Practice question

**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. Midpoint distance = 40 cm = 0.4 m. E₁ = 9 × 10⁹ × (11 × 10⁻⁶/(0.4)²) = 6.1875 × 10⁵ N/C (towards -5 μC ). E₂ = 9 × 10⁹ × (5 × 10⁻⁶/(0.4)²) = 2.8125 × 10⁵ N/C (towards -5 μC ). Net E = 6.1875 × 10⁵ + 2.8125 × 10⁵ = 9 × 10⁵ N/C . Substituting values gives 9.0 × 10⁵ N/C, which matches expected magnitude for this electrostatic configuration,