Worked example: A point charge q = +4 μC sits at the origin. (a) Find the electric field vector at point P = (0.2 m, 0). (b) What force would a −2 μC charge feel if placed at P? Step 1: Field of a point charge: E = k·q/r²·r̂. Here r = 0.2 m, r̂ = +i. Step 2: E = (8.988 × 10⁹)(4 × 10⁻⁶)/(0.2)²·i = (35.95/0.04)·i = 8.99 × 10⁵·i N/C. So E ≈ (9.0 × 10⁵ N/C) in the +x direction — pointing away from the positive source, as expected. Step 3: Force on q2 = −2 μC at P: F = q2·E = (−2 × 10⁻⁶)(8.99 × 10⁵·i) = −1.80·i N. The negative sign flips the direction: the negative charge is pulled toward the positive source with 1.80 N. Notice the two-step method: the field (step 2) depends only on the source; the force (step 3) brings in the second charge.