Physics · Electrostatics

Coulomb’s law: force between point charges

Q: How do you calculate the Coulomb force between two charges?

Use $F=k_e\frac{|q_1q_2|}{r^2}$ for nonnegative force magnitude. Insert charges in coulombs and distance in metres; $ke \approx{}\frac{ 8.99\cdot{}10^{9} N\cdot{}m^{2}}{C^{2}}$ is a rounded vacuum value. Charges +2 μC and −3 μC at 0.30 m give about 0.599 N. Unlike signs mean attraction. Determine direction separately and identify the charge receiving the force. Vector expressions retain charge signs; magnitude expressions use absolute values. With multiple source charges, add directed forces rather than their magnitudes. Convert each microcoulomb to $10^{-6} C$ before multiplying; otherwise the numerical result is wrong by powers of ten.

Coulomb's law describes the electrostatic force between two point charges.

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Formula

Calculate

F = k_{e} \cdot \frac{∣ q _{1} q _{2} ∣}{r ^{2}}

F in N · k_e = 8.99×10⁹ N·m²/C² · q in C (coulombs) · r in m

Variables & units

Variables & units
Symbol	Meaning	Unit
$F$	Electrostatic force magnitude	N
$k_{\mathrm{e}}$	Coulomb constant $(8.99\times{}10^{9})$	N·m²/C²
q₁, q₂	Electric charges	C
$r$	Distance between charge locations; centres in a point approximation	m

LaTeX

F = k_e \cdot \frac{|q_1 q_2|}{r^2}

QUANTA STUDYquanta-study.de

Diagram: two positive charges q₁ and q₂ sit at distance r, from each charge a force arrow F points outward, away from the other charge. — Two like charges q₁ and q₂ at distance r repel each other with the equal force F.

Coulomb's law calculator

Find the force magnitude between two point charges in vacuum, or their separation; the signs of the charges decide between attraction and repulsion.

Find

First charge q₁

First charge with its sign, e.g. +2 µC; an excess of electrons is negative.

Second charge q₂

Second charge with its sign, e.g. −3 µC; like signs repel, unlike signs attract.

Separation r

Distance between the centres of the charges, greater than zero, e.g. 30 cm.

Filled with example values. Type your own and the result updates straight away.

Result: Force magnitude |F|

|F| = 0.5992 N

With q₁ = 2 µC, q₂ = -3 µC and r = 30 cm, |F| = 0.5992 N.

Working · rearrange for |F|, substitute, evaluate

1. Rearrange for |F|
$|F|=k\frac{|q_1q_2|}{r^2}$
2. Substitute values (in SI units)
- q₁ = 2 µC = 2 · 10⁻⁶ C
- q₂ = -3 µC = -3 · 10⁻⁶ C
- r = 30 cm = 0.3 m
$|F|=8987551786.17\cdot \frac{|\left(0.000002\right)\cdot (-0.000003)|}{0.3^2}=0.59917$
3. Result
|F| = 0.5992 N

The charges attract.

Using ε₀ ≈ 8.8541878188·10⁻¹² F/m (CODATA 2022, NIST; standard uncertainty 1.4·10⁻²¹ F/m); k = 1/(4πε₀). The constant and displayed result are approximations.

Quick tips

Double the distance, a quarter of the force: F depends on 1/r². At three times the distance only one ninth is left.
Estimate with k ≈ 9 · 10⁹ N·m²/C²: two charges of 1 µC each, 1 m apart, repel each other with about 9 mN.

Assumptions and limits

Point charges in vacuum with separation r > 0. The result is a magnitude; equal charge signs mean repulsion, opposite signs attraction. It does not determine a spatial direction.

Derivation & background

The empirical electrostatic law follows $\frac1{r^2}$ . Like charges repel and unlike charges attract. F here is nonnegative magnitude; directed components retain sign information. $k_e=\frac1{4\pi\varepsilon_0}$ . With spherical symmetry, Gauss’s law gives the same field description, rather than a purely geometric proof of the physical law.

Exam blueprint

Validity range

Electrostatic point charges in vacuum with $r > 0$ . Extended distributions must actually have spherical symmetry; nearby conductors may redistribute charge. A material permittivity requires an appropriate linear, homogeneous, isotropic model.

Derivation steps

Spherical symmetry and Gauss’s law explain the field expression within electrostatic theory.

1Choose an isolated point charge q₁ in vacuum and $r > 0$ .
2Spherical symmetry: E is radial with constant radial component at equal radius.
3Apply Gauss to a sphere: $E_r\,4\pi r^2=\frac{q_1}{\varepsilon_0}$ . Charge is enclosed and flux counts its signed value.
4Apply the field to the second charge: $\mathbf F_2=q_2\mathbf E_1$ , with $k_e=\frac1{4\pi\varepsilon_0}$ .
5The nonnegative force magnitude is $F=k_e\frac{|q_1q_2|}{r^2}$ . The point model is undefined at $r=0$ .
6Signs determine attraction or repulsion; add multiple forces as vectors.

Rearrangements

Distance from force and charges

r = \sqrt{\frac{k_e |q_1 q_2|}{|F|}}

The sign describes direction; use absolute values for magnitudes.

Task variant

What happens to F if one charge doubles and r doubles?

$\frac{F_{\mathrm{new}}}{F_{\mathrm{old}}}=\frac{2}{2^2}=\frac12$ : F is halved.

Common mistakes

Reading attraction or repulsion from the formula without sign logic.

Like signs repel, unlike signs attract.

Exam context

Classically used in comparisons with gravity or electric-field force tasks.

Formula cluster

Electrostatic force

Connects charge, field, potential and chemical bonding.

F = q(v × B)

Lorentz Force

ℰ = −N dΦB/dt

Faraday’s law of induction

E = E° − (RT/zF)·ln Q

Nernst Equation

E⃗ = F⃗el/q; Ehom = |U|/d

Electric Field Strength

Worked example

A formal point-charge calculation for two $+1.6\cdot{}10^{-19} C$ charges at $r = 10^{-15} m$ gives $F \approx{} 230 N$ . This extrapolation represents neither finite proton charge distributions nor all nuclear interactions. A complete ordinary μC calculation follows below.

Applications

Chemical bonding, solid-state physics, electronics (semiconductors), mass spectrometry

How do you calculate with Coulomb's Law?

Here is how to work through a typical Coulomb's Law (F = ke·|q₁q₂|/r²) task step by step:

1
Task
What happens to F if one charge doubles and r doubles?
Solution path
$\frac{F_{\mathrm{new}}}{F_{\mathrm{old}}}=\frac{2}{2^2}=\frac12$ : F is halved.

Understand and apply the equation

Charged bodies can attract or repel without touching. Coulomb’s law makes this interaction calculable when their charges can be treated as stationary points. First find the nonnegative force magnitude, then determine direction and, for several sources, the vector sum.

From observation to a point charge

A rubbed balloon can attract another charged object. Charge on a balloon is spread out and may redistribute. For our first calculation, imagine two small charged bodies held fixed in vacuum. If their sizes are negligible compared with separation, approximate them by charges at two points. The law initially describes their mutual electric force rather than automatically every force acting on them.[1]

Prior knowledge: charge is measured in coulombs; like signs repel and unlike signs attract.
Mark both charge locations and choose a positive axis. A negative force sign denotes a component, not a negative strength.
Separation means distance between charge locations. In an appropriate point approximation these are the centres, not the surface gap.

[1]

When does the Coulomb force attract and when does it repel?

F=k_e\frac{|q_1q_2|}{r^2},\qquad r>0

F denotes force magnitude on either charge.[1]

The formula for Coulomb force magnitude gives a non-negative value; charge signs determine direction. Like charges repel and unlike charges attract. The two forces have equal magnitude and opposite direction but act on different charges. They do not cancel in the force balance of either individual charge. r is separation of charge locations; for spherical bodies treated as point charges, use centre-to-centre distance.[1]

\vec F_{2\leftarrow1}=k_e\frac{q_1q_2}{r^2}\hat r_{1\to2}

The unit vector points from q₁ to q₂; a negative product reverses the force on q₂.

A negative force component can be meaningful. A negative magnitude instead indicates that magnitude and direction were mixed. First identify the charge receiving the force and the positive axis direction.

Why is separation squared? For a point source, the same total flux is spread over increasingly large spherical surfaces. Their area grows with the square of radius. This explains distance scaling within the electrostatic model. That charges produce this field is a physical law; geometry alone does not establish it.[1][2]

Understanding the field model

Coulomb’s law is an empirical electrostatic law. Explaining it through Gauss’s law uses the same physical theory, rather than an independent purely geometric proof. For an isolated point charge in vacuum, the electric field is radial with equal magnitude everywhere at equal radius.[2]

\oint\vec E\cdot\mathrm d\vec A=\frac{q_1}{\varepsilon_0}\quad\Longrightarrow\quad E_r4\pi r^2=\frac{q_1}{\varepsilon_0}

Spherical symmetry lets the radial field component be taken outside the surface integral.

E_r=\frac{q_1}{4\pi\varepsilon_0r^2},\qquad \vec F_2=q_2\vec E_1

Field and force are related by charge q₂.

This gives $k_e=\frac1{4\pi\varepsilon_0}$ . Without spherical symmetry, $E_r\,4\pi r^2$ is not automatically valid. Gauss’s law still applies, but the field cannot be removed from the integral in that way.

Units and inverse squares

Use coulombs and metres: $1\,\mathrm{\mu C}=10^{-6}\,\mathrm C$ , $1\,\mathrm{nC}=10^{-9}\,\mathrm C$ and $30\,\mathrm{cm}=0.30\,\mathrm m$ . Convert first, then square r. Introductory calculations can use ke ≈ 8.99·10⁹ N·m²/C². This is rounded; CODATA reports vacuum permittivity with measurement uncertainty.[5]

[\frac{k_e q_1q_2}{r^2}]=\frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\frac{\mathrm{C^2}}{\mathrm{m^2}}=\mathrm N

Charge units and area units cancel.

With fixed charges, doubling distance gives one quarter of the force, not one half. Halving distance quadruples it. Doubling both charge magnitudes and distance leaves force unchanged because numerator and denominator both grow by four.

Double separation, one quarter of the force

QUANTA STUDYquanta-study.de

The charges are those of the first example: about 0.599 N at 30 cm and 0.150 N at 60 cm. The curve shows magnitude; attraction towards the other charge is determined separately.[1]

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Two complete worked examples

Attraction with microcoulombs

Given: $q_1=+2\,\mathrm{\mu C}$ on the left, $q_2=-3\,\mathrm{\mu C}$ on the right; $r=30\,\mathrm{cm}$ in vacuum.
Find: Find force magnitude and direction on q₂.

Approach: For stationary vacuum point charges, find Coulomb force magnitude; a negative charge product gives leftward attraction on the right charge.
$\begin{aligned}F&=k_e\frac{|q_1q_2|}{r^2}\\F_{2,x}&=-F\end{aligned}$
Convert charges and distance to SI.
$\begin{aligned}q_1&=2\cdot10^{-6}\,\mathrm C\\{} q_2&=-3\cdot10^{-6}\,\mathrm C\\{} r&=0.30\,\mathrm m\end{aligned}$
Use the product magnitude and square the distance.
$F=8.99\cdot10^9\frac{6\cdot10^{-12}}{0.30^2}\approx0.599\,\mathrm N$
Use charge signs to determine direction.
$q_1q_2<0\quad\Longrightarrow\quad F_{2,x}\approx-0.599\,\mathrm N$

Result: Force on q₂ is about 0.599 N and points left toward q₁.

Check: Force on q₁ is equally large and points right. At 60 cm separation, magnitude would be about 0.150 N.

Finding separation from force

Given: Two fixed point charges of +4 nC each; force magnitude 2.0·10⁻⁵ N in vacuum.
Find: Calculate separation.

Approach: In the vacuum point-charge model, solve the measured force magnitude for positive separation.
$\begin{aligned}F&=k_e\frac{|q_1q_2|}{r^2}\\r&=\sqrt{\frac{k_e|q_1q_2|}{F}}\end{aligned}$
Solve for positive separation.
$r=\sqrt{\frac{k_e|q_1q_2|}{F}}$
Substitute SI values.
$r=\sqrt{\frac{8.99\cdot10^9\cdot16\cdot10^{-18}}{2.0\cdot10^{-5}}}=\sqrt{0.007192}\,\mathrm m$
Take the square root and convert to centimetres.
$r\approx0.0848\,\mathrm m=8.48\,\mathrm{cm}$

Result: Separation is about 8.48 cm; the charges repel.

Check: Substituting $r^2=0.007192\,\mathrm{m^2}$ reproduces the given force. Do not omit the square root.

Multiple charges: how do you calculate the net force?

Calculate each force on the same selected charge and add components. For extended charge distributions, the sum becomes an integral over small charge elements. Adding magnitudes alone is valid only when all forces point in the same direction. If the summed components are perpendicular, the Pythagorean theorem gives the magnitude of the net force. For charges on a straight line, as in the following example, instead add signed components along that line.[1][3]

Original symmetry problem: two +2 nC charges lie at $x=-0.20\,\mathrm m$ and $x=+0.20\,\mathrm m$ . Each exerts 4.495·10⁻⁷ N on +1 nC at the origin. The left source pushes right and the right source pushes left, giving zero net force. Doubling the right source to +4 nC gives a net 4.495·10⁻⁷ N leftward.

Two sources: equal distances, unequal charges

Given: In vacuum, +2 nC is at $x=-0.20\,\mathrm m$ and +4 nC at $x=+0.20\,\mathrm m$ . A positive +1 nC probe is at the origin; both sources remain fixed.
Find: Find both force components and the net force on the probe.

Approach: each source repels the probe. The left source pushes right and the right source pushes left. Add signed components.
$\begin{aligned}F_{x,L}&=+k_e\frac{|Q_Lq|}{a^2}\\F_{x,R}&=-k_e\frac{|Q_Rq|}{a^2}\\F_{x,\mathrm{net}}&=F_{x,L}+F_{x,R}\end{aligned}$
Substitute $a=0.20\,\mathrm m$ and charges in coulombs.
$\begin{aligned}F_{x,L}&=\frac{(8.99\cdot10^9)(2\cdot10^{-9})(1\cdot10^{-9})}{0.20^2}\,\mathrm N\\&=+4.495\cdot10^{-7}\,\mathrm N\\F_{x,R}&=-8.990\cdot10^{-7}\,\mathrm N\\F_{x,\mathrm{net}}&=-4.495\cdot10^{-7}\,\mathrm N\end{aligned}$

Result: The net force points left and has magnitude 0.4495 µN.

Check: At equal separation the right source produces twice the force. Setting both sources to +2 nC makes the opposite contributions equal and their sum zero.

Explore further: Material response and bonding

For an ideal homogeneous, isotropic linear dielectric, an appropriate macroscopic model uses $\varepsilon=\varepsilon_r\varepsilon_0$ . The corresponding force is reduced by εr compared with the same free point charges in vacuum. Dimensionless εr differs from ε measured in F/m.[4]

An arbitrary material number does not describe every real environment. Interfaces, anisotropy, conductors, nonlinear polarisation and local microscopic structure can require another model. Distinguish ε₀ for vacuum from ε for an explicitly appropriate material model.

Salt dissolution cannot be explained by material permittivity alone. Interactions with the solvent and solution conditions also matter. A salt may have low solubility even when its small dissolved fraction largely separates into ions. Solubility and degree of dissociation are distinct questions.[6]

Limits of point charges

An extended charged body is not automatically a point charge. Charge distribution and observation distance matter. Far from a body compared with its size, a point model may approximate its field. An isolated spherical distribution has the exact external field of a central point charge. Nearby conductors can redistribute charge.[3][2]

The expression is undefined at r = 0. In atomic or nuclear problems, classical electrostatic force is not a complete account of all interactions or stable trajectories. Moving charges can additionally require magnetic effects and time-dependent fields. Coulomb’s electrostatic model alone does not answer those questions.

Five problems with solutions

Calculate · Foundation

$q_1=+5\,\mathrm{nC}$ , $q_2=-8\,\mathrm{nC}$ , $r=0.10\,\mathrm m$ in vacuum. Find F and attraction or repulsion.

Show hint

Use the magnitude of the product.

Learning goal: Calculate Coulomb magnitude and identify attraction.

Show solution and calculation

Model and approach: Stationary point charges in vacuum; first convert separation and charges to SI.
$F=k_e\frac{|q_1q_2|}{r^2}$ $\begin{aligned}F&=\frac{8.99\cdot10^9\cdot40\cdot10^{-18}}{0.01}\,\mathrm N\\{} &=3.596\cdot10^{-5}\,\mathrm N\text{.}\end{aligned}$ The charges attract.

Check: $F(2r)=\frac{F(r)}4$ .

[1]

Interpret · Application

One charge doubles and distance halves. By what factor does F change?

Show hint

Consider the numerator and squared denominator separately.

Learning goal: Combine inverse-square distance and charge scaling.

Show solution and calculation

Model and approach: Keep the other quantities fixed; compare charge and distance factors separately.
$\frac{F_{\mathrm{new}}}{F_{\mathrm{old}}}=\frac{2}{(\frac12)^2}=8$ The factor is $2\cdot4=8$ .
Force becomes eight times larger.

F_{\mathrm{new}}=8F_{\mathrm{old}}

[1]

Calculate · Application

Two electrostatic forces on the same charge are perpendicular: 3 mN right and 4 mN up. Find the resultant magnitude.

Show hint

Combine perpendicular components with the Pythagorean theorem.

Learning goal: Combine perpendicular force components.

Show solution and calculation

Model and approach: Both forces act on the same charge and are perpendicular.
$|\mathbf F|=\sqrt{F_x^2+F_y^2}$ $|\mathbf F|=\sqrt{3^2+4^2}\,\mathrm{mN}=5\,\mathrm{mN}$ .
Adding magnitudes to obtain 7 mN would be incorrect.

Check: $4\,\mathrm{mN}<|\mathbf F|<7\,\mathrm{mN}$ .

|\vec F|=5\,\mathrm{mN}

[3]

Interpret · Application

Two identical positive point charges sit in vacuum at x=−a and x=+a, a>0. A small negative test charge at x=0 does not disturb them. Describe the two forces on the probe and their sum. Does each individual force vanish?

Show hint

Both sources are equally distant; their attractions point to opposite sides.

Learning goal: Use symmetry and superposition with nonzero individual forces.

Show solution and calculation

Approach: both Coulomb magnitudes are $k_e\frac{|qQ|}{a^2}$ .
The left force points along −x, the right along +x, so they cancel.
Each remains nonzero for nonzero charges.

Check: cancellation relies on equal distances and charge magnitudes, not absence of interaction. Moving away from the midpoint changes the symmetry.

\begin{aligned}F_{x,L}&=-k_e\frac{|qQ|}{a^2}\\F_{x,R}&=+k_e\frac{|qQ|}{a^2}\\F_{x,\mathrm{net}}&=0\end{aligned}

[1][3]

Error analysis · Foundation

Two free positive point charges repel in vacuum. A solution claims: “Action equals reaction, so each charge has zero net force.” Identify the bodies on which the Coulomb forces act. Do unequal masses change their force magnitudes?

Show hint

Only forces acting on the same body enter its motion equation.

Learning goal: Avoid cancelling interaction pairs in a single-body balance.

Show solution and calculation

Approach: the opposite forces act on different charges.
Each experiences the other’s force and can accelerate.
With no external forces, only the sum for the combined system vanishes.
Unequal masses do not change the common Coulomb magnitude for fixed charges and separation; they change accelerations.

Check: the interaction-pair vector relation still holds for unequal masses.

\begin{aligned}\mathbf F_{1\leftarrow2}&=-\mathbf F_{2\leftarrow1}\\\mathbf a_1&=\frac{\mathbf F_{1\leftarrow2}}{m_1}\\\mathbf a_2&=\frac{\mathbf F_{2\leftarrow1}}{m_2}\end{aligned}

[1]

What to take away

Calculate magnitude, then direction. Charge signs remain relevant even when the magnitude expression uses absolute values.
Add forces on the same probe as vectors. An interaction pair acts on two different bodies.
The point approximation and material model must fit the arrangement.

[1][3][4]

Sources for this explanation

Quanta · Updated October 1, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Distinguished Coulomb force, signs and non-negative magnitude directly; linked the electric-field explanation and Pythagoras only for perpendicular summed components, and made the material model discoverable through a question.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Quanta. Coulomb's Law. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/coulomb-gesetz

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Frequently asked questions about Coulomb's Law

How do you calculate the Coulomb force between two charges?+

Use $F=k_e\frac{|q_1q_2|}{r^2}$ for nonnegative force magnitude. Insert charges in coulombs and distance in metres; $ke \approx{}\frac{ 8.99\cdot{}10^{9} N\cdot{}m^{2}}{C^{2}}$ is a rounded vacuum value. Charges +2 μC and −3 μC at 0.30 m give about 0.599 N. Unlike signs mean attraction. Determine direction separately and identify the charge receiving the force. Vector expressions retain charge signs; magnitude expressions use absolute values. With multiple source charges, add directed forces rather than their magnitudes. Convert each microcoulomb to $10^{-6} C$ before multiplying; otherwise the numerical result is wrong by powers of ten.

How do you tell whether two charges attract or repel?+

Like charges repel and unlike charges attract; force magnitude remains nonnegative. For force on q₂, choose a unit vector from q₁ toward q₂. A positive charge product points along that vector and a negative product points opposite it. Force on q₁ points the other way. These forces act on different bodies and cannot cancel in the force balance of an individual charge. Without a defined axis, a sign alone does not specify a direction such as left or right. First draw the charge positions and label the positive coordinate direction.

What is the difference between Coulomb law and the law of gravitation?+

Both classical point models contain a product of source quantities and a factor $\frac1{r^2}$ . Newtonian gravity between positive masses attracts, whereas positive and negative electric charges permit attraction and repulsion. For two protons, $\frac{k_ee^2}{Gm_p^2}$ is roughly $10^{36}$ . This compares the two model forces, not all nuclear interactions. Gravity is measurable between small masses too. Many large bodies are nearly electrically neutral, so strong force between individual charges does not imply electric dominance in every macroscopic situation. The inverse-square comparison assumes the same centre-to-centre distance and point-source approximations.

How does the Coulomb force change if you double one charge and double the distance?+

Doubling one charge multiplies the numerator by two. Doubling distance multiplies the squared denominator by four. Together $\frac{F_{\mathrm{new}}}{F_{\mathrm{old}}}=\frac24=\frac12$ , so magnitude halves. Charge signs and therefore attraction or repulsion remain unchanged when multiplying charge by a positive factor. This ratio assumes the same field or material model. If both charges and distance double, the factor is $\frac44=1$ and magnitude is unchanged. Write the changed quantities symbolically before inserting values; this keeps the squared distance factor visible and prevents confusing twice the distance with twice the force.

Why must you use a different constant in a medium?+

A dielectric can polarise and modify the macroscopic electric field. An appropriate homogeneous, isotropic linear model uses $\varepsilon=\varepsilon_r\varepsilon_0$ , giving $\frac{k_e}{\varepsilon_r}$ in the corresponding point-charge calculation. This is an assumption about the particular environment. Interfaces, conductors, anisotropic or nonlinear materials and local microscopic fields can require a more detailed description. A single material number does not by itself explain salt solubility: lattice and solvation contributions and thermodynamic conditions also matter. Not every salt readily dissolves in water. Dissociation of the dissolved fraction and the total amount that dissolves are separate questions.

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Subject references and further reading

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[1]OpenStax, University Physics Vol. 2, §5.3 Coulomb’s Law

More formulas: Electricity and magnetism

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

Gauss's Law for Electrostatics

U = R \cdot I

Ohm's Law

P = U \cdot I

Electric Power

R_{ges} = R_1 + R_2 \qquad \frac{1}{R_{ges}} = \frac{1}{R_1} + \frac{1}{R_2}

Resistors in Series and Parallel

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Quanta Study editorial team

Symbol

Meaning

Unit

F

Electrostatic force magnitude

k_{\mathrm{e}}

Coulomb constant

(8.99\times{}10^{9})

N·m²/C²

q₁, q₂

Electric charges

r

Distance between charge locations; centres in a point approximation

Derivation & background

Exam blueprint

Validity range

Derivation steps

Spherical symmetry and Gauss’s law explain the field expression within electrostatic theory.

1Choose an isolated point charge q₁ in vacuum and $r > 0$ .
2Spherical symmetry: E is radial with constant radial component at equal radius.
3Apply Gauss to a sphere: $E_r\,4\pi r^2=\frac{q_1}{\varepsilon_0}$ . Charge is enclosed and flux counts its signed value.
4Apply the field to the second charge: $\mathbf F_2=q_2\mathbf E_1$ , with $k_e=\frac1{4\pi\varepsilon_0}$ .
5The nonnegative force magnitude is $F=k_e\frac{|q_1q_2|}{r^2}$ . The point model is undefined at $r=0$ .
6Signs determine attraction or repulsion; add multiple forces as vectors.

From observation to a point charge

Prior knowledge: charge is measured in coulombs; like signs repel and unlike signs attract.
Mark both charge locations and choose a positive axis. A negative force sign denotes a component, not a negative strength.
Separation means distance between charge locations. In an appropriate point approximation these are the centres, not the surface gap.

[1]

When does the Coulomb force attract and when does it repel?

F=k_e\frac{|q_1q_2|}{r^2},\qquad r>0

F denotes force magnitude on either charge.[1]

\vec F_{2\leftarrow1}=k_e\frac{q_1q_2}{r^2}\hat r_{1\to2}

The unit vector points from q₁ to q₂; a negative product reverses the force on q₂.

Understanding the field model

\oint\vec E\cdot\mathrm d\vec A=\frac{q_1}{\varepsilon_0}\quad\Longrightarrow\quad E_r4\pi r^2=\frac{q_1}{\varepsilon_0}

Spherical symmetry lets the radial field component be taken outside the surface integral.

E_r=\frac{q_1}{4\pi\varepsilon_0r^2},\qquad \vec F_2=q_2\vec E_1

Field and force are related by charge q₂.

Units and inverse squares

[\frac{k_e q_1q_2}{r^2}]=\frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\frac{\mathrm{C^2}}{\mathrm{m^2}}=\mathrm N

Charge units and area units cancel.

Double separation, one quarter of the force

QUANTA STUDYquanta-study.de

The charges are those of the first example: about 0.599 N at 30 cm and 0.150 N at 60 cm. The curve shows magnitude; attraction towards the other charge is determined separately.[1]

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Two complete worked examples

Attraction with microcoulombs

Given: $q_1=+2\,\mathrm{\mu C}$ on the left, $q_2=-3\,\mathrm{\mu C}$ on the right; $r=30\,\mathrm{cm}$ in vacuum.
Find: Find force magnitude and direction on q₂.

Approach: For stationary vacuum point charges, find Coulomb force magnitude; a negative charge product gives leftward attraction on the right charge.
$\begin{aligned}F&=k_e\frac{|q_1q_2|}{r^2}\\F_{2,x}&=-F\end{aligned}$
Convert charges and distance to SI.
$\begin{aligned}q_1&=2\cdot10^{-6}\,\mathrm C\\{} q_2&=-3\cdot10^{-6}\,\mathrm C\\{} r&=0.30\,\mathrm m\end{aligned}$
Use the product magnitude and square the distance.
$F=8.99\cdot10^9\frac{6\cdot10^{-12}}{0.30^2}\approx0.599\,\mathrm N$
Use charge signs to determine direction.
$q_1q_2<0\quad\Longrightarrow\quad F_{2,x}\approx-0.599\,\mathrm N$

Result: Force on q₂ is about 0.599 N and points left toward q₁.

Check: Force on q₁ is equally large and points right. At 60 cm separation, magnitude would be about 0.150 N.

Finding separation from force

Given: Two fixed point charges of +4 nC each; force magnitude 2.0·10⁻⁵ N in vacuum.
Find: Calculate separation.

Approach: In the vacuum point-charge model, solve the measured force magnitude for positive separation.
$\begin{aligned}F&=k_e\frac{|q_1q_2|}{r^2}\\r&=\sqrt{\frac{k_e|q_1q_2|}{F}}\end{aligned}$
Solve for positive separation.
$r=\sqrt{\frac{k_e|q_1q_2|}{F}}$
Substitute SI values.
$r=\sqrt{\frac{8.99\cdot10^9\cdot16\cdot10^{-18}}{2.0\cdot10^{-5}}}=\sqrt{0.007192}\,\mathrm m$
Take the square root and convert to centimetres.
$r\approx0.0848\,\mathrm m=8.48\,\mathrm{cm}$

Result: Separation is about 8.48 cm; the charges repel.

Check: Substituting $r^2=0.007192\,\mathrm{m^2}$ reproduces the given force. Do not omit the square root.

Multiple charges: how do you calculate the net force?

Two sources: equal distances, unequal charges

Given: In vacuum, +2 nC is at $x=-0.20\,\mathrm m$ and +4 nC at $x=+0.20\,\mathrm m$ . A positive +1 nC probe is at the origin; both sources remain fixed.
Find: Find both force components and the net force on the probe.

Approach: each source repels the probe. The left source pushes right and the right source pushes left. Add signed components.
$\begin{aligned}F_{x,L}&=+k_e\frac{|Q_Lq|}{a^2}\\F_{x,R}&=-k_e\frac{|Q_Rq|}{a^2}\\F_{x,\mathrm{net}}&=F_{x,L}+F_{x,R}\end{aligned}$
Substitute $a=0.20\,\mathrm m$ and charges in coulombs.
$\begin{aligned}F_{x,L}&=\frac{(8.99\cdot10^9)(2\cdot10^{-9})(1\cdot10^{-9})}{0.20^2}\,\mathrm N\\&=+4.495\cdot10^{-7}\,\mathrm N\\F_{x,R}&=-8.990\cdot10^{-7}\,\mathrm N\\F_{x,\mathrm{net}}&=-4.495\cdot10^{-7}\,\mathrm N\end{aligned}$

Result: The net force points left and has magnitude 0.4495 µN.

Check: At equal separation the right source produces twice the force. Setting both sources to +2 nC makes the opposite contributions equal and their sum zero.

Explore further: Material response and bonding

Limits of point charges

Five problems with solutions

Calculate · Foundation

$q_1=+5\,\mathrm{nC}$ , $q_2=-8\,\mathrm{nC}$ , $r=0.10\,\mathrm m$ in vacuum. Find F and attraction or repulsion.

Show hint

Use the magnitude of the product.

Learning goal: Calculate Coulomb magnitude and identify attraction.

Show solution and calculation

Model and approach: Stationary point charges in vacuum; first convert separation and charges to SI.
$F=k_e\frac{|q_1q_2|}{r^2}$ $\begin{aligned}F&=\frac{8.99\cdot10^9\cdot40\cdot10^{-18}}{0.01}\,\mathrm N\\{} &=3.596\cdot10^{-5}\,\mathrm N\text{.}\end{aligned}$ The charges attract.

Check: $F(2r)=\frac{F(r)}4$ .

[1]

Interpret · Application

One charge doubles and distance halves. By what factor does F change?

Show hint

Consider the numerator and squared denominator separately.

Learning goal: Combine inverse-square distance and charge scaling.

Show solution and calculation

Model and approach: Keep the other quantities fixed; compare charge and distance factors separately.
$\frac{F_{\mathrm{new}}}{F_{\mathrm{old}}}=\frac{2}{(\frac12)^2}=8$ The factor is $2\cdot4=8$ .
Force becomes eight times larger.

F_{\mathrm{new}}=8F_{\mathrm{old}}

[1]

Calculate · Application

Two electrostatic forces on the same charge are perpendicular: 3 mN right and 4 mN up. Find the resultant magnitude.

Show hint

Combine perpendicular components with the Pythagorean theorem.

Learning goal: Combine perpendicular force components.

Show solution and calculation

Model and approach: Both forces act on the same charge and are perpendicular.
$|\mathbf F|=\sqrt{F_x^2+F_y^2}$ $|\mathbf F|=\sqrt{3^2+4^2}\,\mathrm{mN}=5\,\mathrm{mN}$ .
Adding magnitudes to obtain 7 mN would be incorrect.

Check: $4\,\mathrm{mN}<|\mathbf F|<7\,\mathrm{mN}$ .

|\vec F|=5\,\mathrm{mN}

[3]

Interpret · Application

Two identical positive point charges sit in vacuum at x=−a and x=+a, a>0. A small negative test charge at x=0 does not disturb them. Describe the two forces on the probe and their sum. Does each individual force vanish?

Show hint

Both sources are equally distant; their attractions point to opposite sides.

Learning goal: Use symmetry and superposition with nonzero individual forces.

Show solution and calculation

Approach: both Coulomb magnitudes are $k_e\frac{|qQ|}{a^2}$ .
The left force points along −x, the right along +x, so they cancel.
Each remains nonzero for nonzero charges.

Check: cancellation relies on equal distances and charge magnitudes, not absence of interaction. Moving away from the midpoint changes the symmetry.

\begin{aligned}F_{x,L}&=-k_e\frac{|qQ|}{a^2}\\F_{x,R}&=+k_e\frac{|qQ|}{a^2}\\F_{x,\mathrm{net}}&=0\end{aligned}

[1][3]

Error analysis · Foundation

Two free positive point charges repel in vacuum. A solution claims: “Action equals reaction, so each charge has zero net force.” Identify the bodies on which the Coulomb forces act. Do unequal masses change their force magnitudes?

Show hint

Only forces acting on the same body enter its motion equation.

Learning goal: Avoid cancelling interaction pairs in a single-body balance.

Show solution and calculation

Approach: the opposite forces act on different charges.
Each experiences the other’s force and can accelerate.
With no external forces, only the sum for the combined system vanishes.
Unequal masses do not change the common Coulomb magnitude for fixed charges and separation; they change accelerations.

Check: the interaction-pair vector relation still holds for unequal masses.

\begin{aligned}\mathbf F_{1\leftarrow2}&=-\mathbf F_{2\leftarrow1}\\\mathbf a_1&=\frac{\mathbf F_{1\leftarrow2}}{m_1}\\\mathbf a_2&=\frac{\mathbf F_{2\leftarrow1}}{m_2}\end{aligned}

[1]

What to take away

Calculate magnitude, then direction. Charge signs remain relevant even when the magnitude expression uses absolute values.
Add forces on the same probe as vectors. An interaction pair acts on two different bodies.
The point approximation and material model must fit the arrangement.

[1][3][4]

Sources for this explanation

Quanta · Updated October 1, 2026

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Change history

October 1, 2026Distinguished Coulomb force, signs and non-negative magnitude directly; linked the electric-field explanation and Pythagoras only for perpendicular summed components, and made the material model discoverable through a question.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible entry and prerequisites, explained model steps, and added a complete example and quantitative diagram; retained existing examples, five exercises and sources.

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Frequently asked questions about Coulomb's Law

How do you calculate the Coulomb force between two charges?+

How do you tell whether two charges attract or repel?+

What is the difference between Coulomb law and the law of gravitation?+

How does the Coulomb force change if you double one charge and double the distance?+

Why must you use a different constant in a medium?+

Retain Coulomb's Law for exams

Create a curated FSRS exam set for F = ke·|q₁q₂|/r²: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Coulomb’s law: force between point charges

Variables & units

Coulomb's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatic force

Worked example

Applications

How do you calculate with Coulomb's Law?

From observation to a point charge

When does the Coulomb force attract and when does it repel?

Understanding the field model

Units and inverse squares

Double separation, one quarter of the force

Two complete worked examples

Attraction with microcoulombs

Finding separation from force

Multiple charges: how do you calculate the net force?

Two sources: equal distances, unequal charges

Explore further: Material response and bonding

Limits of point charges

Five problems with solutions

q1=+5 nCq_1=+5\,\mathrm{nC}q1​=+5nC, q2=−8 nCq_2=-8\,\mathrm{nC}q2​=−8nC, r=0.10 mr=0.10\,\mathrm mr=0.10m in vacuum. Find F and attraction or repulsion.

One charge doubles and distance halves. By what factor does F change?

Two electrostatic forces on the same charge are perpendicular: 3 mN right and 4 mN up. Find the resultant magnitude.

Two identical positive point charges sit in vacuum at x=−a and x=+a, a>0. A small negative test charge at x=0 does not disturb them. Describe the two forces on the probe and their sum. Does each individual force vanish?

Two free positive point charges repel in vacuum. A solution claims: “Action equals reaction, so each charge has zero net force.” Identify the bodies on which the Coulomb forces act. Do unequal masses change their force magnitudes?

What to take away

Sources for this explanation

Frequently asked questions about Coulomb's Law

Subject references and further reading

More formulas: Electricity and magnetism

Retain Coulomb's Law for exams

Coulomb’s law: force between point charges

Variables & units

Coulomb's law calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatic force

Worked example

Applications

How do you calculate with Coulomb's Law?

From observation to a point charge

When does the Coulomb force attract and when does it repel?

Understanding the field model

Units and inverse squares

Double separation, one quarter of the force

Two complete worked examples

Attraction with microcoulombs

Finding separation from force

Multiple charges: how do you calculate the net force?

Two sources: equal distances, unequal charges

Explore further: Material response and bonding

Limits of point charges

Five problems with solutions

q1=+5 nCq_1=+5\,\mathrm{nC}q1​=+5nC, q2=−8 nCq_2=-8\,\mathrm{nC}q2​=−8nC, r=0.10 mr=0.10\,\mathrm mr=0.10m in vacuum. Find F and attraction or repulsion.

One charge doubles and distance halves. By what factor does F change?

Two electrostatic forces on the same charge are perpendicular: 3 mN right and 4 mN up. Find the resultant magnitude.

Two identical positive point charges sit in vacuum at x=−a and x=+a, a>0. A small negative test charge at x=0 does not disturb them. Describe the two forces on the probe and their sum. Does each individual force vanish?

Two free positive point charges repel in vacuum. A solution claims: “Action equals reaction, so each charge has zero net force.” Identify the bodies on which the Coulomb forces act. Do unequal masses change their force magnitudes?

What to take away

Sources for this explanation

Frequently asked questions about Coulomb's Law

Subject references and further reading

More formulas: Electricity and magnetism

Retain Coulomb's Law for exams

$q_1=+5\,\mathrm{nC}$ , $q_2=-8\,\mathrm{nC}$ , $r=0.10\,\mathrm m$ in vacuum. Find F and attraction or repulsion.

$q_1=+5\,\mathrm{nC}$ , $q_2=-8\,\mathrm{nC}$ , $r=0.10\,\mathrm m$ in vacuum. Find F and attraction or repulsion.