \mu\text{C}, -5 \, \mu\text{C}, +2 \, \mu\text{C} \) are at the vertices of an
equilateral triangle of side 1.5 m. What is the force magnitude on \( +5 \, \mu\text{C} \)?
Explanation
**Vector addition of forces** underlies multi-charge analysis. Each pair contributes independent Coulomb force, resultant obtained by resolving components along axes. Equilibrium occurs when vector sum vanishes, often at symmetric points where contributions balance. F₁ = 9 × 10⁹ × (5 × 5 × 10⁻¹²/(1.5)²) = 0.1 N (attractive). F₂ = 9 × 10⁹ × (5 × 2 × 10⁻¹²/(1.5)²) = 0.04 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.1² + 0.04² + 0.004) = 0.129 N . Substituting values gives 0.129 N, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.
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