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Electric field at point Gauss law

  1. Dec 12, 2014 #1
    1. The problem statement, all variables and given/known data

    We wont to find out the electric field at a point p due to a point charge q placed at o as shown in the figure

    IMG_11300006.jpg


    Consider a sphere of radius r passing through the point p. Let the electric field at p be E then by Gauss law

    IMG_113006.jpg


    my problem is how we get the last step? I dont understand how he integrate it.

    3. The attempt at a solution

    i tried it and found a solution but not sure if its correct although i got the same result

    IMG_1134.jpg


    sorry that i don't used LATEX but I'm not used with it and i cant write the whole question with LATEX
     
  2. jcsd
  3. Dec 12, 2014 #2

    Doc Al

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    You need to learn the rules for integration (and differentiation) of ##x^n##. Try this: Power Rule
     
  4. Dec 12, 2014 #3

    BvU

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    Looks as if the power rule is known, but the concept of integration causes difficulties: substitution ##x=r^{-3}## is carried out ok initially (*), but the much better (or rather, the 'right') substitution ##x = r^{-2}##, leading to ##dx =-2\; r^{-3}dr## and hence ##\int r^{-3} dr = -{1\over2}\int dx## is overseen.

    (*) Later on, the swap ##\int {x\over r^{-4}}\; dx = {1\over r^{-4}} \int x\; dx## is unforgivable: r is not a constant but depends on x !!
     
  5. Dec 12, 2014 #4
    If r is not constant and depends on x. Therefore i cant put the ##r^4## outside the integration. Right??
     
  6. Dec 12, 2014 #5

    BvU

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    Indeed you can not !
     
  7. Dec 12, 2014 #6
    Ok so how to get this solution?
    Any idea??
    I want the the same result ##E=\frac{q}{4 \pi \epsilon r^2}##
     
  8. Dec 12, 2014 #7

    BvU

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    OK, we have to make a step back. Your "Consider a sphere of radius r passing through the point p. Let the electric field at p be E then by Gauss law " continues with a picture with formulas that seem to be out of context: at P there is no charge, so div E is zero there !

    Where does this snippet come from? It also integrates from ## r=-\infty## to ## r=\infty## which is very weird, since r can't be negative. Then it equates that to ##2\int_0^\infty\ ##, then inserts a ##- {1\over 3 }## from nowhere (re-check the power rule)!

    So, stepping back and trying to find the correct path:

    Gauss law replaces a surface integral of E by a volume integral of rho. Indeed ##\nabla \vec E = {\rho\over \epsilon_0} ##, but ## \rho \ne q\big /{4\over 3}\pi r^3\ ## ! That is ##\rho## for a uniformly charged sphere with radius r !

    In this simple case the volume integral of rho simply yields q. Frome there use rotation symmetry and gauss law to get to E.


    Another thing: note that div E is not ## dE\over dr## !

    Divergence in spherical coordinates tells us that

    for rotational symmetry ##\nabla\cdot\vec E = {1\over r^2}\; {\partial (r^2E_r)\over \partial r}##

    (
    as you can check: div E = 0 for the desired result for E

    and conversely: with rotational symmetry, Er2 is constant outside the area where charge is present.​
    ).

    [edit] forgot the divergence link, inserted it.
     
  9. Dec 12, 2014 #8

    Doc Al

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    Sure. Learn how to integrate ##x^n##; in particular, ##\frac{1}{x^3} = x^{-3}##. (No fancy substitutions needed.)

    See the link I gave about the power rule.
     
  10. Dec 12, 2014 #9
    I know how to use the power rule.
    If you see my solution ;)


    @BvU
    Thank you for your help. Looks like this question is a mess. The whole question is wrong
    I will now use your information and maybe i will get a right solution.
     
  11. Dec 12, 2014 #10

    Doc Al

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    Then why did you say the following?
    The last step was integrating 1/x3.

    I guess I missed something. (Sorry about that!)
     
  12. Dec 12, 2014 #11
    Yes thats right but if you try to integrate the last step you will not get the same result. Therefore i said that i dont understand it. But as BvU told me the whole question has mistakes.
     
  13. Dec 12, 2014 #12

    BvU

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    So is the picture with the formulas from a different exercise ? Or did you do a part of the solution in TeX and then switch to handwritten ?
     
  14. Dec 12, 2014 #13
    No the two pictures ( figure and Tex ) are from exercise. Just the handwriting is mine. But the exercise it is not from one source cause the exercise has the soltuion from a screenshot. It has text and image ( the soultion and the figure ) are screenshots.
     
  15. Dec 13, 2014 #14

    BvU

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    So, do we have the "all clear"now, or is there still some thing unfinished ?
     
  16. Dec 13, 2014 #15
    Thank you very much. Its clear now
     
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