# Why does Coulomb's constant have units?

I think of Coulomb's constant as a conversion factor (not sure if this is correct). Kind of like how you would do calculations in kg and then times it by the conversion constant to convert your answer to pounds. The conversion factor would be $2.2\: \mathrm{lbs/kg}$.

Since the units for Coulomb's constant is $\mathrm{N \cdot m^2/C^2}$, would it make sense to define the Newton as:

$1\:\text{Newton} = \frac{1}{1/1\: \mathrm{meter^2} \cdot 1\: \mathrm{Coulomb^2}}$

Would the above definition be valid?

EDIT: So if $k$ is not a conversion factor since the above definition for a Newton is invalid and $k$ is not just a scaling factor, since it has units, then what is it? If its just a proportionality constant to adjust the magnitude then why does it have units? Shouldn't it be a unit less constant?

EDIT: So $k$ is not just a scaling factor (since it has units) and its not a conversion factor since a Newton can't be expressed as the other units. So if its unit just exists so that things cancel out "nicely" doesn't this make dimensional analysis useless since you can add in random constants and units to cancel out whatever you want?

My question is not about the meaning of $k$. Its about its units.

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Your definition seems to be dimensionally incorrect ! –  Rijul Gupta Jan 14 at 8:57
@rijulgupta Why? –  dfg Jan 14 at 13:37
Newton is unit of $force$ you have equated it with $meter^2/Coulomb^2$ or maybe $1/{meter^2 × Coulomb^2}$ , which are both wrong. –  Rijul Gupta Jan 14 at 13:43
But doesn't that follow from Coloumb's Law since Coloumb's constant is a "conversion factor"? I.e. See the pound kg example I gave. –  dfg Jan 14 at 13:50
Similar question with temperature: physics.stackexchange.com/questions/60830/… –  jinawee Jan 16 at 19:07

## 8 Answers

When the electrostatic force was originally being studied, force, mass, distance and time were all fairly well understood, but the electrostatic force and electric charge were new and exotic. In the cgs system, the charge was defined in relation to the resulting electrostatic force (it's called a Franklin (Fr) an "electrostatic unit" (esu or) sometimes a statCoulomb (statC)).

In that system, we express the force on one charged particle by another as $F_E=\frac{q_1 q_2}{r^2}$ where the unit of charge is the esu, the unit of force is the dyne and the unit of distance is the centimeter. In the MKS system (now called SI), we would write $F_E = k_e\frac{q_1 q_2}{r^2}$ where the unit of charge is the coulomb, the unit of force is the newton, and the unit of distance the meter. It would seem that if things are equivalent, then $k_e$ is indeed just a conversion factor, but things are definitely not equivalent.

A little history is probably useful at this point. Before 1873, when the cgs system was first standardized, it finally made a clear distinction between mass and force. Before that, it was common to express both in terms of the same unit, such as the pound. So if you think of it, people still say things like "I weigh 72 kg" rather than "I weigh 705 N here on the surface of Earth" and they also say $1 \mathrm { kg} = 2.2\mathrm{ lb}$ confusing mass and weight (the cgs Imperial unit of mass is actually the slug).

This is important, because there is a direct analogy to the issue of units of charge and to your question about the units of $k_e$. The Franklin is defined as "that charge which exerts on an equal charge at a distance of one centimeter in vacuo a force of one dyne." The value of $k_e$ is assumed to be 1 and is dimensionless in the cgs system.

In cgs, the unit of charge, therefore, already implictly has this value of $k_e$ built in. However in the SI units, they started with Amperes and derived Coulombs from that and time ($C=It$). The resulting units of $k_e$ are a result of that choice.

So although the physical phenomenon is the same, it is the choice of units that either gives $k_e$ dimension or not.

See this paper for perhaps a little more detail on how this works in practice.

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So If I understand correctly, it is a conversion factor between unit systems, so $Newton = k∗Coloumb^2/m^2$. Combinining this with the definition of a Newton as $Newton=G∗kg^2/m^2$ (from the equation for gravity) can't you show that you can convert from kg to charge and consequently use kilograms to measure charge? –  dfg Jan 16 at 22:23
@dfg: it's only a conversion in the context of this particular phenomenon. For example, I might have an automobile that gets 32 miles per gallon (of fuel), which is the common way to express fuel economy in the United States. It doesn't mean that I can redefine miles or gallons generally except in the context of this particular vehicle. They are not generally related units; they are simply both factors in this particular phenomenon. Does that help? –  Edward Jan 16 at 22:54
@ dfg : You can also start calling conventional red yellow, but you will be living in your own world. Are you attmepting to make a new units system ? –  Rijul Gupta Jan 16 at 22:59
@Edward I don't really understand when you can extend a definition only valid in a specific context to general cases. For example, in $F = ma$, a newton is defined as $m/s^2$ in that specific context. But this definition is also extended to the general case (a newton is always equal to $m/s^2$). So when is it okay to extend a definition to all cases? –  dfg Jan 24 at 3:08
@dfg: It's a fair question. Generally speaking, definitions are the province of international standards bodies. So for example, the definition of a meter has evolved over time and other units continue to be refined for various reasons; most often due to improvements in measurement technology. One could even imagine proposing replacement systems for electromagnetic units, but uptake of any system is more social than technical. –  Edward Jan 24 at 16:53

Systems of units are in some sense flexible and optional.

The relationship

$$\text{Electrostatic force twixt two point-like charges} \propto \frac{(\text{one charge})\times (\text{the other charge})}{(\text{distance between them})^2} \tag{1}$$

is an experimental fact.

In SI, we have units for Force, distance and charge such that (1) is not dimensionally consistent with a dimensionless constant of proportionality. So, $k$ must have dimensions of $\frac{\mathrm{N} \cdot \mathrm{m}^2}{\mathrm{C}^2}$ as well as having a numeric value.

But we could do it another way. Consider the "Statcoulomb". In Gaussian units the unit of charge is defined such that Coulomb's Law has a dimensionless, unit constant of proportionality. $$F = \frac{q_1 \, q_2}{r^2} \,.$$ This is a perfectly valid way to do physics. In essense we have folded $\sqrt{k}$ into each of the charges: $$\text{Statcoulombs} \sim \sqrt{k} \,\text{Coulombs} \,.$$ That makes a Statcoulomb a pretty funny unit when expressed in SI terms, but then the Coulomb is a pretty odd unit expressed in Guassian terms. Each system should be understood in it's own context.

A great many words have been spilled arguing that one set of units is better than another or vice-versa.

In my business (particle physics) it is common to work in units where $c = \hbar = 1 \,\text(dimensionless)$. This gives energy, mass, and momentum the same units (inverse distance, actually) and loses many of the checks that help young physicists keep track of the difference between these quantities, but keeps the scribbling down and simplifies the form of many equations. (By the way, cosmologist often add $G = 1\,\text(dimensionless)$ to the mix.

The moral of the story is, 'Don't read too much significance into the units of "constants", because they depend on the system of units you chose.'

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So according to your definition, $Newton = Statcoloumbs^2/m^2$. Combinining this with the definition of a Newton as $Newton = G*kg^2/m^2$ (from the equation for gravity) can't you show that you can convert from kg to charge and consequently use kilograms to measure charge? –  dfg Jan 16 at 22:15
No. $G$ is also a dimensional constant in SI units (and in Gaussian units). These constant (and infact all physically meaningful quantities) have both magnitude and dimension (you have to allow that "dimensionless" is a dimension for that sentence to hold, but it saves on verbal gymnastics). Two properties and you can't interchange them even though changing one changes the other. –  dmckee Jan 16 at 22:29
But can't you get rid of $G$ by defining a "Statkilogram" and then do what I said above to show a Statkilogram = Statcoloumb? –  dfg Jan 16 at 22:33
Also Gaussian units use a cgm base, so the natural units of Coulomb's law in that system are dynes, statcoulombs and cm. This is one part of why I used the similarity symbol $\sim$ rather than equality in the third block equation. –  dmckee Jan 16 at 22:34
Frankly, I don't know what you are trying to achieve, but I am sure that you are confusing yourself. Yes, you can often define a system of units such that different physical quantities have formally equivalent units (as with energy, mass and momentum in natural units), but they remain different things. That is why we insist that students learn to work in a system of units that emphasizes the differences before letting them play in the notationally convenient but dimensionally conflating systems favored by professionals. The distinction between different types of measurements are important. –  dmckee Jan 16 at 22:40

Force is a vector quantity defined mathematically as the rate of change of momentum, that is
$\vec F = \dfrac{d\vec p}{dt} \tag{1}$ where $\vec p= m\vec v$ in classical mechanics.
The unit of force in Si is "newton". One "newton" is the amount of force requires to accelerate one kilogram mass at a rate of one meter per second squared. You could make your own set of units by saying : I define one newton as the force required to accelerate 2 Kg of mass through 1m/s^2 then you will have to modify eqn 1 as $\vec F=\dfrac{1}{2}m \vec a$ In general Newton's second law can be stated as $\vec F=k \ m\vec a$ where $k$ depends upon the units of measurement.you could also say $\vec F \propto d\vec p/dt$ and $k$ appears as a constant of proportionality. It should be noted here that '$k$' is dimensionless constant.
Let our system of units be Si for the sake of simplicity and eqn 1 be valid. Let us measure the force by a spring scale
The spring obeys Hooke's law which states that the magnitude of applied force is directly proportional to the displacement of the spring, that is $F=k X$ where $x$ is the displacement and $k$ is the proportionality constant. This time $k$ is not dimensionless why it is so? it is so because force is not measured in metres it is measured in newtons. Suppose the spring is manufactured in such a way that a displacement of one metre represent one newton of force then $k$ will have $1$ magnitude. Would it be appropriate to define to define one newton = one metre so as to make $k$ dimensionless? NO! doing this will make all other equations involving '$F$' dimensionally incorrect e.g $F=kma$ here k will have to be dimensional constant.
Similarily if you define "one newton = $1 \text{Newton} = \frac{1}{1/1 meter^2 * 1 Coloumb^2}$ then $F=ma$ will have to be changed to $F=\dfrac{s^2}{Kg m} \times {1/meter^2 *coloumb^2} \times mass \times acceleration$

In coloumb's law the constant $k_e$ has units and magnitude.
$1.$ It has units to make the equation dimensionally correct.
$2.$ it is not $1$ in magnitude because the magnitude of $\dfrac{q_1q_2}{r^2}$ when $q_1$,$q_2$ and $r$ all are one the force in newtons is found to be $9\times 10^9$ in magnitude so $k$ is give this value for proper callibration. $k$ can be seen as a proportionality constant or a scaling factor or a dimensional constant or all. $k$'s magnitude is 9*10^9 only because we defined one newton as 1Kg m/s^2. If you say 9*10^9 newtons= one dgp then in dgp system $k$ will attain one magnitude.

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But then why let $1 newton = 1 m/s^2$? Why not have a constant of 1 with the appropriate units to cancel out the units on the right side? What I'm getting at I guess is why do we in some cases choose to let units be equal when theyre on opposite signs of an equal sign, but in other cases use a conversion factor to cancel units? –  dfg Jan 16 at 22:29
@dfg In Si system we ought to represent every derived unit in terms of base units. You can have a constant of 1 with appropriate units but doing this will make all other equations dimensionally incorrect e.g setting $1N=1metre$ will change units of $k$ in $F=k \ Qq/r^2$. you can do so for your ease of calculations sometimes as Dmckee explained by changing units of charge to stattcoloumbs. –  user31782 Jan 17 at 3:43

$K$ carries the unit of $\mathrm{Nm^2/C^2}$ because the SI has already defined a unit for charge - the Coulomb.

In addition to the units for length,time and mass, the SI makes the somewhat redundant definition of Ampère, the unit for current. It is defined via the attractive force between to parallel wires. It also defines the Coulomb as $C\equiv A\cdot s$

That mean, for three of the four quanties appearing in the Coulomb-Force law $$F = K\frac{q_1q_2}{r^2}$$ the SI fixes their units ($F,q,r$). The unit of $K$ must be adjusted to produce Newtons on both sides. The numerical value is essentially fixed by measuring the force of attraction between know charges.

Compare this to the case of the Gaussian (cgs) unit system. No definition of charge is made a priori and $K\equiv 1$ is chosen. It follows that the unit of charge is $$[Q] = \left([F][r]^2\right)^{1/2}=\frac{\sqrt{g\cdot \mathrm{cm^3}}}{s} \equiv 1\: \mathrm{esu}\quad\text{electro-static-unit}$$

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Yes your definition is invalid, it fails on dimensional terms alone.

So if K is not a conversion factor since the above definition for a Newton is invalid and K is not just a scaling factor, since it has units, then what is it?

K is simply a constant, for example look at the formula for resistance. We know that :
1. Resistance is directly proportional to length of conductor.
2. Resistance is inversely proportional to Area of cross section of conductor.
Thus we conclude that : $R \propto l/A$ . To remove the sign of proportionality we use a constant, which balances the equation. The equation now becomes $R = \rho l/A$ . Here $\rho$ is neither a scaling nor a conversion factor, it is simply a constant. in the same way we have: $F \propto Q^2/R^2$ This is solved by using coulombs constant which is now represented as $\frac{1}{4 \pi \epsilon_0}$

If its just a proportionality constant to adjust the magnitude then why does it have units? Shouldn't it be a unit less constant?

Nobody said that proportionality constant have to be unitless and/or dimensionless we have proportionality constants of all types. For ex :
1. $\rho$ called specific resistance has both dimensions and units.
2. $\mu$ called coefficient of friction has neither units nor dimensions.
3. $\theta$ called angle, this one is not a constant of proportionality but you can notice it has no dimensions but units.

So k is not just a scaling factor (since it has units) and its not a conversion factor since a Newton can't be expressed as the other units.

Newton can be expressed as other units both with and without scaling for ex :
1. $1 N = 1 kg m s^{-2}$
2. $1 N = 10^5 dynes$

So if its unit just exists so that things cancel out "nicely" doesn't this make dimensional analysis useless since you can add in random constants and units to cancel out whatever you want?

It most certainly does not render dimensional analysis useless. Constants such as Coulomb's constant, specific resistance are added to remove sign of proportionality and to derive equations from emperical formulae, you cannot just add them whimsically wherever you want and cancel units.
You perform dimensional analysis over an existing equation ( not an emperical formula stating proportionality rather than equivalence).

My question is not about the meaning of k. Its about its units.

It's units can be several depending on which system you use, but in SI we use $N m^2 C^{-2}$, it is dimensionally $[ML^3A^{-2}T^{-4}]$.

Furthermore, from wikipedia history of coulombs law states that it indeed was first an emperical law and then converted to an equation by using constant of proportionality $k$

Finally, in 1785, the French physicist Charles-Augustin de Coulomb published his first three reports of electricity and magnetism where he stated his law. This publication was essential to the development of the theory of electromagnetism.[12] He used a torsion balance to study the repulsion and attraction forces of charged particles, and determined that the magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

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Short answer: Yes it can be seen as a conversion factor, but the numerical value will depend on the choice of units used.

Depends on the units we use. Coulomb's law is usually given by $$F=k_e\frac{q_1q_2}{r^2}$$ where one should use such units that the equations holds by dimensional analasys.

If we for example take $q_1$ and $q_2$ to be in Coulombs and $r^2$ to be in meters then $k_e=8.987...\cdot10^9 \frac{Newton\cdot Meters}{Coulomb^2}$.

If we take $q_1$ and $q_2$ to be in microCoulombs and $r^2$ to be in centimeters then $k_e=8.987...10^{-1} \frac{Newton\cdot centimeters}{microCoulomb^2}$. (Notice that $10^9$ became $10^{-1}$.)

If we take some other units the numerical value and it's dimensions will change to something other.

Can we really interpret it as an conversion factor?

Yes we can. I will take a simpler example first. Einsteins most famous equations is given by $E=m\cdot c^2$ where $m$ is the mass, $E$ is the energy and $c$ is the speed of light. Here $c^2$ is the conversion factor telling us how much energy we have if someone gives us $x$ kg. Or if we have $y$ Joule we can convert it to mass.

Your question is of the same kind. Given two charges and a distance between them Coulomb's law tells you how to arrange the given quantites (multiply the charges and divide by the distance squared) and Coulomb's constant tells you how to convert the answer you get from [multiply the charges and divide by the distance squared] to a force.

So one calculates the quantity $\frac{q_1q_2}{r^2}$, then $k_e$ is used to convert this quantity to Newtons.

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So Newton = k∗Coloumb2/m2. Now combining this definition of a Newton, along with the definition from the equation for gravity Newton=G∗kg2/m2, can't the definitions be equated to prove that kg=a∗Coloumb for some constant a? But this isn't true, so it can't be a conversion factor? –  dfg Jan 16 at 21:42
It is true since your first equation $Newton = k∗Coloumb^2/m^2$ means that the dimensions of $k$ are $k=Newton\cdot m^2/coloumb^2$. And for the second one we have that $G=Newton\cdot m^2/kg^2$. And for the third one to make sense the conversion factor $a$ must be of the dimensions $a=kg/Coloumb$. If this seams strange, just think about the speed of light $c$. One uses this all the time in General Relativity to convert time to length and length to time since $c=length/time$. –  Natanael Jan 17 at 8:37

Well, I think this is about the ineludible constants.

Experimental laws represents the reality, and this reality doesn't depend on unit measurement systems; it's also independent of system's changes. To explain the reality, we use physical theories, which are based on fundamental equations; often they're expressed as proportion, like in

$$[F]\propto[m][a]$$

This is a relation between magnitudes, not between quantities. If I want the quantity relation, I have to use an unit system. To achieve this, is neccesary to introduce a constant

$$F= C ma$$

Well, you can try to make $C=1$, choosing an adecuate unit system. This constant is often called "conversion factor" because you use it to change your unit system, but its proper name if "superflux constant. However, if you try to eliminate ALL physical constant by choosing certain unit system, you find that you can't. There's some constants that you can't eliminate from the equations, no matter what you do. This constants are:

• Universal Gravitation, $G$
• Planck constant, $h$
• Light speed in vacuum, $c$
• Boltzmann constant, $k_B$
• Avogadro number, $N_A$
• And vacuum permitivity, $\epsilon _0$

Your Coulomb constant depends on this last constant, and the permitivity of the material you're using (and other adimensional factors). As you can see, all this are ineludible constants. These constants are not a "conversion factor" since everything is in the same system. But they're is neccesary to make the equations dimensionally correct.

You can try to find some book of Dimensional Analysis and basic theory of physics: how physic models are constructed, the importance of dimensional homogenus equations, etc.

EDIT: I've said that "ineludible constants cannot be eliminated". Well, if you use natural units, then this constants are eliminated, but you will find some new consntants (Planck length, Planck mass...) so in fact, at the the end, you cannot eliminate all the constants from all the equations, which is I wanted to say.

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Your definition when simplified is equivalent to:

$$1 \text{Newton} = \frac{1}{1/1 meter^2 * 1 Coloumb^2} = \frac{c^2}{m^2}$$

This is only incorrect because the unit name $Newton$ is already taken however this would be perfectly valid:

$$1 \text{Your_unit_name} = \frac{c^2}{m^2}$$

In this case $\text{Your_unit_name}$ would be then unit of force in a unit system where $k=1$ much like Plank unit system has $ħ=G=c=k_e=k_B=1$

Then the value of $k$ is the conversion factor between $\text{Newtons}$ and $\text{Your_unit_names}$ and can be thought of having units $\frac{\text{Newtons}}{\text{Your_unit_names}}$ which if your substitute from the definition of $\text{Your_unit_names}$ you get the familiar form $N*m^2/C^2$

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So Newton = $k*Coloumb^2/m^2$. Now combining this definition of a Newton, along with the definition from the equation for gravity $Newton = G*kg^2/m^2$, can't the definitions be equated to prove that $kg = a*Coloumb$ for some constant $a$? But this isn't true, so it can't be a conversion factor? –  dfg Jan 16 at 21:37
Yes they can, but in that case $a$ will have units $kg/C$ so the units workout. –  Michal Jan 16 at 21:44
Right but you then have a way to convert between kg and Coloumbs, so you could use kg to measure charge. But you can't use one unit to measure two different quantities (mass and charge) unless there equal. –  dfg Jan 16 at 22:13
Strictly speaking $a$ would be a constant in very specific circumstances, where all charges are a distance of 1 apart, this wouldn't be a very useful constant. However on the other hand $k$ isn't actually a constant it is $k=\frac{1}{4 \pi \epsilon_0}$ which makes it a constant only in free space, other materials have different values of $\epsilon$ and therefore $k$ –  Michal Jan 16 at 22:33