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Question about calculating electric field made by finite point charges

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi!
    I have a question about calculating electric field made by finite point charges
    q[itex]_{1}[/itex],q[itex]_{2}[/itex],..., q[itex]_{n}[/itex].
    From the book "introduction to electrodynamics", you can see that the electric field E at a point P made by the finite point charges can be calculated by the below equation;
    E(P) = [itex]\frac{1}{4\pi\epsilon_{0}}[/itex][itex]\sum[/itex][itex]^{n}_{i=1}[/itex][itex]\frac{q_{i}}{r_{i}}[/itex][itex]\hat{r[itex]_{i}[/itex]}[/itex] where r[itex]_{i}[/itex]'s are the distances between a point charge and the point P.

    I can see that above electric field makes sense if P is located at a different position from each point charge q[itex]_{i}[/itex].
    However, what if P is located at one of those places at which the point charges are located? For example, what is the electric field at the position at which q[itex]_{1}[/itex] is located?? As you can see, the electric field function has a singular point at the point, so that I think it is impossible to calculate it. Am I right?? Does it mean that the electric field doesn't exist at the point??



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 17, 2012 #2
    Re: Electric field

    Then you have an infinite electric field at that point :cry:

    Really though, I don't think that this situation would be physically possible. The field is representative of the force that a charge would experience at a given location. Since that location is occupied, you can't put a charge there. At least that's my understanding.

    I believe that if you put a charge infinitely close to another charge, the whole idea of coulombs law would break down, since now you are entering the domain of the nuclear force, and I am positive other laws apply.
     
  4. Apr 18, 2012 #3
    Re: Electric field

    Thanks for your responding!

    I actually asked you above one because of some equations.
    I attach it here.
    div.3.jpg
    The divergence of E.jpg
    Div.2.jpg

    As you can check in the first equation, it represents the electric field at a point r occured by continuously distributed electric charges.

    Here's my question in the second picture...
    It is said that because the charge density out of the electric charges is zero everywhere, the integral of the equation in the first picture can be extended over the whole space....
    But, in this case, you can see that the integrand inevitably contains the singularity at r=r'. What does it mean? I can't understand the meaning of the integral;;
    Furthermore, if r indicates a point within the volume of the distributed charges, it would be more complicating...
     
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