Worked example: A uniform field E = 200 N/C points in the +x direction. Point A is at x = 0, point B at x = 0.5 m. (a) Find V_B − V_A. (b) How much work must an external agent do to move q = +3 μC from A to B slowly? Step 1: ΔV = −∫_A^B E·dl = −E·Δx = −(200)(0.5) = −100 V. Moving with the field, potential drops. Step 2: External work W_ext = q·ΔV = (3 × 10⁻⁶)(−100) = −3 × 10⁻⁴ J. The negative sign means the field itself does +3 × 10⁻⁴ J of work — the agent actually holds the charge back rather than pushing it. Step 3 (check): Potential energy change ΔU = qΔV = −3 × 10⁻⁴ J; the charge loses potential energy moving down the potential hill, as expected for a positive charge moving with the field.