Practice question
Question
Two charges \( +8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are 40 cm apart. What is the electric
field magnitude at the midpoint?
Explanation
**Coulomb's law** gives force between point charges as F = k·|q₁q₂|/r², k = 1/(4π ε₀) = 9×10⁹ N·m²/C², directed along line joining charges. Like charges repel, opposite attract, magnitude scales with product of charges and inverse square of separation r². Midpoint distance = 20 cm = 0.2 m. E₁ = 9 × 10⁹ × (8 × 10⁻⁶/(0.2)²) = 1.8 × 10⁶ N/C (towards -4 μC ). E₂ = 9 × 10⁹ × (4 × 10⁻⁶/(0.2)²) = 9 × 10⁵ N/C (towards -4 μC ). Net E = 1.8 × 10⁶ + 9 × 10⁵ = 2.7 × 10⁶ N/C . Substituting values gives 2.7 ×
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.