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Coulomb's Law Calculator: Force, Electric Field, Potential Energy

Calculate electric forces, fields, and potential energy between charged particles with vector analysis and superposition.

Last Updated: February 2026

Charles-Augustin de Coulomb published the inverse-square force law for static electric charges in 1785, measuring it with a torsion balance he built himself. Newton had argued by analogy that the gravitational form should also describe electricity, but Coulomb's experiment is what made the relationship quantitative. The modern statement: F = k q₁q₂ / r², where k = 8.99 × 10⁹ N·m²/C² and the force points along the line between the charges (repulsive for like signs, attractive for opposite). Coulomb's constant relates to vacuum permittivity through k = 1/(4πε₀), the form Maxwell would later adopt for the SI version. The 1/r² scaling is exact in vacuum, tested to better than 10⁻¹⁶ deviation by modern null-result experiments.

Solve-For Summary

UnknownRearranged FormulaRequired Inputs
FF = k|q₁q₂|/r²q₁, q₂, r
q₁ or q₂q = Fr²/(k|q_other|)F, r, other charge
rr = √(k|q₁q₂|/F)F, q₁, q₂
E (field)E = kq/r²q, r

Defining Polarity and Current Direction Before You Solve

For Coulomb's law, polarity isn't a current convention. It's the sign attached to each charge and the direction the resulting force points. F = k|q₁q₂|/r² gives a magnitude. The sign of the product q₁q₂ tells you whether the force on each particle pushes the two apart or pulls them together. Same signs repel. Opposite signs attract. That's the only physics in the sign rule, and you can't skip it.

Pick a coordinate system before you start adding anything. Put one charge at the origin, the second along +x, and decide which way is positive. Now the force on charge 2 from charge 1 either points +x (repulsion if both signs match) or −x (attraction if signs differ). Write the unit vector r̂₁₂ pointing from 1 to 2. The vector form F⃗_(on 2) = k(q₁q₂/r²) r̂₁₂ carries the sign automatically: if q₁q₂ > 0, F⃗ aligns with r̂; if q₁q₂ < 0, F⃗ flips.

The same logic applies to the electric field from each source charge. E⃗ from a positive charge points radially outward. From a negative charge, radially inward. The force on a test charge is F⃗ = qE⃗, so if q is negative, F⃗ flips relative to E⃗. Students lose points for mixing the field direction with the force direction. They're only the same when the test charge is positive.

Sign convention tip: Some textbooks define force as a signed quantity along the line between charges (positive = repulsive, negative = attractive). Others use magnitude plus direction separately. Check which convention your course uses before submitting answers. Free-body diagrams should show force arrows, not field arrows, unless the problem asks for the field.

Vector Superposition: Adding Forces from Multiple Charges

Coulomb's law was written for two charges. Three or more, and you need superposition: compute each pairwise force separately, then add them as vectors. The principle works because Maxwell's equations are linear. The field from charge A doesn't care that charge B is also present. They're independent contributions you sum afterward.

For charges in a line, vector addition collapses to signed scalar arithmetic along that axis. If charge X sits between Y and Z, the force on X from Y points one way and the force from Z points the other, so you subtract magnitudes. Same direction? Add. Opposite directions? Subtract. The trap is forgetting that the test charge's position matters: move X to the outside of the line and now both forces from Y and Z push the same way.

For charges in a plane, decompose. Compute each F⃗_i magnitude with k|q_i q_test|/r_i². Find the angle θ_i each force makes with the +x-axis using arctan(Δy/Δx). Then F_x = F cos θ and F_y = F sin θ. Sum component-wise: ΣF_x and ΣF_y. The net magnitude is √(ΣF_x² + ΣF_y²) and the direction is arctan(ΣF_y/ΣF_x), with quadrant adjustment.

Watch out: When three charges form a triangle, the angles between force vectors aren't 90°. You'll need cos θ and sin θ to resolve components correctly. Sketch the geometry first, write each angle relative to a single fixed axis, and check that mirror-image symmetry (if any) zeroes one of the component sums.

Time Constants and Steady-State: What the Circuit Looks Like at t = 0 vs. t = ∞

Electrostatics is the t = ∞ limit. Charges that started moving have either come to rest or settled into steady drift. Coulomb's law assumes that ideal: each charge is fixed, the geometry isn't evolving, and you're reading the force right now without worrying about how the system got there. That's why the textbook calls it "static."

The contrast with the t = 0 case shows up the moment any charge is free to move. Release a positive test charge near another positive charge and it accelerates outward. The force is largest at t = 0 (smallest separation) and decays as the charge flies off. The integral of that decaying force is the kinetic energy gained, which equals the drop in potential energy kq₁q₂(1/r₀ − 1/r_∞). Pin the charges down and that exchange never happens.

Real conductors reach static equilibrium fast. Drop net charge on a copper sphere and the redistribution settles in roughly 10⁻¹⁹ seconds, set by the Drude relaxation time. After that, the field inside the conductor is zero and all excess charge sits on the surface. That's why Coulomb's law experiments use insulated charge carriers: a metal rod with charge on it equilibrates faster than you can measure, so you can't isolate a point charge inside a metal.

For a calculation, ask whether the geometry is fixed during the time you care about. Charge sitting on a charged balloon you're holding still? Static, Coulomb's law applies. Charge on a moving sphere being pulled toward another? Use Coulomb's law instantaneously and then update r as the motion proceeds. The force is always F = k|q₁q₂|/r², but r is no longer constant.

Real Components and Tolerances (the ±10% Resistor Problem)

For a circuit, the "±10% resistor problem" is about how a tolerance on one component propagates to the answer. For Coulomb's law it's the same idea applied to charge, distance, and the constant 1/(4πε₀). Each input has measurement uncertainty, and they all show up in F.

Distance dominates because it's squared. A 5% error in r becomes a 10% error in F. Measuring r between two charged spheres with a ruler good to ±1 mm at a 10 cm separation gives you a 1% distance uncertainty, so 2% in F. Drop to a 1 cm separation with the same ruler and the uncertainty climbs to 20% in F. That's why Coulomb's original torsion balance ran at large separations: he traded force magnitude for distance precision.

Charge is harder. Modern undergrad labs use electron-pair calibration: a Faraday cup or an integrating electrometer counts elementary charges directly. You can hit ±0.1% on q if the cup is shielded. Without that, you're guessing from contact potential or capacitance, both of which leak. The constant k = 8.9875517923 × 10⁹ N·m²/C² is now defined exactly through ε₀ and the redefined elementary charge, so it contributes no uncertainty.

Propagation rule: For F = k q₁ q₂ / r², relative uncertainty satisfies (δF/F)² ≈ (δq₁/q₁)² + (δq₂/q₂)² + (2 δr/r)². The factor of 2 on the distance term is what makes r the limiting input for most setups. Cut your distance uncertainty in half and the force uncertainty drops by a factor of four.

Static vs. Dynamic Cases: When Charges Move

Coulomb's law strictly applies when the charges are at rest. Once they move, the field carries information at the speed of light, and a moving charge produces magnetic field too. For slow motion (v ≪ c), Coulomb's law is a good approximation: just use the instantaneous positions and update them as the system evolves. That's how you'd simulate a hydrogen atom non-relativistically, or two ions in a Penning trap below the cyclotron frequency.

For relativistic motion you can't. The retarded potential matters: the force on a charge depends on where the source was a time r/c ago, not where it is now. Synchrotron radiation, the Liénard-Wiechert potentials, and the radiation-reaction problem all live here. Griffiths covers this in Chapter 11 of Introduction to Electrodynamics, and it's the cleanest place to see why the simple kq₁q₂/r² form is a non-relativistic limit.

Continuous distributions are another "dynamic" case in a different sense. F = kq₁q₂/r² assumes point charges. For a charged rod, disk, or sphere, you split the distribution into infinitesimal dq elements, apply Coulomb's law to each, and integrate. A uniformly charged sphere happens to look exactly like a point charge at its center for any external observer. Inside the sphere, the field grows linearly with r (Gauss's law result), nothing like 1/r².

The practical upshot: if your charges are tiny and slow compared to the apparatus, Coulomb's law is exact. If they're large, distributed, or fast, you're using it as a building block inside a larger calculation, not as the final answer.

Worked Example: Net Force on a Charge in an Equilateral Triangle

Three point charges sit at the corners of an equilateral triangle with side 5 cm. q₁ = +2 nC at the top vertex. q₂ = −3 nC at the bottom-left vertex. q₃ = +4 nC at the bottom-right vertex. Find the net Coulomb force on q₃.

Geometry and coordinates

Place q₂ at the origin and q₃ at (0.05, 0). q₁ sits directly above the midpoint of q₂q₃ at (0.025, 0.0433), since the triangle's height is 0.05 × √3/2.

Step 1: Force on q₃ from q₂

r₂₃ = 0.05 m. q₂q₃ = (−3 × 10⁻⁹)(4 × 10⁻⁹) = −12 × 10⁻¹⁸. Negative product means attraction.

|F₂₃| = (8.99 × 10⁹)(3 × 10⁻⁹)(4 × 10⁻⁹)/(0.05)² = 4.32 × 10⁻⁵ N

Direction on q₃: toward q₂, so −x. F⃗₂₃ = (−4.32 × 10⁻⁵, 0) N.

Step 2: Force on q₃ from q₁

r₁₃ = 0.05 m (equilateral). Both positive, so repulsion. Direction: from q₁ to q₃, which is the unit vector (0.025 − 0.025, 0 − 0.0433)/0.05 wait, recompute. Vector from q₁ at (0.025, 0.0433) to q₃ at (0.05, 0) is (0.025, −0.0433); divide by 0.05 to get unit vector (0.5, −0.866).

|F₁₃| = (8.99 × 10⁹)(2 × 10⁻⁹)(4 × 10⁻⁹)/(0.05)² = 2.88 × 10⁻⁵ N

F⃗₁₃ = 2.88 × 10⁻⁵ × (0.5, −0.866) = (1.44 × 10⁻⁵, −2.49 × 10⁻⁵) N.

Step 3: Net force

F_x = −4.32 × 10⁻⁵ + 1.44 × 10⁻⁵ = −2.88 × 10⁻⁵ N

F_y = 0 + (−2.49 × 10⁻⁵) = −2.49 × 10⁻⁵ N

|F_net| = √(F_x² + F_y²) = 3.81 × 10⁻⁵ N

Direction: arctan(F_y/F_x) = arctan(−2.49 / −2.88) in the third quadrant = 180° + 40.8° ≈ 221° measured counter-clockwise from +x. The net force on q₃ points down and to the left, dominated by the attraction toward q₂.

Sanity check: q₂ is the largest magnitude charge and it's attractive on q₃, so F_x is negative. q₁ is above q₃ and repulsive, so its contribution to F_y is negative (pushes q₃ down). Both signs match the result. If F_y came out positive, you'd know the q₁ direction was flipped.

References

For derivations of the inverse-square law from Maxwell's equations, the relativistic correction, and the formal treatment of continuous distributions, the standard textbook is Griffiths. For the experimental side and the modern CODATA constants, NIST is canonical. The 2019 SI redefinition fixed e exactly, which is worth knowing if you're writing up a lab report.

  • Griffiths, D. J. (2017). Introduction to Electrodynamics, 4th ed. Cambridge University Press. Chapter 2 covers electrostatics; Chapter 11 covers radiation from moving charges.
  • CODATA 2022  Fundamental physical constants, NIST: physics.nist.gov/cuu/Constants
  • Halliday, Resnick & Walker (2018). Fundamentals of Physics, 11th ed. Wiley. Chapters 21 and 22 walk through Coulomb's law and the electric field with worked problems.
  • OpenStax College Physics  Free peer-reviewed textbook: openstax.org
  • Jackson, J. D. (1999). Classical Electrodynamics, 3rd ed. Wiley. The graduate-level reference for retarded potentials, multipole expansions, and the limits of the Coulomb form.

Constants Used by This Calculator (CODATA 2022)

ConstantSymbolValue
Coulomb constantk8.9875517923 × 10⁹ N·m²/C²
Vacuum permittivityε₀8.8541878128 × 10⁻¹² F/m
Elementary chargee1.602176634 × 10⁻¹⁹ C (exact)
Relationk = 1/(4πε₀)derived

Educational Use Notice

This calculator is built for physics coursework, homework verification, and conceptual exploration. It uses the point-charge approximation and ideal vacuum conditions. Real engineering applications (ESD protection, capacitor design, high-voltage systems) require additional factors such as material properties, geometry effects, and safety margins that this tool doesn't model. For professional applications, consult domain-specific engineering software and standards.

Fixing Common Coulomb's Law Errors

Real questions from students who got stuck on unit conversions, sign conventions, and inverse-square traps.

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. The constant k = 8.99 × 10⁹ N·m²/C² in vacuum, also written as 1/(4πε₀) where ε₀ is the permittivity of free space. The force is attractive when charges have opposite signs and repulsive when they share the same sign. Force is a vector pointing along the line connecting the charges, but the formula gives only magnitude. For multiple charges, use superposition: compute the force from each source charge separately, then add them as vectors. Two protons separated by 1 fm (about a nuclear diameter) feel a Coulomb repulsion of F = (8.99 × 10⁹)(1.6 × 10⁻¹⁹)² / (10⁻¹⁵)² = 230 N. That's why nuclei need the strong force to hold together. The inverse-square dependence catches a lot of students. Doubling the distance cuts the force to one-quarter, not one-half. The same form shows up in Newton's gravity (with masses replacing charges and G replacing k), so the algebra carries over. In materials with relative permittivity εᵣ, replace k with k/εᵣ. Water (εᵣ ≈ 80) screens charges that effectively, which is why ionic compounds dissolve.

The calculator says 'attractive' but I expected repulsion—why?

Nine times out of ten, the issue is sign convention. Opposite signs (+ and −) always attract; same signs (both + or both −) always repel. If you entered one charge as positive and one as negative, the force is attractive regardless of magnitudes. Also verify you didn't accidentally flip a sign when entering values. The direction depends only on signs, not on which charge is larger.

My answer is off by a factor of a million—what happened?

The most common cause is treating microcoulombs (µC) as plain coulombs. 1 µC = 10⁻⁶ C, so if your problem says 5 µC, you need to enter 5 × 10⁻⁶ (or 5e-6) into the calculator. Entering 5 directly makes the charge a million times larger, which makes the force a trillion times larger (since force depends on the product of two charges). This is the single most common error in Coulomb's law problems.

How do I solve for the unknown charge when I know the force?

Rearrange Coulomb's law: q = Fr²/(k|q_other|). Multiply the force by distance squared, then divide by Coulomb's constant times the magnitude of the known charge. Select 'Solve for Unknown' mode, choose which variable you want, and enter the other three values. The calculator shows each algebraic step so you can follow the rearrangement.

I doubled the distance and expected half the force—why is it one-quarter?

Coulomb's law has an inverse-square dependence: F ∝ 1/r². When distance doubles, force decreases by 2² = 4×, not 2×. Triple the distance and force drops to 1/9. This catches a lot of students on exams because we're used to linear relationships. The same inverse-square law applies to gravity, so once you internalize it for charges, you'll recognize it everywhere.

Can I use this for three or more charges?

Yes, using superposition. Calculate the force from each charge on your target charge separately (each is a two-charge Coulomb problem), then add the forces as vectors. For collinear charges, this means adding or subtracting magnitudes depending on direction. For non-collinear charges, break each force into x and y components, sum components, then find the resultant. The calculator's superposition mode automates this.

What's the difference between force and electric field?

Force (F) is the actual push or pull between two specific charges, measured in newtons. Electric field (E) is force per unit charge at a location, measured in N/C. A charge q creates a field E = kq/r² around it. If you place another charge Q in that field, it feels force F = QE. Fields let you separate 'what the source charge creates' from 'what the test charge experiences.'

My problem uses permittivity instead of Coulomb's constant—how do I convert?

Coulomb's constant k and vacuum permittivity ε₀ are related by k = 1/(4πε₀). In vacuum, k = 8.99 × 10⁹ N·m²/C² and ε₀ = 8.85 × 10⁻¹² F/m. Some textbooks write Coulomb's law as F = q₁q₂/(4πε₀r²), which is the same formula with k expanded. For materials with relative permittivity εᵣ, replace ε₀ with ε₀εᵣ, or equivalently use k_eff = k/εᵣ.

How do I find where to place a third charge so net force on it is zero?

Set the forces from the other two charges equal in magnitude and opposite in direction. If both source charges have the same sign, the equilibrium point lies between them. If they have opposite signs, it lies outside (beyond the smaller charge). Write F₁ = F₂, cancel common factors, and solve for position. The equilibrium point is closer to the smaller charge because you need to be nearer to compensate for its weaker force.

Does this work for electrons and protons, or only macroscopic charges?

Coulomb's law applies at all scales, from elementary particles to Van de Graaff generators. For electrons and protons, use the elementary charge e = 1.60 × 10⁻¹⁹ C. At very short distances (below ~10⁻¹⁵ m), quantum effects and the strong nuclear force dominate, but for atomic and molecular scales (10⁻¹⁰ m), Coulomb's law remains accurate.

Why does water reduce the force between charges by ~80×?

Water has a high relative permittivity (εᵣ ≈ 80) because its polar molecules partially align with the electric field, creating opposing fields that screen the charges. The effective force becomes F = kq₁q₂/(εᵣr²), about 80× weaker than in vacuum. This is why ionic compounds dissolve in water—the medium weakens ion-ion attraction, allowing ions to separate and disperse.

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