Coulomb's Law

Beginner

Two charged objects push or pull on each other without touching. Coulomb's law tells us which way, and how hard.

Step 1: which way. Like charges (+ and +, or − and −) push each other apart; opposite charges (+ and −) pull each other together.

Step 2: how much charge. The more charge, the stronger the force. Double the charge on one object and the force doubles; double the charge on both and it becomes four times as strong.

Step 3: how far apart. Distance matters even more. Move two charges twice as far apart and the force drops to a quarter; three times as far and it drops to a ninth. Bring them twice as close and it becomes four times stronger. Gravity follows exactly the same "inverse square" pattern.

Step 4: the unit. Charge is measured in coulombs. One coulomb is enormous, the charge of about six billion billion electrons, so the charge on a rubbed balloon is less than a millionth of a coulomb (under one microcoulomb).

Step 5: fair shares. Both objects feel a force of the same size, in opposite directions, even if one carries much more charge than the other.

Check yourself: two small charged balls 10 cm apart push each other with a force of 8 units. How strong is the push when they are 20 cm apart? Answer: 2 units, because twice the distance gives a quarter of the force.

In the animation, two charged pucks slide closer together and then apart along a track. In the first half both are positive and the green arrows push them apart; in the second half one turns negative and the arrows pull them together. Watch how the arrows stay tiny when the pucks are far apart but shoot out when they come close.

Intermediate

What we measure: charge and distance

A rubbed balloon carries less than a millionth of the unit in which charge is measured. Coulomb's law links two charges and the distance between them. Charge is measured in coulombs (C) and has a sign. It comes in whole multiples of the elementary charge $e=1.602\times 10^{-19}\mathrm{C}$, the charge of a proton; an electron carries $-e$. A coulomb is huge, so everyday static charges are quoted in microcoulombs ($1\mu \mathrm{C}=10^{-6}\mathrm{C}$) and nanocoulombs ($1\mathrm{nC}=10^{-9}\mathrm{C}$). Each puck in the animation carries a charge of size 2 µC, about $1.25\times 10^{13}$ elementary charges.

The distance $r$ is measured between the centres of the two bodies. The law is exact for point charges, bodies much smaller than their separation, and for uniformly charged spheres, which act from outside as if all their charge sat at the centre. Convert everything to SI units first: $2\mu \mathrm{C}=2\times 10^{-6}\mathrm{C}$ and $15\mathrm{cm}=0.15\mathrm{m}$.

Coulomb's law

Two point charges $q_1$ and $q_2$ a distance $r$ apart push or pull on each other with a force of size
$F=\frac{k|q_1{q}_2|}{r^2}$
where $k$ is Coulomb's constant:
$k=\frac{1}{4\pi {\epsilon }_0}\approx 8.99\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$

The force acts along the line joining the charges: like charges repel, opposite charges attract. Its size is proportional to each charge: doubling one doubles the force, doubling both quadruples it. It also varies inversely with the square of the distance: twice as far gives a quarter of the force, three times as far a ninth, half as far four times as much. Gravity follows the same inverse-square pattern.

Here ${\epsilon }_0\approx 8.85\times 10^{-12}F/m$ is the permittivity of free space. The size of $k$ shows how strong the electric force is: two charges of 1 C, 1 m apart, would repel with $9\times 10^9\mathrm{N}$, roughly the weight of 900 000 tonnes.

Equal and opposite forces

Each charge acts on the other, and the two forces form a Newton's third law pair. With ${\vec{F}}_{12}$ the force that $q_1$ exerts on $q_2$ and ${\vec{F}}_{21}$ the force that $q_2$ exerts on $q_1$,
${\vec{F}}_{12}=-{\vec{F}}_{21}$
The forces are equal in size even when the charges differ, because both contain the same product $q_1{q}_2$: charges of 1 µC and 4 µC, 30 cm apart, each feel $8.99\times 10^9\cdot 4\times 10^{-12}/0.30^2\approx 0.40\mathrm{N}$.

Equal forces do not mean equal motions: by Newton's second law each body accelerates at $a=F/m$, so a light charged bead is flung away from a heavy charged sphere that barely moves. The sign of the product gives the direction: $q_1{q}_2>0$ pushes the charges apart, $q_1{q}_2<0$ pulls them together. That is why the size formula uses $|q_1{q}_2|$.

Worked example: the pucks in the animation

Each puck carries a charge of size 2 µC, and their separation swings between 40 cm and 15 cm. First work out the product that never changes:
$k|q_1{q}_2|=8.99\times 10^9\cdot (2\times 10^{-6}{)}^2=0.03596\mathrm{N}\cdot {\mathrm{m}}^2$
At the closest approach, $r=0.15\mathrm{m}$, so
$F=\frac{0.03596}{0.15^2}=\frac{0.03596}{0.0225}\approx 1.60\mathrm{N}$

At the widest separation, $r=0.40\mathrm{m}$, the force is $0.03596/0.16\approx 0.22\mathrm{N}$. These are the extreme values on the $r$ and $F$ readouts. The distance changes by a factor of $40/15\approx 2.67$ and the force by $2.67^2\approx 7.1$, so the green arrows are about seven times longer at the closest approach. At 30 cm the readout passes 0.40 N, a quarter of the value at 15 cm.

After 6 s the right puck turns negative. The readouts repeat exactly, since the size depends only on $|q_1{q}_2|$ and $r$, but both arrows turn round to point towards each other. Labelled $F_{12}$ and $F_{21}$, they always have the same length.

Worked example: two charged balls on threads

Two identical small balls of mass 2.0 g hang from one point on insulating threads, 40 cm from the point of suspension to each centre. Given equal charges, they push each other apart and settle with their centres 12 cm apart. What is the charge on each ball?

Each ball is in equilibrium under its weight $mg$, the tension of the thread and the horizontal Coulomb push. The thread makes an angle $\theta$ with the vertical, with $\sin \theta =0.06/0.40=0.15$, so $\theta \approx 8.6^{\circ }$ and $\tan \theta \approx 0.152$. The vertical part of the tension balances the weight and its horizontal part the electric force, so
$F=mg\tan \theta =0.0020\cdot 9.81\cdot 0.152\approx 2.98\times 10^{-3}\mathrm{N}$

Solving Coulomb's law for the charge, with $q_1=q_2=q$, gives
$q=r\sqrt{\frac{F}{k}}=0.12\cdot \sqrt{\frac{2.98\times 10^{-3}}{8.99\times 10^9}}\approx 6.9\times 10^{-8}\mathrm{C}$
Each ball carries about 69 nC, or $4.3\times 10^{11}$ elementary charges. The electric push is only 15 % of the weight, yet it visibly holds the balls apart. If half of each charge leaked away, the force at 12 cm would drop to a quarter and the balls would sag closer together.

Several charges: forces add

With more than two charges, work out the force from each one separately, as if the others were absent, and add the results as vectors. This is the superposition principle:
${\vec{F}}_{\mathrm{net}}=\sum {\vec{F}}_i\text{(superposition)}$

Put a small charge of $-0.5\mu \mathrm{C}$ midway between the pucks when they are 30 cm apart. Each puck acts on it with $8.99\times 10^9\cdot 2\times 10^{-6}\cdot 0.5\times 10^{-6}/0.15^2\approx 0.40\mathrm{N}$. If both pucks are positive, they pull it in opposite directions and the net force is zero. If the right puck is negative, one pulls and the other pushes, both to the left: 0.80 N in all, towards the positive puck. The expert layer moves the small charge off the line, where the sum must be drawn as a vector.

Common mistakes

• Forgetting to convert units: writing 2 for each 2 µC charge makes the force $10^{12}$ times too big, and writing 15 for 15 cm makes it 10 000 times too small.
• Not squaring the distance: dividing by $r$ instead of $r^2$ turns the 1.60 N at 15 cm into 0.24, which is not even a force (its unit is N·m).
• Thinking that the bigger charge feels the bigger force: charges of 1 µC and 4 µC, 30 cm apart, both feel 0.40 N.
• Measuring $r$ between the surfaces: two spheres of radius 2 cm with surfaces 11 cm apart have centres 15 cm apart, and using 11 cm overestimates the force $(15/11{)}^2\approx 1.9$ times.
• Reading the sign as a size: $+2$ and $-2\mu \mathrm{C}$ at 15 cm give $k{q}_1{q}_2/r^2=-1.60\mathrm{N}$, but the minus sign only means attraction; the size is 1.60 N.

Where you meet it

Coulomb's law is at work wherever charge is moved on purpose. Laser printers and photocopiers lay a pattern of charge on a drum that picks up toner; electrostatic precipitators charge smoke particles in chimneys and collect them on plates of the opposite sign; car factories charge paint droplets so that they wrap around the car body.

Far more important is what you do not see: the electric force holds atoms and molecules together. In a hydrogen atom the proton pulls the electron, about $5.3\times 10^{-11}\mathrm{m}$ away, with $8.2\times 10^{-8}\mathrm{N}$, more than $10^{39}$ times their gravitational attraction. Chemistry and the stiffness of solids come from this force, tamed by the almost perfect balance of positive and negative charge in matter. Electric Field and Potential describes the same force through the field that each charge creates, and Capacitors shows how separated charges store energy.

Advanced

The law in vector form

The Intermediate level gives the size of the force, $F=k|q_1{q}_2|/r^2$, and a separate rule for its direction: like charges repel, opposite charges attract. As soon as several charges act at once, forces have to be added as vectors, so it pays to put size and direction into one formula. All this needs is a way to name positions.

Legend

$q_1$	First charge.	$\mathrm{C}$
$q_2$	Second charge.	$\mathrm{C}$
$O$	Origin: any fixed point from which positions are measured.
${\vec{r}}_1$	Position of $q_1$, an arrow from $O$.	$\mathrm{m}$
${\vec{r}}_2$	Position of $q_2$, an arrow from $O$.	$\mathrm{m}$
${\vec{r}}_2-{\vec{r}}_1$	Arrow from $q_1$ to $q_2$; its length is the distance $r$ between the charges.	$\mathrm{m}$
${\hat{r}}_{12}$	Unit vector from $q_1$ towards $q_2$: the arrow ${\vec{r}}_2-{\vec{r}}_1$ divided by its length $r$.
${\vec{F}}_{12}$	Force that $q_1$ exerts on $q_2$.	$\mathrm{N}$
${\vec{F}}_{21}$	Force that $q_2$ exerts on $q_1$: the same size as ${\vec{F}}_{12}$, opposite direction.	$\mathrm{N}$

In Figure 1 the charges $q_1$ and $q_2$ sit at the positions ${\vec{r}}_1$ and ${\vec{r}}_2$, arrows drawn from an origin $O$. Look at the triangle formed by the three orange arrows. Going from $O$ to $q_1$ and then on to $q_2$ ends where the direct arrow ${\vec{r}}_2$ ends, so ${\vec{r}}_1+({\vec{r}}_2-{\vec{r}}_1)={\vec{r}}_2$ and the third side, ${\vec{r}}_2-{\vec{r}}_1$, is an arrow from $q_1$ to $q_2$. It does not depend on where $O$ is, only on where the charges are, as a force between them should.

The length of this arrow is the distance between the charges, $r=|{\vec{r}}_2-{\vec{r}}_1|$. Dividing the arrow by its own length keeps its direction and makes its length 1. The result is the unit vector (a hat marks a vector of length 1)
${\hat{r}}_{12}=({\vec{r}}_2-{\vec{r}}_1)/r$
drawn as the short blue arrow, pointing from $q_1$ towards $q_2$. With it, the force that $q_1$ exerts on $q_2$ is
${\vec{F}}_{12}=\frac{k{q}_1{q}_2}{r^2}{\hat{r}}_{12}$
where $k\approx 8.99\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$ is Coulomb's constant. The charges now enter with their signs, not as sizes.

The signs then take care of the direction. Both charges in Figure 1 are positive, so $q_1{q}_2>0$: the number in front of ${\hat{r}}_{12}$ is positive, and ${\vec{F}}_{12}$, the green arrow on $q_2$, points along ${\hat{r}}_{12}$, away from $q_1$. That is repulsion. If one charge were negative, $q_1{q}_2<0$ and the same formula would turn the force round, towards $q_1$: attraction. No separate rule is needed.

Newton's third law is built in as well. The force on $q_1$ is the same formula with the labels swapped. Swapping turns ${\hat{r}}_{12}$ into ${\hat{r}}_{21}=-{\hat{r}}_{12}$ but leaves the product $q_1{q}_2$ and the distance $r$ unchanged, so ${\vec{F}}_{21}=-{\vec{F}}_{12}$. In Figure 1 the green arrow ${\vec{F}}_{21}$ on $q_1$ is as long as ${\vec{F}}_{12}$ and points the other way.

The force is also central: it acts along the line joining the charges, and its size depends only on their distance. Such a force is conservative, because the work it does between two separations does not depend on the path taken, so it has a potential energy, found in the section on energy below.

In SI units Coulomb's constant is written $k=1/(4\pi {\epsilon }_0)$, where ${\epsilon }_0\approx 8.85\times 10^{-12}F/m$ is the permittivity of free space. The factor $4\pi$ is the area of a sphere of radius 1. Placing it in $k$ looks clumsy here, but it makes the $4\pi$ disappear from Gauss's law and from the formulas for capacitors, which are used far more often.

Charles-Augustin de Coulomb established the law in 1785 by measuring the twist of a torsion balance. Until 2019 the SI fixed $k$ at exactly $10^{-7}{c}^2\approx 8.98755\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$, where $c$ is the speed of light. Since then the elementary charge $e$ has an exact value and ${\epsilon }_0$ is measured instead; the measurement agrees with the old exact value to better than a part in a billion.

Superposition and the electric field

With only two charges the vector form just repackages what we already knew. Its value shows when a third charge joins them. Experiment shows that electric forces simply add: the force on a charge is the vector sum of the forces that each of the others would exert on it alone, unchanged by the presence of the rest. This principle of superposition is a second experimental fact, as basic to electrostatics as the inverse-square law itself.

Legend

$q_1$	Charge of the left puck, positive in both panels.	$\mathrm{C}$
$q_2$	Charge of the right puck: positive in the left panel, negative in the right.	$\mathrm{C}$
$q_3$	Fixed negative charge below the pucks.	$\mathrm{C}$
${\vec{F}}_{13}$	Force of the left puck on $q_3$.	$\mathrm{N}$
${\vec{F}}_{23}$	Force of the right puck on $q_3$.	$\mathrm{N}$
${\vec{F}}_3$	Net force on $q_3$: the vector sum ${\vec{F}}_{13}+{\vec{F}}_{23}$.	$\mathrm{N}$
$r$	Distance from each puck to $q_3$, the same for both.	$\mathrm{m}$
$h$	Depth of $q_3$ below the line of the pucks.	$\mathrm{m}$
$a$	Distance from each puck to the midpoint between them: half their separation.	$\mathrm{m}$

Figure 2 uses the set-up of the animation's expert layer. Two pucks, each carrying a charge of size $q=2\mu \mathrm{C}$, slide along a straight rail. A fixed charge $q_3=-0.5\mu \mathrm{C}$ sits a depth $h=14\mathrm{cm}$ below the rail, under the midpoint between the pucks. The figure shows the moment of closest approach, when the pucks are 15 cm apart and each is $a=7.5\mathrm{cm}$ from the midpoint.

Start with the force of one puck on $q_3$. The puck, the midpoint and $q_3$ form a right-angled triangle with the sides $a$ and $h$, so each puck is the same distance from $q_3$:
$r=\sqrt{a^2+h^2}=\sqrt{0.075^2+0.14^2}\approx 0.159\mathrm{m}$
Coulomb's law gives the size of each force, which acts along the dashed line joining the puck to $q_3$:
$F=\frac{8.99\times 10^9\cdot 2\times 10^{-6}\cdot 0.5\times 10^{-6}}{0.159^2}\approx 0.356\mathrm{N}$
These are the gold arrows ${\vec{F}}_{13}$ and ${\vec{F}}_{23}$. They are equally long because the distances are equal; only their directions differ.

To add them, split each force into a part along the rail and a part at right angles to it. The force lies along the slanted side $r$ of the triangle, so its part along the rail is $Fa/r$ and its part at right angles to the rail is $Fh/r$.

In the left panel both pucks are positive, so each pulls the negative $q_3$ towards itself, up and to one side. The sideways parts, $Fa/r$ each, point in opposite directions and cancel. The upward parts, $Fh/r$ each, add. The sum ${\vec{F}}_3={\vec{F}}_{13}+{\vec{F}}_{23}$, the purple diagonal of the dashed parallelogram, therefore points straight up, with the size
$F_3=2\cdot 0.356\cdot 0.14/0.159\approx 0.63\mathrm{N}$

In the right panel the right puck is negative and pushes $q_3$ away, so ${\vec{F}}_{23}$ now points down and to the left. This time the vertical parts cancel and the sideways parts add. The sum, $F_3=2\cdot 0.356\cdot 0.075/0.159\approx 0.34\mathrm{N}$, points along the rail, towards the positive puck.

The same sums change as the pucks move. For the like pair the upward pull grows steadily as the pucks close in, from 0.17 N when they are 40 cm apart to 0.63 N at 15 cm. For the unlike pair the sideways pull, $2Fa/r=2kq|q_3|a/(a^2+h^2{)}^{3/2}$, is largest not at the closest approach but at $a=h/\sqrt{2}$, where its derivative with respect to $a$ is zero. That is a separation of $\sqrt{2}\cdot 0.14\approx 0.20\mathrm{m}$, and the pull there is 0.35 N. Closer pucks pull harder, but their forces also turn towards the vertical, where the unlike pair's forces cancel.

Superposition also lets us describe what the charges do to a point before anything is placed there. Put a small test charge $q_0$ at the point and divide the total force on it by $q_0$. The result, $\vec{E}=\vec{F}/q_0$, no longer depends on the test charge: it is the electric field, measured in newtons per coulomb.

For charges $q_j$ at the positions ${\vec{r}}_j$, the field at the point $\vec{r}$ is
$\vec{E}(\vec{r})={\sum }_j\frac{k{q}_j(\vec{r}-{\vec{r}}_j)}{|\vec{r}-{\vec{r}}_j{|}^3}$
Each term is Coulomb's law in vector form: the vector $\vec{r}-{\vec{r}}_j$ divided by the cube of its length equals the unit vector from $q_j$ to the point divided by the square of the distance. For charge spread out with a density $\rho$ (charge per unit volume) the sum becomes a volume integral. Once the charges and their positions are known, this formula settles every electrostatic force; Gauss's law, potentials and field lines are shortcuts to the same answers.

Seen from a distance $d$ much larger than their separation, the like pair acts as one charge of 4 µC, whose field falls as $1/d^2$. The unlike pair has no net charge: its two fields nearly cancel, and what remains falls faster, as $1/d^3$. A pair of equal and opposite charges like this is called an electric dipole.

Why the exponent is exactly 2

Nothing in the vector form forces the power of $r$ to be 2; a law with $1/r^{2.01}$ would look much the same. The exact 2 matters because it turns into Gauss's law, one of the basic equations of electromagnetism, and because it can be tested with extraordinary precision. The key quantity is the flux. For a small flat patch of area $dA$ that faces the field squarely, the flux is $EdA$, the field strength times the area. If field lines are drawn with a density proportional to $E$, the flux counts the lines that cross the patch.

Legend

$q$	Point charge at the tip of the cone.	$\mathrm{C}$
$d\Omega$	Solid angle of the cone: the area it cuts from a sphere of radius 1.	$\mathrm{sr}$
$r$	Distance from $q$ to the nearer patch.	$\mathrm{m}$
$dA$	Area of the nearer patch, which faces the charge squarely.	${\mathrm{m}}^2$
$E$	Field of $q$ at the nearer patch.	$N/C$
$E/4$	Field at twice the distance: a quarter of $E$.	$N/C$

Figure 3 shows how the exponent and the geometry of space fit together. A narrow cone leaves the point charge $q$. Its opening is measured by the solid angle $d\Omega$: the area the cone cuts from a sphere of radius 1 centred on its tip, in steradians (a whole sphere is $4\pi$). At the distance $r$ the cone cuts a small patch facing the charge, of area $dA=r^{2}d\Omega$. At $2r$ the cone is twice as wide in both directions, so the patch there has four times the area, $4dA$.

Now compare the fields at the two patches, the two blue arrows. Coulomb's law gives $E=kq/r^2$ at the near patch and a quarter of that, $E/4$, at the far one. The flux through the near patch is $EdA=(kq/r^2)(r^{2}d\Omega )=kqd\Omega$, and the flux through the far patch, $(E/4)(4dA)$, is the same. The $r^2$ in the denominator of the law cancels the $r^2$ in the area. So the flux through any patch the cone cuts depends only on the solid angle $d\Omega$, not on how far away the patch is.

Next, surround $q$ with a sphere of radius $r$ centred on it. On this sphere the field has the same size everywhere and crosses the surface at right angles, so the total flux is the field times the whole area:
$\oint \vec{E}\cdot d\vec{A}=\frac{kq}{r^2}\cdot 4\pi r^2=\frac{q}{{\epsilon }_0}$
The circle on the integral sign means the surface is closed. The vector $d\vec{A}$ is a patch of the surface, with length $dA$ and pointing outwards, and the dot product keeps only the part of the field that crosses the patch. The radius has dropped out, so every sphere gets the same flux, and the $4\pi$ placed in $k$ cancels the $4\pi$ of the sphere's area.

The shape of the surface does not matter either. Cut any closed surface around $q$ into patches, each seen from $q$ within some solid angle $d\Omega$. A tilted patch is larger, but the field crosses it at a slant, and the two effects cancel, so each patch still receives the flux $kqd\Omega$. The solid angles add up to $4\pi$, which gives $q/{\epsilon }_0$ again. For a charge outside the surface, each thin cone enters the surface as often as it leaves it, with flux in and flux out equal: the charge contributes nothing. That is Gauss's law: the flux out of any closed surface equals the charge inside divided by ${\epsilon }_0$.

The argument works only for the exponent 2. For a force $F\propto 1/r^{2+\delta }$ the flux through a sphere would vary as $r^{-\delta }$. It would differ between two spheres with no charge between them, as if field lines appeared or vanished in empty space.

Two consequences make the exponent measurable. First, a uniformly charged thin spherical shell acts outside as if all its charge sat at its centre, and exerts no force at all on a charge inside. To see the inside case with the idea of Figure 3, draw a thin double cone through a point inside the shell. It cuts the shell in two patches, at the distances $r_1$ and $r_2$. A straight line through a sphere meets it at the same angle at both ends, so the two patches are tilted alike, and their areas, and so their charges, are in the ratio $r_1^2:r_2^2$. Each patch pulls with a force proportional to its charge divided by the square of its distance, so the small near patch pulls exactly as hard as the large far one, in the opposite direction. Second, the field inside an empty closed conductor of any shape is zero, whatever charge is put on the conductor.

Looking for these zeros is a null test. Instead of measuring the force at many distances, which is hard to do precisely, one charges a closed conductor and looks for any field or charge inside it. If the exponent differed from 2, something would show up, so finding nothing is far more sensitive. Henry Cavendish did this in 1773 and found that the exponent differs from 2 by less than 0.02; a shell experiment of 1971 pushed $\delta$ below a few parts in $10^{16}$.

At atomic distances, around $10^{-10}\mathrm{m}$, the energy levels of hydrogen, computed with the $1/r$ potential plus small quantum corrections, match measurement to extraordinary precision. In modern terms, an exact inverse-square law means that the photon, the particle that carries the electric force, has no mass.

Energy and the force–distance graph

Because the force changes with distance, Newton's second law gives the acceleration at each moment but not, without much work, the speed after a long push. Work and energy give it in one step, and a graph of force against distance shows where the energy comes from.

Legend

$r$	Separation of the charges.	$\mathrm{m}$
$F$	Size of the repulsive force, $k{q}^2/r^2$.	$\mathrm{N}$
$U$	Potential energy of the pair, $k{q}^2/r$, zero at infinite separation.	$\mathrm{J}$
$r_a$	Starting separation.	$\mathrm{m}$
$r_b$	Final separation.	$\mathrm{m}$
$F_a$	Force at $r_a$.	$\mathrm{N}$
$F_b$	Force at $r_b$.	$\mathrm{N}$
$U_a$	Potential energy at $r_a$.	$\mathrm{J}$
$U_b$	Potential energy at $r_b$.	$\mathrm{J}$
$W$	Work done by the repulsion from $r_a$ to $r_b$: the shaded area, equal to the drop $U_a-U_b$.	$\mathrm{J}$

Figure 4 has two graphs for the two equal charges $q=2\mu \mathrm{C}$ of the pucks, both against their separation $r$. The upper graph shows the size of the force, $F=k{q}^2/r^2$. The lower graph shows the potential energy of the pair, $U=k{q}^2/r$, which the next paragraphs derive. Here $k{q}^2=8.99\times 10^9\cdot (2\times 10^{-6}{)}^2\approx 0.03596\mathrm{N}\cdot {\mathrm{m}}^2$. The pucks move between $r_a=15\mathrm{cm}$ and $r_b=40\mathrm{cm}$. On the upper graph the dots mark $F_a\approx 1.60\mathrm{N}$ and $F_b\approx 0.22\mathrm{N}$; on the lower graph they mark $U_a\approx 0.240\mathrm{J}$ and $U_b\approx 0.090\mathrm{J}$.

The force curve is steepest at small $r$. Differentiating $k{q}^2{r}^{-2}$ gives the slope $dF/dr=-2F/r$: about $-1.1N/m$ at 40 cm and $-21N/m$ at 15 cm, a factor of $(40/15{)}^3\approx 19$. The expert graph of the animation plots the same curve on ordinary axes. As the pucks swing between 40 and 15 cm, its green dot slides along it between 0.22 N and 1.60 N and shoots up at the left end. On logarithmic axes the same curve would be a straight line of slope $-2$, the exponent of the law.

Now the work. When the pucks move apart by a small step $dr$, the repulsion does the work $Fdr$: a thin strip of area under the upper curve. Adding the strips from $r_a$ to $r_b$ gives the shaded area $W$ of Figure 4:
$W={\int }_{r_a}^{r_b}\frac{k{q}_1{q}_2}{r^2}dr=k{q}_1{q}_2\left(\frac{1}{r_a}-\frac{1}{r_b}\right)$

The answer is one function of $r$ taken at the start minus the same function at the end. Calling that function the potential energy, $U(r)=k{q}_1{q}_2/r$, the work is $W=U(r_a)-U(r_b)$: the force does work by using up potential energy. A constant could be added to $U$ without changing any difference; the form above makes $U$ zero at infinite separation, where the force vanishes. Read the other way, the force is minus the slope of the energy, $F=-dU/dr$ along the line joining the charges.

Between 15 and 40 cm the shaded area is $W=0.03596\cdot (1/0.15-1/0.40)\approx 0.150\mathrm{J}$. On the lower graph the same 0.150 J appears as the drop from $U_a$ to $U_b$, marked by the bracket $W$: $0.240-0.090=0.150\mathrm{J}$.

The animation shows both signs. In the repelling phase, the hand that pushes the like pucks together from 40 to 15 cm does 0.150 J of work against the repulsion, and the pair stores it. In the attracting phase $q_1{q}_2<0$, so $U$ is negative: it drops from $-0.090$ to $-0.240\mathrm{J}$ as the pucks approach from 40 to 15 cm, and now the field does the 0.150 J of work instead.

Worked problem: letting the pucks go

Suppose each puck has a mass of 20 g and the like-charged pair is released from rest, 15 cm apart, on the frictionless rail. Find the initial acceleration of each puck, and its speed when the pucks are 40 cm apart and when they are very far apart.

At the start each puck feels the force $F_a\approx 1.60\mathrm{N}$, so by Newton's second law its acceleration is $1.60/0.020\approx 80m/{\mathrm{s}}^2$, about eight times $g$. The force weakens as the pucks separate, so the acceleration is not constant and the formulas for constant acceleration do not apply. The speeds come from energy instead.

Two facts are needed. The forces on the two pucks are equal and opposite, so the total momentum stays zero; with equal masses, the pucks move with equal speeds in opposite directions about a fixed midpoint. And energy is conserved: the kinetic energy of both pucks together equals the drop of the lower curve of Figure 4 from $r_0=0.15\mathrm{m}$ to the current separation $r$:
$2\cdot \frac{1}{2}m{v}^2=m{v}^2=U(r_0)-U(r)$

At 40 cm the drop is the 0.150 J found above, so $m{v}^2=0.150\mathrm{J}$ and $v=\sqrt{0.150/0.020}\approx 2.74m/s$. Very far away $U\to 0$, so the whole 0.240 J becomes kinetic energy: $m{v}^2=0.240\mathrm{J}$ and $v=\sqrt{0.240/0.020}\approx 3.46m/s$.

By the time the gap has grown from 15 to 40 cm, the pair has released the fraction $1-U_b/U_a=1-15/40=0.625$ of its stored energy. Speed grows as the square root of kinetic energy, so the pucks already have $\sqrt{0.625}\approx 0.79$ of their final speed: an inverse-square push acts hard and briefly. The attracting pair runs the same film backwards: released from rest 40 cm apart, the pucks pass the 15 cm mark at 2.74 m/s each, moving towards each other.

Limits of validity

Coulomb's law is a law of charges at rest, in empty space, at ordinary distances. Each of these conditions marks a limit.

Moving charges also exert magnetic forces. For two charges moving side by side at the speed $v$, the magnetic force is a fraction of order $v^2/c^2$ of the electric one. In addition, when one charge moves, the change reaches the other only after the travel time $r/c$, 0.5 ns across 15 cm. For the pucks of the animation, slower than 0.07 m/s, $v^2/c^2\approx 5\times 10^{-20}$; even at the 3.46 m/s of the worked problem it is only about $10^{-16}$. In an accelerator beam, by contrast, the electrons move so close to $c$ that the magnetic attraction between them nearly cancels their electric repulsion. Charge itself is the same at any speed.

In an insulating medium the molecules polarise: their charges shift slightly, and the shifted charges partly screen the two charges from each other. The force is divided by the relative permittivity ${\epsilon }_r$ of the medium, about 80 for water. This is why salt dissolves. Separating a sodium ion and a chloride ion that touch, 0.28 nm apart, takes about 5 eV in vacuum but, crudely, $5/80\approx 0.06\mathrm{eV}$ in water, only a few times the thermal energy at room temperature (0.026 eV).

Air limits how much charge a body can hold: it breaks down and sparks in fields near $E_{\max }=3MV/m$. The field at the surface of a sphere of radius $R$ carrying the charge $Q$ is $kQ/R^2$, so the charge cannot exceed about $E_{\max }{R}^2/k$: 33 nC for $R=1\mathrm{cm}$. Microcoulomb charges like the animation's need large spheres (2 µC needs a radius of about 8 cm) or a vacuum.

At nuclear distances the force is enormous: two protons 2 fm apart ($1\mathrm{fm}=10^{-15}\mathrm{m}$) repel with about 58 N, the weight of a 6 kg mass, acting on a single particle. Only the short-range strong force holds nuclei together; in heavy nuclei, with many protons, the electric repulsion gains ground, which makes fission possible. Nor is the proton a point: its charge is spread over a radius of about 0.84 fm. Finally, in quantum electrodynamics, the quantum theory of the electric force, short-lived electron–positron pairs partly screen every charge. Below about $10^{-12}\mathrm{m}$ this screening thins out and the force grows slightly faster than $1/r^2$, an effect seen in the hydrogen spectrum.

Where this leads

The field introduced above is the main tool of Electric Field and Potential, where the energy $U=k{q}_1{q}_2/r$ becomes the potential $V=kQ/r$, energy per unit charge, and Gauss's law gives fields without summing. Capacitors uses ${\epsilon }_0$ and ${\epsilon }_r$. Gravitation has the same inverse-square form, so what Gravitation and Orbits shows about circular orbits carries over to an electron circling a nucleus in Atomic Models: the Bohr Atom. Magnetic Force on Moving Charges adds the force between moving charges, which relativity ties to this one, and Electromagnetic Waves shows how changes spread at $c=1/\sqrt{{\epsilon }_0{\mu }_0}$, where ${\mu }_0$ is the magnetic constant. Particles and Forces: the Standard Model describes the electric force as the exchange of photons.