# Coulombs constant

1. Feb 23, 2012

### acpower89

Hi

This may sound like a trivial question, but I'm a beginner physics student and need some help understanding what the k constant is from Coloumb law.

$$F_e = k_e\frac{k_1 k_1}{r^2}$$

where

$$k_e = \frac{1}{4\piε_0}$$

I did read a bit about it and I have two questions:

1. Why is k, the proportionality constant, necessary and what is its physical meaning? And where does the /pi come from?
2. How can you measure it?

Thank you

2. Feb 23, 2012

### vanhees71

The electrostatic Coulomb force of a charge $Q_1$ at the origin and $Q_2$ at position $\vec{x}=r \vec{e}_r$ is given by

$$\vec{F}=k_e \frac{Q_1 Q_2}{r^2} \vec{e}_r.$$

This is an innocent looking formula, but physically it has a lot of content.

First of all it's on the very foundation of electromagnetic phenomena. It tells you that there is an intrinsic quantity of matter, called electric charge, i.e., a piece of matter is not only characterized by its mass but also by its electric charge. Then Coulomb's Law tells you that the force betweent two charged point particles is proportional to the charges of these particles and that it depends on distance between the particles by an inverse-square law. Last but not least it acts along the straight line connecting the two charges.

In other words the Coulomb law defines how to measure charges and also includes the observed behavior of the electrostatic with distance.

On the other hand, there is no more content to it than this, and we have not even mentioned the constant $k_e$ in the explanation of the physical meaning of the symbols in this formula. This tells you that in principle you do not need this constant at all, and indeed in the good old days of the Gaussian system of units, one has just set $k_{e}^{(\text{Gauss})}=1$.

This choice is, however not very convenient from a practical point of view since then charge has funny units, when in terms of the basis units of mechanics (in the Gaussian system, these were centimetres, grams, and seconds for length, mass, and time).

Thus, later one has taken a forth basis unit into the game, when one extends mechanics to the realm of electromagnetic phenomeno. In the modern system of units, the SI (Systeme international de poids et messures), this is the Ampere as the unit of electric current (Charge per time flowing through a given area). The price to pay for that is to introduce the factor $k_e=1/(4 \pi \epsilon_0)$. I guess, this will be the units you'll be using most probably in your studies.

If it comes to theoretical high-energy physics, you'll stumble over still another system of units, which is very close to Gauss units, the socalled Heaviside-Lorentz system, which boils down to set $k_e=1/(4 \pi)$. The reason for this choice is that then the fundamental equations governing all electromagnetic phenomena, Maxwell's equations, become most simple: There are no "artificial" unit-conversion factors as $\epsilon_0$ as in the SI and also no factors $4 \pi$ in these fundamental equations.

This answers also your final questions: The $4 \pi$ in the denominator make Maxwell's equations simpler, and the $\epsilon_0$ is needed for conversion of charge units to mechanical units in Coulomb's Law of the electrostatic force. You can, in principle, measure $\epsilon_0$ using Colomb's Law. Then you have to use the very definition of the Ampere, given by the magnetostatic force between two straight current-conducting wires, to charge bodies with a well quantified amount of electric charge (measured in the unit $1 \text{C}=1 \text{A} \text{s}$.

3. Feb 23, 2012

### rbj

it is necessary to convert units that have been defined independently. it has no other physical meaning than that. it is not an intrinsic constant of nature in the same sense as the dimensionless constants, like the fine-structure constant, are.

the $4 \pi$ comes from the fact that the surface area of a sphere is $4 \pi r^2$. a better, more natural, representation of any inverse-square law is

$$F_e = k \frac{q_1 q_2}{4 \pi r^2}$$

$$F_g = G \frac{m_1 m_2}{4 \pi r^2}$$

but both Coulomb and Newton seemed to have missed this. this inverse-square relationship comes from the concept of flux and flux density and is the basis behind Gauss's Law.

well, you could measure it directly by measuring the force applied to known charges at various known distances. but since

$$c^2 = \frac{1}{\epsilon_0 \mu_0}$$

and since $\mu_0$ is defined because of how the ampere is defined, you know what $\epsilon_0$ is from accurately measuring the speed of light. nowadays even $c$ is defined, so all three symbols have defined values.

Last edited: Feb 23, 2012
4. Feb 23, 2012

### Dadface

The constant does have a meaning.It expresses the relevant electrical properties of the medium in which the charges are situated.Two charges separated by a certain distance in air will experience a different force to the same two charges separated by the same distance in,for example,water,each one having its own constant.
Expressing the constant by means of the second method above is usually preferred,a good reason being that for many problems the pi in the constant cancels out other pi's.Epsilon is a constant of the medium called its permittivity.Epsilon zero is known as the permittivity of free space which can be considered as a vacuum or,to a good approximation,the atmosphere.

5. Feb 23, 2012

### rbj

relative permittivity and relative permeability have meaning, as properties of the material medium of propagation. but the base permittivity and permeability ($\epsilon_0$ and $\mu_0$) that these material $\epsilon$ and $\mu$ are relative to, are not intrinsic properties of the vacuum. they are reflections purely of the units that people (or the aliens on the planet Zog) choose to use to measure physical quantity with. if you use Heaviside-Lorentz natural units (a lot like Planck units), then $\epsilon_0 = 1$ and $\mu_0 = 1$ and they go away.

but the relative permittivity and permeability remain and are properties of the material medium.

6. Feb 23, 2012

### Dadface

And since the OP is using SI units he will need to know the permittivity(or relative permittivity )of the medium in order to carry out calculations.The constant is different for different media.

7. Feb 23, 2012

### rbj

so far, i haven't read the OP ask anything about media. just about the basic Coulombs law with $k_e$ or $\epsilon_0$ and he/she asked about "physical meaning".

but, just in case there is a question, the relative permittivity (which is the dimensionless $\epsilon/\epsilon_0$) of different media, does have physical meaning.

8. Feb 23, 2012

### roboticmehdi

it was found that the force is proportional to q1 and q2, and inversely proportional to r^2. So, F~(q1*q2)/(r^2). It was also found that F always is k times bigger than (q1*q2)/(r^2). So F=k*(q1*q2)/(r^2). it is as simple as that. k is constant. and every constant can be represented by other constants. for example if k was 10, then i could say that 10=2*5 and call some w as w=5. so k=2w. now our equation becomes F=2*w*(q1*q2)/(r^2). In order to measure these things you need to have very precise apparatus. this is how k was found.

9. Feb 24, 2012

### acpower89

Thank you for all the answers, makes sense now (mostly).

Last edited: Feb 24, 2012
10. Feb 24, 2012

### Hassan2

This is true if we already had units of electric charges independent of this equation. I'm not sure but I think electric charge was defined based on this equation, therefore we can't measure the constant of proportionality as simple as you mentioned.

11. Feb 24, 2012

### Dadface

There is a lot of relevant information out there and below is a reference to just one of the sources"

type "www.newton.dep.anl.gov"

go to archives and type "what is the significanceof the permittivity of free space?"

An answer is given and I quote the last sentence only:

"in this way it becomes a property of space" (end of quote)

No matter how we define the units used we find that the force between say,two protons,depends amongst other things upon the medium in which they are situated.Permittivity is medium dependant and dependant upon the relevant properties of that medium.Like relative permittivity it does have physical meaning.Take for example the relative permittivity of air.This might be quoted as having a numerical value given by:

E=1.0006 times Ezero (approximately) but a more detailed data source will also quote other relevant information for which this data applies,such as the pressure and temperature(eg stp).

One assumed property of the classical vacuum is that the pressure is zero and this is inherent in the "presently" defined value of E zero.

Last edited: Feb 24, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook