Skip to content

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?

Options

Choose one · Correct answer highlighted

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

0 comments

No comments yet. Be the first to start the discussion.