Coulomb's law and electric field calculator

Place point charges on the plane and get the force on each one by Coulomb's law, the electric field and potential at any point, with the field-line and equipotential map and the vector sum explained step by step.

What you can do

Net force by vector superposition.
Field and potential at a point.
Field lines and equipotentials.

Frequently asked questions

What does Coulomb's law say?

The force between two charges is F = k·q1·q2/r², with k = 8.99×10⁹ N·m²/C², along the line joining them.

Guide: Coulomb's law, electric field and potential

Two electric charges attract if their signs are opposite and repel if they are the same. Coulomb's law says by how much, and from it come the electric field, the potential and the energy of a set of charges. This guide reviews those four ideas and the superposition principle, and works an example.

Coulomb's law

F = k · |q₁ · q₂| / r², k = 1/(4πε₀) ≈ 8.988 × 10⁹ N·m²/C²

The force acts along the line joining the charges: attractive if the signs are opposite and repulsive if they are the same. It goes as the inverse square of the distance: at twice the distance, a quarter of the force.

Superposition

With several charges, the force on one of them is the vector sum of the forces each of the others exerts on it, computed separately. You add components (x, y, z), not magnitudes.

Field, potential and energy

Electric field at a point: the force per unit charge a positive test charge would feel, E = k·q/r² for a point charge, pointing away from positive charges and toward negative ones. It is measured in N/C (or V/m).
Potential at a point: V = Σ k·q_i/r_i. It is a scalar, so it is added with signs, without components. It is measured in volts.
Potential energy of the system: U = Σ k·q_i·q_j/r_ij, counting each pair once. It is negative if attractions dominate.

Worked example

q1 = +2 µC at the origin and q2 = −3 µC at x = 0.30 m.

Force between them: F = 8.988 × 10⁹ · (2 × 10⁻⁶ · 3 × 10⁻⁶) / 0.30² ≈ 0.599 N, attractive: on q1 it points toward +x, toward q2.
Field at the midpoint P (0.15 m, 0): the positive charge pushes away from itself (toward +x) with E₁ = k·2 × 10⁻⁶/0.15² ≈ 7.99 × 10⁵ N/C; the negative one pulls toward itself (also toward +x) with E₂ = k·3 × 10⁻⁶/0.15² ≈ 1.198 × 10⁶ N/C. Total: E ≈ 1.997 × 10⁶ N/C toward +x.
Potential at P: V = k·(2 − 3) × 10⁻⁶ / 0.15 ≈ −59.9 kV.
Energy of the system: U = k·(2 × 10⁻⁶)(−3 × 10⁻⁶)/0.30 ≈ −0.180 J. Negative, because the charges attract.

The tool gives the same values. Note that the field at P is not zero even though there is a charge on each side: since the signs are opposite, both fields point the same way.

How to use the tool

Place the charges on the canvas (in 2D or 3D) and type their values with prefixes (µ, n…). Choose the units of the coordinates.
Read the net force on each charge, the potential energy of the system and the step-by-step Coulomb's law with the sum of components.
Place a point P to see the field and the potential there, and turn on the field lines or the vector grid.

Common mistakes

Adding force magnitudes instead of vectors.
Forgetting to convert distances to meters and charges to coulombs (1 µC = 10⁻⁶ C).
Putting the charge's sign into the magnitude formula and then not knowing which way the force points: the direction is decided separately, by attraction or repulsion.
Treating the potential as a vector. It is a scalar: add it with signs.

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