A +3.00 μC point charge is fixed at x = 0 and a −1.00 μC charge at x = 0.400 m. At the midpoint x = 0.200 m, calculate the net electric field and electric potential. Find the external work required to bring a +2.00 nC test charge slowly from infinity to the midpoint. State the initial direction of its acceleration if released, and explain why zero potential would not necessarily imply zero field.

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E = 8.99 × 10⁵ N C⁻¹ to +x; V = 8.99 × 10⁴ V; external work = 1.80 × 10⁻⁴ J; a positive test charge accelerates toward +x. At the midpoint, the positive source repels a positive test charge to the right. The negative source attracts it to the right, so the two field magnitudes add rather than subtract. E = k(3.00 × 10⁻⁶ + 1.00 × 10⁻⁶)/(0.200)² = 8.99 × 10⁵ N C⁻¹ in the +x direction. Potential is scalar and includes charge signs: V = k[(3.00 × 10⁻⁶)/0.200 − (1.00 × 10⁻⁶)/0.200] = 8.99 × 10⁴ V. Slow transfer changes electric potential energy by qV, so the external work is (2.00 × 10⁻⁹)(8.99 × 10⁴) = 1.80 × 10⁻⁴ J. The released positive charge accelerates with E toward +x. Potential can cancel algebraically at a point while field vectors do not; field depends on the spatial gradient of potential, not simply its value.

Field, potential and work between two charges

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