How can I find the net electric field from two idealized point charges?

First mark the observation point—the location where you want the field—and choose a positive direction, such as right along a line. For each charge, measure its distance to that point. Its field contribution is proportional in magnitude to |q|/r²; its direction is away from a positive charge and toward a negative charge. Give each contribution a positive or negative sign based on your chosen axis, then add them. This is a vector sum; on one straight line, signed numbers handle the direction. [S1]

Keep the source-to-point distance separate for each charge. Don’t add the charges first: their fields can point in opposite directions, and the field depends on where the observation point is.

Draw a small direction sketch for each charge at the observation point. A positive source charge’s field points away from it; a negative source charge’s field points toward it. Compare each arrow with your chosen positive axis: an arrow along that axis gets a plus sign, and one against it gets a minus sign. [S1]

For example, if right is positive and both arrows point right, both contributions are positive—even if one source charge is negative. The source charge’s sign determines the arrow direction; the arrow’s direction relative to your axis determines the contribution’s sign.

Assume a one-dimensional setup: q₁ = +2.0 μC at x = 0 m, q₂ = −3.0 μC at x = 0.50 m, and the observation point P is at x = 0.20 m. Choose right as positive. These are illustrative values, not measurement results.

1. Distances to P: r₁ = 0.20 m and r₂ = 0.30 m.
2. Directions: q₁ is positive and P is to its right, so its field points right. q₂ is negative and P is to its left, so its field points toward q₂, also right. Both contributions are positive. [S1]
3. Compare their magnitudes using the proportional factor |q|/r²:

For q₁: (2.0 μC)/(0.20 m)² = 50 μC/m². For q₂: (3.0 μC)/(0.30 m)² ≈ 33.3 μC/m².

1. The signed factors add to about 83.3 μC/m², in the rightward direction. This is a comparison factor, not a complete electric-field value: the field is proportional to it, and the proportionality constant is not included here. [S1]

This shows how the distances, units, and directions enter the calculation without assigning an unsupported numerical field strength.

Check each distance from the observation point to a charge, rather than the distance between the charges. Confirm that your direction sketch uses the same positive axis throughout, and that you squared each distance in the denominator. Then check the signed sum: contributions in the same direction reinforce; contributions in opposite directions subtract through their signs. [S1]

A negative result does not mean the field has a negative strength; it means the net direction is opposite your chosen positive axis. This one-dimensional shortcut applies when the contributions lie along the same line. For fields pointing in different directions, use vector components rather than adding their magnitudes as signed one-dimensional values.