Two positive point charges are fixed on the x axis: +Q = +1.800 × 10^-9 C at x = 0 and +4Q = +7.200 × 10^-9 C at x = 0.38 m. Use k = 8.99 × 10⁹ N m² C⁻² and neglect every other field. Find the finite point on the line where the net electric field is zero, calculate the magnitude of each individual field there, and explain why no zero-field point lies outside the two charges.

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Zero net field at x = 0.1267 m from +Q, between the charges; each individual field has magnitude 1009 N C⁻¹. Each positive charge produces a field pointing away from itself. Between the charges, the field from +Q points right and that from +4Q points left, so cancellation is possible there. Let x be the distance from +Q with 0 < x < d. Equal magnitudes require kQ/x² = 4kQ/(d − x)². Both distances are positive, giving d − x = 2x. Thus x = d/3 = 0.1267 m. The point is closer to the weaker charge, which compensates for that charge's smaller source strength through the inverse-square dependence. Each field magnitude is kQ/x² = 1009 N C⁻¹. Their equal magnitudes and opposite directions produce zero resultant field, not an absence of either individual field. To the left of both positive charges, both fields point left; to their right, both point right. Such same-direction contributions cannot cancel at a finite exterior point. This directional check prevents accepting extraneous algebraic roots.

Locate a zero-field point between charges · set 4

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