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Coulomb's Law is wrong!

  1. Feb 26, 2005 #1

    AKG

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    Okay, not really, but suppose it were. This is what my assignment wants me to suppose. Suppose that the actual force of interaction between two piont charges is found to be:

    [tex]\mathbf{F} = \frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{s^2}\left (1 + \frac{s}{\lambda}\right )e^{-s/\lambda}\mathbf {\hat {s}}[/tex]

    Where [itex]\lambda[/itex] is a new physical constant which is very large, [itex]\mathbf{s}[/itex] is the separation vector between the two charges, and [itex]s = |\mathbf{s}|[/itex].

    I have found that:

    The electric field of a charge distribution [itex]\rho[/itex] is:

    [tex]\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi \epsilon _0}\int _{\mathcal{V}} \frac{\rho (\mathbf{r'})}{s^2}\mathbf{\hat{s}}\left (1 + \frac{s}{\lambda}\right )e^{-s/\lambda}\, d\tau '[/tex]

    I have found that the potential of a point charge [itex]q[/itex] is:

    [tex]V(\mathbf{r}) = \frac{q}{4\pi \epsilon _0 r e^{r/\lambda}}[/tex]

    Where, of course, [itex]r = |\mathbf{r}|[/itex].

    For a point charge [itex]q[/itex] at the origin, I found that:

    [tex]\oint _{\mathcal{S}} \mathbf{E}\cdot d\mathbf{a} + \frac{1}{\lambda ^2}\int _{\mathcal{V}}V\, d\tau = \frac{q}{\epsilon _0}[/tex]

    where [itex]\mathcal{S}[/itex] is the surface of a sphere [itex]\mathcal{V}[/itex] centered at the origin.

    I now need to show that:

    [tex]\oint _{\mathcal{S}} \mathbf{E}\cdot d\mathbf{a} + \frac{1}{\lambda ^2}\int _{\mathcal{V}}V\, d\tau = \frac{Q_{enclosed}}{\epsilon _0}[/tex]

    Where [itex]Q_{enclosed}[/itex] is the charge enclosed within [itex]\mathcal{V}[/itex]. How do I do this?

    I can rewrite the things as:

    [tex]\int _{\mathcal{V}} (\mathbf{\nabla} \cdot \mathbf{E}) + \frac{V}{\lambda ^2}\, d\tau = \int _{\mathcal{V}}\frac{\rho}{\epsilon _0}\, d\tau[/tex]

    Where [itex]\rho : \mathbb{R}^3 \to \mathbb{R}[/itex]. If I could show the integrands to be equal, that would be good. The second equation in this post gives an expression for E in terms of [itex]\rho[/itex], and I could, I believe, get an expression for V in terms of [itex]\rho[/itex], and then possibly use that to show that the integrands are equal, but that sounds like an enormous task. Is there a better way to do it?

    For the real Gauss's Law, my book says that:

    [tex]\oint \mathbf{E}\cdot d\mathbf{a} = \frac{q}{\epsilon _0}[/tex]

    For a charge centered at the origin, and the surface of integration being the surface of a sphere centered at the origin. The book then says that any shape enclosing the charge will do by appealing to the diagram that shows the field lines that go through the surface, saying that "any closed surface, whatever its shape, would trap the same number of field lines." It then says that if we had a bunch of charges scattered about, rather than one at the origin, the total field is the sum of all individual fields, so a surface that encloses them all will have:

    [tex]\oint \mathbf{E}\cdot d\mathbf{a} = \oint \left (\sum _{i = 1} ^n \mathbf{E}_i\cdot d\mathbf{a}\right ) = \sum _{i = 1} ^n \left ( \oint \mathbf{E}_i \cdot d\mathbf{a}\right ) = \sum _{i = 1} ^n \left (\frac{q_i}{\epsilon _0}\right ) = \frac{Q_{enclosed}}{\epsilon _0}[/tex]

    The problem with that is that:

    1) To me, the argument that says that it's okay for the shape to be something other than a sphere just sounds like hand-waving to me.
    2) It assumes that [itex]\oint \mathbf{E}_i \cdot d\mathbf{a} = \frac{q_i}{\epsilon _0}[/itex], even if the charge is not found at the origin.
    3) This only deals with discrete charge distributions, but what if [itex]Q_{enclosed}[/itex] is continuously distributed?

    So on the one hand, I can't see how to prove the proposition myself, at least not without too much computation (some of which may not be possible), on the other hand, if I do it the way that the book appears to have done it, it doesn't seem like a proof at all. Suggestions please.
     
    Last edited: Feb 26, 2005
  2. jcsd
  3. Feb 26, 2005 #2

    dextercioby

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    1.Coulomb's law is right when it applies in its domain.
    2.Coulomb's law is a logical consequence of the axioms of classical electromagnetism,namely Maxwell equations.
    3.Guass' law (in vacuum) is expressed under the integral form:
    [tex] \oint\oint_{\Sigma} \vec{E}\cdot \vec{n}dS=\frac{1}{\epsilon_{0}}\iiint_{V_{\Sigma}} \rho dV [/tex]

    and differential:
    [tex] \nabla\cdot\vec{E}=\frac{\rho}{\epsilon_{0}} [/tex]

    and that's that...For macroscopical level,we have no reason to use other equations...

    Daniel.

    P.S.For a point charge "q",the electric field is always ~1/r^{2}
     
  4. Feb 26, 2005 #3

    AKG

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    I guess you didn't read my post. I edited to avoid confusion, but the point was that I have an assignment where we have a hypothetical universe where Coulomb's Law is different. Given that, I am asked to find various expressions and prove various things. Given that, can you help me?
     
  5. Feb 26, 2005 #4

    dextercioby

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    You may wanna check again the expression for the electric field.I have a bad feeling about it.

    Daniel.
     
  6. Feb 26, 2005 #5

    AKG

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    I don't see anything wrong with it. My book gives the equation:

    [tex]\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi \epsilon _0}\int _{\mathcal{V}} \frac{\rho (\mathbf{r'})}{s^2}\mathbf{\hat{s}}\, d\tau '[/tex].

    [itex]\mathbf{r'}[/itex] is the position vector of some point in the region [itex]\mathcal{V}[/itex], [itex]\tau '[/itex] is the infinitessimal volume element at that point, so [itex]\rho (\mathbf{r'})\, d\tau '[/itex] is the infintessimal charge there. [itex]s[/itex] is the separation between [itex]\mathbf{r'}[/itex] and [itex]\mathbf{r}[/itex].
     
  7. Feb 26, 2005 #6

    dextercioby

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    Okay.Your notation is weird.It leaves the impression that the round bracket & the exp are not under the volume integral.

    Where/How did u get that eq.with E and V under two kinds of integrals...?

    Daniel.
     
  8. Feb 26, 2005 #7

    AKG

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    I see, fixed it.
    It was part of the question, the book asked me to prove it, which I did. "How" I proved it would be too much work to write up in LaTeX and post here, but I don't believe that how I proved it affects the problem that I'm having. For this problem, we can just take that equation as given. That is, the equation with q is given, the equation with Qenclosed is the one I'm trying to prove.
     
  9. Feb 26, 2005 #8

    dextercioby

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    I reckon the solution u found as writing everything under the same volume integral and comparing the two sides of the equation is the only one which would work.It doesn't matter how difficult the calculations may be (you aleady said u made a few trying to prove the eq.i had previously mentionsed),it's important IF they yield the same thing...

    Daniel.

    P.S.I don't see other way.U gave "rho" in both expressions for E and V.U're asked to prove smth.with a variable not mentioned in the problem b4,but which,fortunately,can, be expressed in terms of 'rho'...
     
  10. Feb 26, 2005 #9

    AKG

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    [tex]\int _{\mathcal{V}} (\mathbf{\nabla} \cdot \mathbf{E}) + \frac{V}{\lambda ^2}\, d\tau = \int _{\mathcal{V}}\frac{\rho}{\epsilon _0}\, d\tau[/tex]

    I want to prove that, so for that, I need to prove:

    [tex](\mathbf{\nabla} \cdot \mathbf{E}) + \frac{V}{\lambda ^2} = \frac{\rho}{\epsilon _0}[/tex]

    [tex]\left [\mathbf{\nabla} \cdot \left (\frac{\rho}{4\pi s^2}(1 + s/\lambda)e^{-s/\lambda}\right \mathbf{\hat{s}})\right ] + \frac{\rho}{4\pi \s \lambda ^2}e^{-s/\lambda} = \rho[/tex]

    Now I remember why I said that some of these calculations might be impossible. I don't know what the partial derivatives of [itex]\rho[/itex] would be. Can I do the computation without knowing those derivatives, and just deal with its partials in general?
    Well what about the way similar to how the book did it?
     
  11. Feb 26, 2005 #10

    AKG

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    Actually, I have an idea. In this universe, [itex]E = E'f_{\lambda}[/itex], where E' is the electric field in our universe, and [itex]f_{\lambda} = (1 + s/\lambda)e^{-s/\lambda}[/itex]. Then:

    [tex](\mathbf{\nabla} \cdot \mathbf{E}) + \frac{V}{\lambda ^2} = \frac{\rho}{\epsilon _0}[/tex]

    [tex](\mathbf{\nabla} \cdot \mathbf{E'}f_{\lambda}) + \frac{V}{\lambda ^2} = \frac{\rho}{\epsilon _0}[/tex]

    [tex]f_{\lambda}(\mathbf{\nabla} \cdot \mathbf{E'}) + \mathbf{E'}\cdot(\mathbf{\nabla} f_{\lambda})+ \frac{V}{\lambda ^2} = \frac{\rho}{\epsilon _0}[/tex]

    [tex]f_{\lambda}\rho + \mathbf{E'}\cdot(\mathbf{\nabla} f_{\lambda})+ \frac{\rho}{4\pi \s \lambda ^2}e^{-s/\lambda} = \rho[/tex]

    Am I on the right track?
     
  12. Feb 27, 2005 #11

    AKG

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    Oops, the above is wrong. It only sort of makes sense when we're dealing only with a point charge, not a continuous distribution over some region. Even if that were the case, it wouldn't be totally right, only partially. However, since that's not even the case, it's entirely wrong. The best I could actually get it to was the following:

    [tex]\mathbf{\nabla}\cdot \left (\int _{\mathcal{U}} \frac{\rho (\mathbf{r'})}{s^2}\mathbf{\hat{s}}\left (1 + \frac{s}{\lambda}\right )e^{-s/\lambda}\, d\tau '\right ) - \int _{\mathcal{O}} ^{\mathbf{r}} \frac{1}{\lambda ^2}\left (\int _{\mathcal{U}} \frac{\rho (\mathbf{r''})}{(s')^2}\mathbf{\hat{s'}}\left (1 + \frac{s'}{\lambda}\right )e^{-s'/\lambda}\, d\tau ''\right )\cdot d\mathbf{l'} = 4\pi \rho (\mathbf{r})[/tex]

    Where [itex]\mathcal{U}[/itex] is the region over which the charge is distributed, [itex]\mathbf{s} = \mathbf{r} - \mathbf{r'}[/itex], [itex]\mathbf{s''} = \mathbf{r'} - \mathbf{r''}[/itex], and [itex]\mathcal{O}[/itex] is a reference point taken at infinity. Unless anyone knows of an easy way to do this, I doubt it is worth doing, so I'm just going to use the book's method. :cry:
     
    Last edited: Feb 27, 2005
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