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

  1. Apr 1, 2005 #1
    Hi,

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

    Thanks.
     
  2. jcsd
  3. Apr 1, 2005 #2

    Doc Al

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    Staff: Mentor

    K is about [itex]9.0 \times 10^9\; [Nm^2C^{-2}][/itex] in MKS units.
     
  4. Apr 1, 2005 #3

    rbj

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    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 [tex] G, c, \hbar, \epsilon_0 [/tex].

    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 [tex] \frac{1}{4 \pi \epsilon_0} [/tex].

    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 [tex] F = \alpha \frac{dp}{dt} [/tex] and [tex] \alpha [/tex] 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 ([tex] \epsilon_0 = \frac{1}{4 \pi} [/tex] 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 [tex] \frac{1}{4 \pi \epsilon_0} [/tex] 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
     
    Last edited: Apr 1, 2005
  5. Apr 1, 2005 #4
    Thanks for that, rbj, that's exactly the kind of answer I was looking for.

    Ok, so you say in CGS units [tex]\epsilon_0[/tex] is dimensionless. But then, what are the dimensions of [tex]\mu_0[/tex]? Because if [tex]c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}[/tex] (and I assume this still holds in CGS units) then [tex]\mu_0[/tex] must have a dimension of (time/length)2, but I know that in MKS units [tex]\mu_0[/tex] is inductance per length. So how does this work out...?

    Thanks again. :smile:
     
  6. Apr 1, 2005 #5

    rbj

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    yer welcome.
    it holds in any units.

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

    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: Apr 1, 2005
  7. Apr 1, 2005 #6

    Doc Al

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    Excellent answer, r b-j.

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

    So I just assumed Chen was asking about the MKS system. :wink:
     
  8. Apr 2, 2005 #7
    On the same subject, what's the reason for the difference in equations for the magnetic force? In MKS: [tex]q\vec v \times \vec B[/tex], in cgs: [tex]\frac{q}{c}\vec v \times \vec B[/tex].

    Thanks. :smile:
     
  9. Apr 2, 2005 #8
    Thanks rbj, that helped a lot.
     
  10. Apr 2, 2005 #9

    rbj

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    scaling and definitions. that's all.

    in MKS, the Lorentz force Equation is

    [tex] \vec F = q ( \vec E + \vec v \times \vec B ) [/tex]

    in cgs, the Lorentz force Equation is

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

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

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

    r b-j
     
    Last edited: Apr 2, 2005
  11. Apr 3, 2005 #10
    Thank you very much! :smile:
     
  12. Apr 3, 2005 #11

    rbj

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    yer wlecome, FWIW.

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

    r b-j
     
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