Coulomb's Law Calculator: Force, Field & Potential Energy
Calculate electric forces, fields, and potential energy between charged particles with vector analysis and superposition.
Calculate electric forces, fields, and potential energy between charged particles with vector analysis and superposition.
This Coulomb's law calculator handles the algebra so you can focus on physics. A student building a lab report needed to find what charge magnitude would produce a 0.5 N force at 10 cm separation—given the other charge was +3 µC. She kept getting nonsense answers because she forgot to convert centimeters to meters before squaring. The calculator flagged her input as inconsistent with SI units, she fixed it, and the correct answer (q₂ ≈ 1.85 µC) appeared in seconds. Whether you're solving for charge, distance, or force, this tool rearranges F = kq₁q₂/r² for any unknown and shows every algebraic step.
| Unknown | Rearranged Formula | Required Inputs |
|---|---|---|
| F | F = k|q₁q₂|/r² | q₁, q₂, r |
| q₁ or q₂ | q = Fr²/(k|q_other|) | F, r, other charge |
| r | r = √(k|q₁q₂|/F) | F, q₁, q₂ |
| E (field) | E = kq/r² | q, r |
The formula F = k|q₁q₂|/r² gives you the magnitude—always positive. Direction comes from the signs of the charges themselves. Same signs (both positive or both negative) mean repulsion: the force pushes each charge away from the other. Opposite signs mean attraction: the force pulls them together.
When drawing free-body diagrams, place the force vector along the line connecting the two charges. For repulsion, the vector points away from the other charge. For attraction, it points toward the other charge. This sounds obvious until you have three or more charges and need to add vectors—keeping track of which way each force points becomes the whole challenge.
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.
Most homework problems give you three of the four variables (F, q₁, q₂, r) and ask for the fourth. Here are the rearrangements you need:
Solving for force: F = k|q₁q₂|/r²
Plug in both charges (in coulombs) and distance (in meters). Remember: k = 8.99 × 10⁹ N·m²/C².
Solving for charge: q = Fr²/(k|q_other|)
When one charge is unknown, multiply force by distance squared, then divide by k times the known charge magnitude.
Solving for distance: r = √(k|q₁q₂|/F)
Take the square root after dividing the product kq₁q₂ by force. Don't forget the square root—this is a common exam mistake.
The calculator handles these rearrangements automatically. Select "Solve for Unknown" mode, specify which variable you want, enter the known values, and it shows the algebraic steps leading to your answer.
With two charges, there's one force to compute. With three charges, there are three pairs, and each charge experiences forces from the other two. The superposition principle says: calculate each pairwise force separately using Coulomb's law, then add them as vectors.
For charges arranged in a straight line, vector addition simplifies to adding or subtracting magnitudes depending on direction. If charge A is between B and C, the forces from B and C on A point in opposite directions—you subtract magnitudes to get the net force.
For charges in a plane (not collinear), break each force into x and y components, add all x-components together, add all y-components together, then compute the resultant magnitude and direction using Pythagoras and arctan.
Watch out: When three charges form a triangle, the angles between force vectors are not 90°. You'll need trigonometry (cos θ and sin θ) to resolve components correctly.
Here's a worked example showing the full process for a test charge influenced by two other charges.
Problem Setup
Charge A: +4 µC at origin (0, 0)
Charge B: −2 µC at (0.3 m, 0)
Test charge C: +1 µC at (0.15 m, 0)—midway between A and B
Step 1: Force from A on C
r_AC = 0.15 m, q_A = 4 × 10⁻⁶ C, q_C = 1 × 10⁻⁶ C
F_AC = (8.99 × 10⁹)(4 × 10⁻⁶)(1 × 10⁻⁶)/(0.15)² = 1.60 N
Direction: Away from A (rightward, +x) since both are positive.
Step 2: Force from B on C
r_BC = 0.15 m, q_B = −2 × 10⁻⁶ C, q_C = 1 × 10⁻⁶ C
F_BC = (8.99 × 10⁹)(2 × 10⁻⁶)(1 × 10⁻⁶)/(0.15)² = 0.80 N
Direction: Toward B (rightward, +x) since opposite signs attract.
Step 3: Net Force
Both forces point in +x direction: F_net = 1.60 + 0.80 = 2.40 N rightward
The test charge gets pushed to the right by A (repulsion) and pulled to the right by B (attraction). Same direction means add magnitudes. If B were on the left of C, the forces would oppose and you'd subtract.
Coulomb's law F = kq₁q₂/r² applies strictly to point charges— objects whose physical size is negligible compared to the distance between them. For extended charge distributions (a charged rod, a charged disk, a uniformly charged sphere), you can't treat the whole object as a single point.
Instead, you divide the distribution into infinitesimal charge elements dq, apply Coulomb's law to each element, and integrate over the entire distribution. This is where calculus enters electrostatics. The integral gives you the net force or field from a continuous distribution.
One important exception: a uniformly charged sphere behaves exactly like a point charge located at its center—but only for points outside the sphere. Inside, the field follows a different pattern (proportional to r, not 1/r²). This result, proven using Gauss's law, explains why planets and spherical shells can be treated as point masses or charges in many calculations.
When charges don't lie on a straight line, force vectors point in different directions. You need to break each force into components.
Component method:
Be careful with signs when computing angles. If a charge is in the third quadrant relative to the test charge, both x and y components of the force vector will be negative (for repulsion) or positive (for attraction). A sketch showing all charges and force arrows before computing helps avoid sign errors.
A classic exam problem: where should you place a third charge so the net force on it is zero? This requires the forces from the other two charges to cancel exactly.
Problem
Charge Q₁ = +9 µC is at x = 0. Charge Q₂ = +4 µC is at x = 0.3 m. Where along the x-axis can you place a third charge Q₃ so the net electrostatic force on Q₃ is zero?
Analysis
Since Q₁ and Q₂ are both positive, they repel Q₃ regardless of Q₃'s sign. For forces to cancel, Q₃ must be between Q₁ and Q₂, where the two repulsive forces point in opposite directions.
Setup
Let Q₃ be at position x (where 0 < x < 0.3 m).
Distance from Q₁: r₁ = x
Distance from Q₂: r₂ = 0.3 − x
Force Balance
F₁ = F₂ → k(9)(Q₃)/x² = k(4)(Q₃)/(0.3−x)²
Cancel k and Q₃:
9/x² = 4/(0.3−x)²
Cross-multiply and take square roots:
3/x = 2/(0.3−x) → 3(0.3−x) = 2x → 0.9 = 5x
x = 0.18 m
Result: The equilibrium position is at x = 0.18 m, or 18 cm from the larger charge. Notice that the equilibrium point is closer to the smaller charge—this makes sense because the smaller charge exerts less force, so you need to be closer to it for the forces to balance.
Electrostatics calculations depend on precisely defined constants. Here are the values this calculator uses, based on the most recent CODATA recommendations:
| Constant | Symbol | Value |
|---|---|---|
| Coulomb constant | k | 8.9875517923 × 10⁹ N·m²/C² |
| Vacuum permittivity | ε₀ | 8.8541878128 × 10⁻¹² F/m |
| Elementary charge | e | 1.602176634 × 10⁻¹⁹ C (exact) |
| Relation | k = 1/(4πε₀) | — |
Since 2019, the elementary charge e is defined exactly as 1.602176634 × 10⁻¹⁹ C, with zero uncertainty. The Coulomb constant k is derived from ε₀, which is now measured rather than defined. For most homework problems, k = 8.99 × 10⁹ N·m²/C² (rounded to three significant figures) is sufficient.
This calculator is designed 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 (material properties, geometry effects, safety margins) that this tool does not model. For professional applications, consult domain-specific engineering software and standards.
Real questions from students who got stuck on unit conversions, sign conventions, and inverse-square traps.
You probably entered microcoulombs (µC) as 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.
Check the signs of your charges. 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.
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.
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.
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.
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.'
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.
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.
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.
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.
Select a mode and enter your charge values to calculate electric forces, fields, or potential energy.
F = k · |q₁q₂| / r²
k ≈ 8.99 × 10⁹ N·m²/C² (Coulomb constant)
E = k · q / r²
Field points away from positive, toward negative charges
U = k · q₁q₂ / r
Positive for like charges, negative for opposite charges
F_net = Σ F_i (vector sum)
Net force is the vector sum of individual forces
Coulomb's Law describes the electrostatic force between two charged particles. The force is proportional to the product of the charges and inversely proportional to the square of the distance between them: F = k·|q₁q₂|/r². The force is attractive if charges have opposite signs and repulsive if they have the same sign. The constant k ≈ 8.99×10⁹ N·m²/C² in vacuum.
Electric force (F) is the actual force exerted between two charges, measured in Newtons. Electric field (E) is the force per unit charge at a point in space, measured in N/C or V/m. The field exists even without a test charge present. The relationship is F = qE, where q is the test charge experiencing the field. A positive charge creates a field pointing away from it, while a negative charge creates a field pointing toward it.
The superposition principle states that the net electric force (or field) from multiple charges is the vector sum of the individual forces (or fields) from each charge. Each charge contributes independently, and we add them as vectors. This means we calculate Fₓ and Fᵧ components separately for each charge, then sum them: F_net = F₁ + F₂ + F₃ + ... This principle allows us to analyze complex charge distributions by breaking them into simple two-charge interactions.
Electric potential energy (U = k·q₁q₂/r) is the work needed to bring two charges from infinite separation to distance r. It's positive when like charges (both + or both −) are brought together, meaning work is required to overcome repulsion. It's negative when opposite charges are brought together, meaning the system releases energy as they attract. A system naturally moves toward lower potential energy, so opposite charges spontaneously come together, while like charges naturally separate.
When charges are in a medium other than vacuum (like water or oil), the medium's molecules partially shield the charges, reducing the force. This is quantified by the relative permittivity εᵣ (also called dielectric constant). The effective Coulomb constant becomes k_eff = k/εᵣ. For example, water has εᵣ ≈ 80, so electrostatic forces in water are about 80 times weaker than in vacuum. This is why ionic compounds dissolve in water—the medium significantly reduces the attractive forces between ions.