Superposing Fields & Forces in Two Dimensions

You’ll be able to

Add electric field contributions from several charges as vectors
Locate the point where the net field from two charges is zero
Distinguish the vector addition of fields from the scalar addition of potentials

Fields add as vectors, potentials add as numbers

This is the single most useful distinction in the unit. Electric field is a vector, so contributions from several charges must be added by components and recombined. Electric potential is a scalar, so contributions add as signed numbers with no direction and no components. A configuration can therefore have zero field with nonzero potential, or zero potential with nonzero field, and the exam asks about both.

Superposition

E_net,x = Σ E_i cos θ_i and E_net,y = Σ E_i sin θ_i, then |E| = √(E_x² + E_y²) · V_net = Σ kq_i/r_i (signed, no components)

Use the magnitude kq/r² for each field contribution and let the geometry set the direction. Keep the sign of q for potential.

Where the net field is zero

For two like charges, the zero-field point lies between them, closer to the smaller charge. For two unlike charges, it lies outside the pair, beyond the smaller charge — never between them, because between two opposite charges both fields point the same way and cannot cancel. Recognizing which case you have before doing algebra saves solving for a root that does not exist.

The equilateral and right-angle setups

Exam geometries are almost always chosen so the components are clean. Two equal charges at the base of an equilateral triangle give a net field at the apex directed along the perpendicular bisector, with the horizontal components canceling. Charges at the corners of a square give diagonal contributions at 45°, so each contributes 1/√2 of its magnitude to each axis. Recognizing the geometry is usually faster than grinding through general trigonometry.

Worked example

Two charges of +4.0 nC and −4.0 nC sit 0.20 m apart. Find the magnitude of the net electric field at the midpoint, and the potential there.

1.Each charge is 0.10 m from the midpoint. E from each = kq/r² = (8.99 × 10⁹)(4.0 × 10⁻⁹)/(0.10)² = 3596 N/C.
2.The field from the positive charge points away from it; the field from the negative charge points toward it. Both point the same way at the midpoint.
3.So they ADD: E_net = 2 × 3596 ≈ 7.2 × 10³ N/C, directed from the positive toward the negative charge.
4.Potential: V = kq/r for each, with signs. V = (8.99 × 10⁹)(+4.0 × 10⁻⁹)/0.10 + (8.99 × 10⁹)(−4.0 × 10⁻⁹)/0.10.
5.The two terms cancel exactly: V = 0.

Answer: E ≈ 7.2 × 10³ N/C but V = 0 at the midpoint. This is the standard demonstration that zero potential does not mean zero field — the scalars cancel while the vectors reinforce.

Watch out

Do not put the sign of the charge into the field magnitude formula. Use kq/r² with the absolute value of q, then let the geometry decide the direction — away from positive, toward negative. Signs in the magnitude produce component errors that are hard to spot.

Checkpoint

At the midpoint between equal and opposite point charges:

Checkpoint

Two positive charges of different magnitude are separated by a distance d. The point where the net field is zero lies:

Checkpoint

Three equal positive charges sit at the corners of an equilateral triangle. The net electric field at the center is:

On the exam

Sketch the field vectors at the point of interest before computing anything. Whether they reinforce or oppose is usually visible immediately, and it tells you whether to add or subtract magnitudes.

Answer the 3 checkpoints as you read.