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Homework Help: Help compute integral (electric potential)

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    How to calculate the following integral:

    Integral( 1/ |x-y| dS) where x and y are vectors in R^2, || represents norm.
    So say x= (x1,x2), y=(y1,y2). Then the integral is [ dS/ sqrt ((y1-x1)^2 + (y2-x2)^2)].

    The problem is: I need to integrate this over any plane in R^3.

    3. The attempt at a solution

    So I assume the plane is of the form ax+by+cz+d =0. Then I really have no clue how to proceed.
  2. jcsd
  3. May 20, 2009 #2


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    Can you tell me the original problem? Where is this integral coming from?
  4. May 20, 2009 #3
    Sure, it's a problem which arises from medical physics. The idea is that we may assume that the electrical activity in a certain piece of tissue has a continuous distribution of dipoles with constant density but this tissue is "separated" by the plane which formed by the exterior wall of the tissue and the interior wall of it.

    I know the formula for the electric potential in this case is equal to potential = p * Integral( dS / |x-y| ) where | | represents norm and p density (since its constant we can take it out of the integral).

    Hence the problem reduces to calculate this integral but over an arbitrary plane (which is the plane I explained above). But I'm very confused on which limits of integration should I choose and how would the area element should look. If someone could please explain this I'd really appreciate it.
  5. May 20, 2009 #4


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    Typically when one calculates electric potentials, it involves integrating over the charge distribution, and for a surface with a uniform charge distribution, you encounter an integral of the form:

    [tex]\int_{\mathcal{S}} \frac{d^2x'}{|\vec{x}-\vec{x'}|}[/tex]

    Where [itex]\vec{x}[/itex] is the vector from the origin (in whatever coordinate system you choose) to the field point (the point at which you want to determine the potential), [itex]\vec{x'}[/itex] is the vector from the origin to a point on the surface containing the charge distribution, and the integration is over the source points.

    The way you describe your integral makes no sense to me. You seem to be saying that [itex]\vec{x}[/itex] and [itex]\vec{y}[/itex] are vectors in some plane, and you are integrating over an entirely different plane. Is this really what you mean?
  6. May 25, 2009 #5
    Dear gabbagabbaney, that's exactly the integral I need to calculate.
    Here S is a surface which is bounded by two planes.

    Here's a picture which illustrates the situation:


    Page 41, picture D. You can see the two planes I was referring two. Here "S" is the surface which is exactly in the middle.

    How can you calculate that integral?
    Last edited by a moderator: Apr 24, 2017
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