Electromagnetism · Electrostatics

Electric Field Calculator and Simulator for Point Charges (Coulomb's Law)

Place up to eight point charges, drag them around, and watch the electric field lines and equipotential lines update live. Move the probe to read the electric field vector and potential at any point, see the net Coulomb force on every charge, or use the quick Coulomb's law calculator for two charges in any medium.

Live field linesEquipotentialsE and V at any pointForces on each chargeCODATA 2022 ε₀

By Bimal Ghimire · Published October 30, 2026 · Results verified against an independent superposition calculation.

Key facts

Coulomb's law: F = k q₁q₂/r², with k = 1/(4πε₀) = 8.9875517862 × 10⁹ N·m²/C².
Field of a point charge: E = kq/r² (V/m = N/C), radially outward for q > 0 and inward for q < 0.
Potential of a point charge: V = kq/r (volts); the potential of several charges is the algebraic sum.
Superposition: fields add as vectors, potentials add as scalars; E = −∇V.
Field lines leave positive charges, end on negative charges, never cross, and meet equipotentials at right angles.
Two 1 μC charges 1 m apart repel with about 9 mN; air breaks down near 3 MV/m.

Coulomb's Law Calculator (Two Charges)

Charge q₁

Charge q₂

Separation r

Medium (εr)

Force magnitude

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Electric Field Simulator

Drag charges or the ◎ probe · double-click empty space to add a charge

Field lines Equipotentials Field arrows Potential colour mapView: 10 × 10 cm

Charges (position in cm, charge in nC)

x (cm)y (cm)q (nC)

Theory

Coulomb's Law

In 1785 Charles-Augustin de Coulomb used a torsion balance to show that the force between two small charged spheres varies as the inverse square of their separation. In vector form:

$$\mathbf F_{12}=\frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r^2}\,\hat{\mathbf r}_{12},\qquad k=\frac{1}{4\pi\varepsilon_0}=8.9875517862\times10^{9}\ \text{N·m}^2/\text{C}^2$$

$\hat{\mathbf r}_{12}$ points from charge 2 to charge 1. ε₀ = 8.8541878188 × 10⁻¹² F/m (CODATA 2022). Since 2019, ε₀ and μ₀ are measured quantities linked by $\varepsilon_0\mu_0c^2=1$, with c exact.

The force obeys Newton's third law (equal and opposite) and is enormously stronger than gravity: between two protons the electric repulsion exceeds the gravitational attraction by a factor of about 10³⁶. The law applies to point charges or spherically symmetric charge distributions; for extended conductors, induced charge redistribution changes the result.

Theory

Electric Field and the Superposition Principle

The electric field is the force per unit positive test charge, $\mathbf E=\mathbf F/q_0$. For a set of point charges, the fields simply add as vectors:

$$\mathbf E(\mathbf r)=\sum_i\frac{q_i}{4\pi\varepsilon_0}\,\frac{\mathbf r-\mathbf r_i}{|\mathbf r-\mathbf r_i|^3}$$

This superposition principle holds because Maxwell's equations are linear in vacuum. The simulator evaluates exactly this sum at every point it draws; the charges are three-dimensional point charges viewed in a plane that contains them, so the 1/r² law applies (not the 1/r law of infinite line charges).

Theory

Electric Potential and Equipotential Lines

$$V(\mathbf r)=\sum_i\frac{q_i}{4\pi\varepsilon_0\,|\mathbf r-\mathbf r_i|},\qquad \mathbf E=-\nabla V,\qquad U=\sum_{i<j}\frac{q_iq_j}{4\pi\varepsilon_0\,r_{ij}}$$

Potential is a scalar measured in volts (J/C), taken as zero infinitely far away. Equipotential lines join points of equal V; no work is needed to move a charge along them, so the field is always perpendicular to them. Closely spaced equipotentials mean a strong field. The total potential energy U of the arrangement is the work needed to assemble it from infinity; the simulator reports it below the force table.

Visualising

Rules for Drawing Electric Field Lines

Start and end

Lines start on positive charges and end on negative charges, or extend to infinity when the charges do not balance.

Number ∝ charge

The number of lines leaving or entering a charge is proportional to |q| (the simulator uses this rule).

Density ∝ |E|

Lines crowd where the field is strong and spread out where it is weak.

Never cross

The field has one direction at each point; lines meet only at charges or at null points where E = 0.

Practice

Worked Examples

1. Force between 1 μC and 2 μC at 10 cm (vacuum)

$F=8.98755\times10^9\times(1\times10^{-6})(2\times10^{-6})/0.10^2=$ 1.80 N, repulsive. In water (εr ≈ 80) it drops to about 0.022 N.

2. Field of a 5 nC charge at 3 cm

$E=kq/r^2=8.98755\times10^9\times5\times10^{-9}/0.03^2=$ 4.99 × 10⁴ V/m, pointing away from the charge. The potential there is $V=kq/r=1498$ V.

3. Dipole: ±2 nC separated by 4 cm, field at the midpoint

Both fields point from + toward −, so they add: $E=2kq/(d/2)^2=2\times8.98755\times10^9\times2\times10^{-9}/0.02^2=$ 8.99 × 10⁴ V/m, while V = 0 there. Load the Dipole preset and drag the probe to the centre to check.

Keep Learning

Related Tools and Guides

FAQ

Frequently Asked Questions

What is Coulomb's law?

Coulomb's law states that the electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them: F = k q₁q₂ / r², with k = 1/(4πε₀) ≈ 8.9876 × 10⁹ N·m²/C². Like charges repel and unlike charges attract, and the force acts along the line joining them.

How do you calculate the electric field of a point charge?

E = k q / r², directed away from a positive charge and toward a negative charge. For several charges, add the field vectors of each charge (superposition). For example, 5 nC produces about 4.99 × 10⁴ V/m at 3 cm.

What is the difference between electric field and electric potential?

The electric field E is a vector (V/m or N/C) giving the force per unit charge. The electric potential V is a scalar (volts) giving the potential energy per unit charge. They are related by E = −∇V: the field points downhill in potential and is perpendicular to equipotential lines.

Why do electric field lines never cross?

At any point the electric field has a single direction, so only one field line can pass through it. Field lines start on positive charges, end on negative charges (or at infinity), and their density is proportional to the field strength.

What is the value of the Coulomb constant k?

k = 1/(4πε₀) = 8.9875517862 × 10⁹ N·m²/C², using the CODATA 2022 value ε₀ = 8.8541878188 × 10⁻¹² F/m. The commonly used approximation 9 × 10⁹ is within 0.14%.

How does a medium change the Coulomb force?

In a linear, homogeneous dielectric that surrounds both charges, the force is reduced by the relative permittivity εr: F = k q₁q₂ / (εr r²). Air has εr ≈ 1.0006 and water about 80 at room temperature, which is why water dissolves ionic salts so well.

What is the electric field at the midpoint of a dipole?

For charges +q and −q separated by d, the fields at the midpoint add because both point from + to −: E = 2kq/(d/2)² = 8kq/d². The potential there is zero. For ±2 nC and 4 cm this gives about 8.99 × 10⁴ V/m.

Sources

References

NIST (2024). CODATA Recommended Values of the Fundamental Physical Constants: 2022 (vacuum electric permittivity ε₀). physics.nist.gov/cuu/Constants

Griffiths, D. J. (2017). Introduction to Electrodynamics (4th ed.), Chapter 2: Electrostatics. Cambridge University Press.

Ling, S. J., Sanny, J., & Moebs, W. (2016). University Physics Volume 2, Chapters 5 (Electric Charges and Fields) and 7 (Electric Potential). OpenStax. openstax.org

Purcell, E. M., & Morin, D. J. (2013). Electricity and Magnetism (3rd ed.). Cambridge University Press.

Cite this tool: Ghimire, B. (2026). Electric Field Calculator: Point Charges & Coulomb's Law. Engineers Views. https://engineersviews.com/tools/electric-field-calculator-point-charges/