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Why is the electric field ever defined inside volume charge distributions?

  1. Aug 9, 2012 #1
    Example: A common problem in undergraduate electromagnetic classes is calculating the electric field inside a solid spherical charge distribution with known charge density. The common method of solution in my experience is as follows:

    1) Place the origin of a coordinate system at the center of the sphere.

    2) Using symmetry and a spherical Gaussian surface, calculate E with Gauss' Law.

    Here's the problem... The most IMO rigorous derivation I have seen of Gauss' Law seems to be valid only if there is no charge on the boundary of the Gaussian surface. This made sense to me, because if you calculate the electric field using the "brute-force" method of integration using Coulomb's Law, I don't believe the field would be defined at any point inside the charge distribution. For example, when using this brute-force method, the denominator of the integrand has a (r - r') term in it. Since the integration is with respect to the source point r', it seems that all calculations of this form must implicitly assume that the field point r under consideration cannot be an element of the volume being integrated over (if it was, there would be division by zero).

    So, I am wondering if anyone can explain why Gauss' Law is valid inside volume charge distributions and why the electric field (or potential for that matter)is defined inside volume charge distributions.

    I am mostly looking for a mathematical reason, because this is really a math problem. I know that IRL there is separation between charges, etc. but we are modeling the charge distribution as a continuum in this type of problem. Hope my question makes sense.
     
  2. jcsd
  3. Aug 10, 2012 #2

    mfb

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    The boundary is a 2-dimensional surface. With a 3-dimensional charge distribution, the total charge within this area is 0.
     
  4. Aug 11, 2012 #3
    Thanks for the reply.

    I agree that there is no finite charge in a 2-dimensional region with volume charge density.
    However, that doesn't change the fact that in the brute-force method, we are integrating a vector-valued function of three variables over a region where this function is not everywhere defined.

    Here is Coulomb's law for the case of a volume charge distribution, in Cartesian coordinates.

    [tex]\iiint_{V}\frac{\rho(x',y',z')}{4\pi\varepsilon_{0}((x-x')^{2}+(y-y')^{2}+(z-z')^{2})^{3/2}}(x-x',y-y',z-z')dx'dy'dz'[/tex]

    Here, a prime denotes a "source" point, whereas an unprimed x, y, or z refers to a "field" point. The integration is with respect to the primed coordinates, so for the purpose of evaluating the integral, the values of x, y, and z are constants.

    If the point [tex](x,y,z)[/tex] is allowed to lie within the region V, then clearly the integrand is not defined at the point [tex](x',y',z') = (x,y,z)[/tex] because the denominator of the integrand [tex]((x-x')^{2}+(y-y')^{2}+(z-z')^{2})^{3/2}[/tex] is equal to zero. So I am very confused why it is valid to allow [tex](x,y,z)\in V[/tex]

    Any and all further elucidation would be greatly appreciated. I realize that my first post may have been unclear, so I hope this helps explain what I mean. (This is my first time using Latex, so I apologize if I screwed something up).
     
  5. Aug 11, 2012 #4
    The easiest way, I think, is to look at the integral as the limit of a Riemann sum of charge in the volume element ΔV as ΔV → 0 , which would give the indeterminate form ΔV/Δr2 → 0/0. Since the volume element in the numerator goes as x3 and the denominator goes as x2, that means the numerator goes to zero faster than the denominator and the fraction goes to zero in the limit.
     
  6. Aug 11, 2012 #5
    Thanks for the reply, Jasso!


    1) I thought of considering it as a limit before. One thing that kinda bugs me is: sure, the integral is equivalent to a Riemann sum letting ΔV. But I don't understand why we're letting Δr2→0; The way I was thinking about it, Δr2 IS 0. I thought that as (x',y',z') "range" over all possible possible values within the specified bounds, that if (x,y,z) is a value within this region, then (x',y',z') will take on the EXACT value (x,y,z). Granted I'm no math expert, so I may be thinking about that totally wrong?

    2) Another thing that bugs me is I think that there may be a logical error related to your limit. I think you're treating two completely different quantities as the same. You mentioned that you are taking the limit of ΔV/r2 as ΔV→0. But Δr2 is independent of the size of ΔV. By letting Δr2→0, you're simply considering what happens as you vary the location of your field point; it seems independent of the size of your volume element.

    Here's a one-dimensional analogue:

    [tex]\int_{-1}^{1}\frac{1}{|x|}dx[/tex]

    Just like the integral version of Coulomb's Law, this integrand "blows up" at some point: in this case, x = 0. However, this integral diverges. I think that trying to apply the same limit logic to this integral helps illustrate what I am saying: Δx/|x| as Δx→0 is not really relevant, because Δx = xi+1-xi, which is has no effect on what values |x| will take in the Riemann sum other than setting the discrete "step size" between values of x.

    Thoughts?
     
  7. Aug 12, 2012 #6
    When dealing with indeterminate forms like 0/0, one method is to take the limit as the numerator and denominator go to zero. It's exactly the same as L'Hospital's Rule, since the derivative is the limit of the difference quotient.

    All that it really means is that the contribution of that specific point (or really any other point) to the sum of the integral is infinitesimal, essentially zero. This is because the infinitesimal volume element goes to zero faster than the infinitesimal displacement. I believe that this is called a removable singularity, since the function is defined and well behaved up to the limit of that point; the limit for the function at a position rr as Δr → 0 goes to zero as well.

    In the 1-d example you gave, the limit of the numerator and denominator go to zero equally, turning the indeterminate 0/0 into a non-zero constant. This means that in the region 0 to dx there is an unlimited number of non-infinitesimal values contributing to the integral sum, giving an undefined infinity.
     
    Last edited: Aug 12, 2012
  8. Aug 12, 2012 #7
    Thanks, Jasso. That gave me some useful insight. I think I see what you mean. While the magnitude portion of the integrand

    [tex]\frac{\rho(x',y',z')}{4\pi\varepsilon_{0}((x-x')^{2}+(y-y')^{2}+(z-z')^{2})^{3/2}}[/tex]


    itself is not a function with a removable singularity, I think (like you were saying) the Riemann sum term corresponding to (x,y,z) could basically be written as a function with a removable discontinuity:

    [tex]\frac{\rho(x',y',z')*(x'-x)(y'-y)(z'-z)}{4\pi\varepsilon_{0}((x-x')^{2}+(y-y')^{2}+(z-z')^{2})^{3/2}}[/tex]

    Here I am choosing to let the coordinate of the source point always coincide with a specific corner of the volume element "box". (I think I remember reading that we can use any value of the function evaluated at any point within this box as long as the values don't change too much, so I think that's okay?).

    By plotting the 2-variable case in Matlab, I am convinced that the value of this Riemann sum function does approach zero as (x',y',z')→(x,y,z). So like you said, this term (like all other individual terms) contributes nothing to the Riemann sum.

    I THINK I can see why this is true based on your reasoning. So, for example, I believe the integral

    [tex]\int_{-1}^{1}\frac{1}{|x^{1/2}|}dx = 4[/tex]

    So the contribution to the Riemann sum corresponding to this integral from the rectangle corresponding to x = 0 is zero (I am placing the left side of the rectangle at x = 0, but using the function value from the right side of the interval), because
    (x-0)*(1/|x|1/2)→0 as x→0 (once again using the value the function takes at the right end of the interval [0,x] defined by the base of the rectangle). This makes sense to me as one way of thinking about this integral. (Am I thinking about this one right?)

    Back to my earlier example:

    [tex]\int_{-1}^{1}\frac{1}{|x|}dx [/tex]

    I see what you mean, that the Riemann sum term at x = 0 is finite, because f(x)Δx→1 as (Δx,x)→(0,0) (at least if I write the limit explicitly as |x-0|/(|x|)→1 as x→0).

    Are you saying that we are adding infinitely many of these finite values because not only does f(x)Δx → (finite value) as (Δx,x)→(0,0), but also f(x+Δx)Δx → (finite value) as (Δx,x) →(0,0) , and same with f(x+2Δx)Δx and so on, because f(x+dx) is essentially equal to f(x), and so if f(x) is infinite, so is f(x+dx), and therefore if f(x)dx is finite, so is f(x+dx)dx?

    Incidentally, I read that integrals of the form

    [tex]\int_{0^{+}}^{1}\frac{1}{x^{p}}dx = \frac{1}{1-p}[/tex]

    if 0 ≤ p < 1, and diverge otherwise. I wonder if there is an analogous theorem for triple integrals that would be relevant to my original question?

    I can't thank you enough for your patience so far with my posts, it's already helped a lot!
     
  9. Aug 12, 2012 #8
    "but we are modeling the charge distribution as a continuum in this type of problem"
    Exactly. But sometimes the charge distribution is not a simple continuum. Suppose that the charge distribution piles up at the surface so that the surface, the charge density is effectively infinite. Then one has to use a surface integral, or some mathematical equivalent to a surface integral.
    I think what you are looking at is the use of Green functions with both a volume distributed charge and a surface distributed charge. The Green's function that you are describing approaches zero in the limit of infinite distance. It looks something like this:
    G(r-r')=1/|r-r'|
    where G is the Green's function, r is the vector of the probe point, and r' is the vector that describes the position of an element of charge. It isn't enough to integrate Gρ over the entire volume. You have to treat charges piled up at the surface with an integral over the entire surface.
    The expression for potential in a 3 dimensional region has both a volume integral for charge potentials that are continuous in 3 dimensions. However, continuous charge densities in 3D don't work very well if they are infinite on the surface of the volume. Sometimes, charge piles up at the surface. In those cases, the contribution of the surface charges are best analyzed with a surface integral.
    So most expansions of potential have two contributions. There is a volume integral for the charges distributed in the volume and a surface integral for the charges that are concentrated on the surface.
    I am using as a reference
    "Classical Electrodynamics" 3rd edition by John David Jackson (Wiley, 1999).
    On the bottom of page 36, equation 1.36 shows the full expression for a potential given a charge density function that is continuous and smooth in 3 dimensions. The first term is a volume integral of the Green's function shown above and the continuous charge density, ρ. However, the second term involves a surface integral of the sum of the gradient of potential and the gradient of the Green's function over the entire surface of the the 3 dimensional region. So the potential, inside and outside the 3D region, has a contribution from the charges that are concentrated at the surface.
    There is more than one way to analyze surface charges. It should be pointed out that the boundary conditions on an electric field incorporate surface charge densities. A discontinuity in the electric field on a surface determines the surface charge density. See page 31 of the above reference, equation 122.
    There is a volume charge density that requires a volume integral. There is a surface charge density that requires a surface integral. I suppose there is a line charge density that requires a line integral.
    It should be noted that there is another way to incorporate surface charges in the potential. One can use a volume integral and Green's function alone to calculate the potential. One can use a single volume charge density to describe the distribution of charges. However, you can add a term to the volume charge density that includes a Dirac delta function. At the surface, the charge density approaches infinity in such a way that a volume integral over a surface region is finite. A delta function is described in the above reference on page 26.

    Actually, you don't have to use my reference. Any standard textbook on electromagnetic theory will show you how to calculate the potential from charge densities. One has to handle the surface charge densities and surface current densities a little differently that the volume charge densities and the surface current densities.
     
  10. Aug 12, 2012 #9
    Thanks for the reply, Darwin!

    I don't have Jackson, but I am familiar with the formula you mentioned on page 36. I learned it when I first read Elements of Electromagnetics by Sadiku (I'm really not a fan of the Sadiku book. I am an undergraduate electrical engineering major, and that's the book we used. I'm currently reading Griffiths).

    The formula you mentioned sounds like this formula for potential due to a polarized dielectric:

    [tex]V = \iiint_{v'}\frac{-\nabla'\cdot {\bf P}}{4\pi\varepsilon_{0}|{\bf r-r'}|}dv' + \iint_{\partial v'}\frac{{\bf P\cdot n}}{4\pi\varepsilon_{0}|{\bf r-r'}|}dS'[/tex]

    ...but doesn't this formula run into the same problems? Since integration is with respect to the primed coordinates, wouldn't you still run into the exact same division by zero problems in both integrals UNLESS you assume [itex]{\bf r}\not \in v'[/itex]?

    I don't really see how this remedies the problem.

    Also, Jasso- I thought more about what I said earlier, and I was wondering: why does it just happen to work out that if you evaluate the integral as if there was no singularity, that it gives you the correct answer that this improper integral converges to? Like magic almost...
     
  11. Aug 15, 2012 #10
    Update: According to Vector Calculus by Marsden & Tromba, it appears that Coulomb's Law is an improper integral of a class of functions that are unbounded at isolated points (one in this case).

    So if we accept Coulomb's law for volume charge distributions as being given by an improper integral, my original question is answered.

    But why is Coulomb's Law given by an improper integral? It is essentially like saying, "Okay, the electric field due to a volume charge element is undefined at the location of the charge, because we can't divide by zero. But if we measure the field at a location infinitely close to the volume charge element, the field due to that charge element is zero (the limiting discussion earlier). So it seems maybe Coulomb's law is actually defined for volume charge distributions as an improper integral and not the basic Riemann integral.

    My understanding of improper integrals is as follow: the basic definition of the Riemann integral does not apply to unbounded functions on finite regions or bounded functions on unbounded regions. However, we may define the integral for those two cases using limits (an extension of the Riemann integral). So essentially the value of an improper integral for an unbounded function is the value of the integral due to all points in the domain EXCEPT the undefined point. Whereas the Riemann sum would be the value of the improper integral PLUS (dV)*(undefined value) which is a defined value plus an undefined value which yields an undefined value.

    It seems like the improper integral is a reasonable physical model because there are no real continuums of charge, and I guess any attempt to measure the field at a point would displace whatever charge was at the point anyway, so there is no problem with measuring an undefined (in the basic Riemann integral sense) electric field, because you'd be measuring the small (practically zero) field of a small charge, electron, etc. in addition to the fields of all the other charges.

    I'm sure some people are thinking I am completely overanalyzing this...I just wanted to update with my thoughts. It's confusing for me when any sort of hand-waving arguments are used in textbooks, etc.
     
    Last edited: Aug 15, 2012
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