Two charges \( +9 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 70 cm apart. What is the electric field magnitude at
**Electric field** defined as E = F/q₀, force per unit positive test charge, unit N/C or V/m, direction along force on positive test charge. For point charge, E = k q/r² radially outward for q>0. Field lines start on positive and end on negative, density indicates strength. Midpoint distance = 35 cm = 0.35 m. E₁ = 9 × 10⁹ × (9 × 10⁻⁶/(0.35)²) = 6.61 × 10⁵ N/C (towards -3 μC ). E₂ = 9 × 10⁹ × (3 × 10⁻⁶/(0.35)²) = 2.2 × 10⁵ N/C (towards -3 μC ). Net E = 6.61 × 10⁵ + 2.2 × 10⁵ = 8.81 × 10⁵
Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines