Dimensions of K (the constant in Coulomb's Law)

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SUMMARY

The dimensions of the Coulomb Force Constant, often denoted as K, are defined as 1/(4πε₀) in MKS units, where ε₀ is the permittivity of free space. In CGS electrostatic units, ε₀ is considered dimensionless, leading to the conclusion that K is also dimensionless in that system. The discussion highlights the importance of understanding the physical concept of electric charge and its relationship with fundamental quantities such as mass, length, and time. The dimensions of the permeability of free space, μ₀, are also explored, revealing that it has dimensions of (time/length)² in CGS units.

PREREQUISITES
  • Understanding of MKS and CGS unit systems
  • Familiarity with fundamental physical constants such as ε₀ and μ₀
  • Knowledge of electromagnetic theory and Coulomb's Law
  • Basic grasp of dimensional analysis in physics
NEXT STEPS
  • Research the implications of using different unit systems in electromagnetism
  • Explore the derivation and significance of the permittivity ε₀ and permeability μ₀
  • Study the relationship between electric charge and fundamental physical quantities
  • Investigate the differences in electromagnetic equations between MKS and CGS systems
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Physicists, electrical engineers, and students studying electromagnetism who seek to deepen their understanding of fundamental constants and their dimensional analysis.

Chen
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Hi,

What are the dimensions of K? Is it (Force x Length^2 / Charge^2), or is it dimension-less?

Thanks.
 
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K is about 9.0 \times 10^9\; [Nm^2C^{-2}] in MKS units.
 
Chen said:
What are the dimensions of K? Is it (Force x Length^2 / Charge^2), or is it dimension-less?

Doc, i think there is more to Chen's question than that. 3 or 4 years ago, I've had a lot of discussions on sci.physics.research about dimensions, fundamental constants, etc, particularly of G, c, \hbar, \epsilon_0.

first of all, let's not call the Coulomb Force Constant "k" (because i have used it and gotten mixed up in conversation with people who think "k" means the Boltzmann constant). i would call it \frac{1}{4 \pi \epsilon_0}.

i think the answer to Chen's question depends solely on what we really think that electric charge is.

for instance the dimension of force is mass x length / time^2 because we don't look at the concept of force as anything other than the time-derivative of momentum. We don't say force is proportional to the time derivative of momentum (and toss in some constant of proportionality), we say force is the derivative of momentum w.r.t. time. i s'pose we could have declared force to be a completely different and new physical quantity undefinable by any other physical quantity, independently defined a unit of force (call it a "farg"), observed in experiment that it's proportional to the time derivative of momentum, then said something like F = \alpha \frac{dp}{dt} and \alpha would have dimensions of fargs x time^2 / (mass x length). but there is no need to do that. it isn't like that the concept of force existed independently in nature without time or length or mass, so we could and did define force solely and naturally in terms of time, length, and mass.

here's another for instance (going back to Boltzmann), at one time people didn't understand heat to be "merely" the random motion of molecules in solids, liquid, and gas. heat was a totally different concept of physical quantity existing independently in nature outside of quantities of time or length or mass (or momentum, force, mechanical energy or power derived from the big three: time, length, mass). but now we know different. heat is not some new and separate physical quantity but is about the kinetic energy of the particles of matter. we could define temperature in terms of energy (in Joules) per particle per degree of freedom, instead of Kelvin (and all we would do is multiply T in K by half of the Boltzmann constant, and that's what we would have). so if you let the dimension of temperature of objects be the same as energy, then the Boltzmann constant is dimensionless. but if you define temperature as this "other stuff", not simply energy per particle, and you independently define a unit of temperature, then the dimension of the Boltzmann constant is energy/(degree of temperature). in MKS it's Joule/Kelvin.

so the question is whether or not the physical concept of charge exists independently in nature without time or length or mass, then whether we can define charge solely and naturally in terms of time, length, and mass.

in the CGS electrostatic units they define the Coulomb Force Constant to be the dimensionless 1 (\epsilon_0 = \frac{1}{4 \pi} in cgs). that results in the dimension of electric charge being length / time x sqrt(length x mass) or velocity x (length x mass)^(1/2).

now if you think that is what this stuff that lives on (or is a property of) electrons and protons is, that it's velocity x sqrt(length x mass), then i guess the answer to Chen's question is that the dimension of \frac{1}{4 \pi \epsilon_0} is dimensionless.

but some of us continue to have trouble accepting that. if you understand electromagnetic charge to be this totally different concept of physical quantity than length, time, and mass, then the Coulomb Force Constant has to have dimension (mass x length^3 / (time^2 x charge^2)).

so, i guess it's a matter of which belief system you have.

r b-j
 
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Thanks for that, rbj, that's exactly the kind of answer I was looking for.

rbj said:
in the CGS electrostatic units they define the Coulomb Force Constant to be the dimensionless 1 (\epsilon_0 = \frac{1}{4 \pi} in cgs). that results in the dimension of electric charge being length / time x sqrt(length x mass) or velocity x (length x mass)^(1/2).
Ok, so you say in CGS units \epsilon_0 is dimensionless. But then, what are the dimensions of \mu_0? Because if c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} (and I assume this still holds in CGS units) then \mu_0 must have a dimension of (time/length)2, but I know that in MKS units \mu_0 is inductance per length. So how does this work out...?

Thanks again. :smile:
 
Chen said:
Thanks for that, rbj, that's exactly the kind of answer I was looking for.
yer welcome.
Ok, so you say in CGS units \epsilon_0 is dimensionless. But then, what are the dimensions of \mu_0? Because if c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} (and I assume this still holds in CGS units)

it holds in any units.

\epsilon_0 \mu_0 c^2 = 1 or \mu_0 = \frac{1}{\epsilon_0 c^2}. that must mean, in cgs, that \mu_0 must be in reciprocal units of velocity^2.

then \mu_0 must have a dimension of (time/length)2, but I know that in MKS units \mu_0 is inductance per length. So how does this work out...?

it's still inductance per length, but the E&M unit in this is inductance and that will come down a little different in cgs than in MKS. there is no seaprate unit of charge in cgs, so current, voltage, inductance, etc. are defined in terms of the cm, gram, and second.

r b-j
 
Last edited:
Excellent answer, r b-j.

The reason for my brief (perhaps naive) response was that K appears in the MKS system:
F = Kq_1q_2/r^2​
but not in the cgs system, where:
F = q_1q_2/r^2​

So I just assumed Chen was asking about the MKS system. :wink:
 
On the same subject, what's the reason for the difference in equations for the magnetic force? In MKS: q\vec v \times \vec B, in cgs: \frac{q}{c}\vec v \times \vec B.

Thanks. :smile:
 
Thanks rbj, that helped a lot.
 
Chen said:
On the same subject, what's the reason for the difference in equations for the magnetic force? In MKS: q\vec v \times \vec B, in cgs: \frac{q}{c}\vec v \times \vec B.

scaling and definitions. that's all.

in MKS, the Lorentz force Equation is

\vec F = q ( \vec E + \vec v \times \vec B )

in cgs, the Lorentz force Equation is

\vec F = q ( \vec E + \frac{\vec v}{c} \times \vec B )

which makes the dimensions of the B field the same as the E field in cgs. i actually like that better than the standard MKS. now, of course, in cgs the Maxwell's Equations have to be fixed a little (from their expression in MKS).

first, you put in \frac{1}{4 \pi} for every occasion of \epsilon_0, then you put in \frac{4 \pi}{c^2} for every occasion of \mu_0. so far, nothing really different except that you've chosen your unit of charge to be whatever makes the Coulomb Force constant \frac{1}{4 \pi \epsilon_0} = 1. it's called a "statcoulomb" or "esu of charge".

but the Lorentz force equations of cgs has B scaled down by a factor of c, so to make things equivalent, you have to scale it down by the same factor in the Maxwell's Eqs. you are literally substituting \vec B = \frac{\vec B}{c} everywhere in both sets of equations. that's really just a change in symbols.

r b-j
 
Last edited:
  • #10
Thank you very much! :smile:
 
  • #11
Chen said:
Thank you very much! :smile:

yer wlecome, FWIW.

personally, i wished they had normalized \epsilon_0 in cgs instead of \frac{1}{4 \pi \epsilon_0} (and also for "natural units" such as Placnk units). they way they did it leaves all sorts of factors of 4 \pi lying all over the place in many pretty fundamental physical equations.

r b-j
 

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