Practice question
Question
A point charge \( -10 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a
point 10 m along the x-axis?
Explanation
**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 10 × 10⁻⁶ C , r = 10 m . E = 9 × 10⁹ × (10 × 10⁻⁶/(10)²) = 9 × 10⁹ × (10 × 10⁻⁶/100) = 9 × 10² N/C . Substituting values gives 900 N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.