Practice question
Question
Three charges \( +5 \, \mu\text{C}, -3 \, \mu\text{C}, +4 \, \mu\text{C} \) are at the vertices of an
equilateral triangle of side 1.5 m. What is the force magnitude on \( +5 \, \mu\text{C} \)?
Explanation
**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. F₁ = 9 × 10⁹ × (5 × 3 × 10⁻¹²/(1.5)²) = 0.06 N (attractive). F₂ = 9 × 10⁹ × (5 × 4 × 10⁻¹²/(1.5)²) = 0.08 N (repulsive). Angle 60°. Net F = √(F₁² + F₂² + 2 F₁ F₂ cos 60°) = √(0.06² + 0.08² + 0.0048) = 0.108 N . Substituting values gives 0.108 N, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.
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