Worked example: Two charges, q1 = +3 nC at x = −0.1 m and q2 = −2 nC at x = +0.1 m, lie on the x-axis. Find V at the origin and at x = +0.3 m. Step 1: At the origin, r1 = r2 = 0.1 m. V = kq1/r1 + kq2/r2 = (8.988 × 10⁹)(3 × 10⁻⁹)/0.1 + (8.988 × 10⁹)(−2 × 10⁻⁹)/0.1 = 269.6 − 179.8 = +89.8 V ≈ +90 V. Just signed addition — no vectors. Step 2: At x = 0.3 m: r1 = 0.4 m, r2 = 0.2 m. V = (8.988 × 10⁹)(3 × 10⁻⁹)/0.4 + (8.988 × 10⁹)(−2 × 10⁻⁹)/0.2 = 67.4 − 89.9 = −22.5 V. Step 3 (sanity): Near the negative charge the potential is negative; between them it crosses zero somewhere — find it: k(3)/(.1+d)... setting V = 0 on the axis gives 3/(0.1 + d)... solving 3/(x+0.1) = 2/(x−0.1) for x > 0.1 gives x = 0.5 m. A zero of potential, found with pure algebra.