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Flux through sphere

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    The trick to this problem is the E field is in cylindrical coordinates.
    ##E(\vec{r})=Cs^2\hat{s}##


    2. Relevant equations
    ##\int E \cdot dA##


    3. The attempt at a solution

    I tried converting the E field into spherical coords and I can find the flux that way but it is a complicated answer. The problem suggests keeping the field in cylindrical and doing the integral of the circle in cylindrical instead of spherical. I'm sort of lost on how I would do that. Would I have the limits of s be 0→R and z -R→R and ##\phi## the same as hat it would normally be?

    I doubt it is that simple but since I've never tried to use non-optimal coordinates for an object I'm not entirely sure how I would go about this.
     
  2. jcsd
  3. Nov 13, 2013 #2

    rude man

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    And the problem is ... ?
     
  4. Nov 13, 2013 #3
    The problem is to find the flux through a sphere where the E field is given in cylindrical coordinates. I can't convert the field into spherical as the question specifically asks that I do it the other way. And, I must also finally graph the divergence on the sz plane.
     
  5. Nov 13, 2013 #4
    For example

    ##E=Cs^2\hat{s}##
    ##s=rsin(\theta)## and ##\hat{s}=sin(\theta)\hat{r}+cos(\theta)\hat{\theta}##
    so ##E=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})##
    ##\int E \cdot dA=E4\pi r^2=4\pi r^2(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta})##

    The next step it asks me to calculate the divergence of E and then graph it on the sz plane.

    I can do this with the original equation but I now have answers in two different coordinate systems. Which I suppose sounds fair since they did gave me two also.

    ##∇\cdot E=\frac{1}{s}\frac{∂}{∂s}(sE_s)##
    ##=\frac{C}{s}(3s^2)=3Cs##

    Finally, it asks that I now do the integral ##\int (∇\cdot E) dV## to show that the two methods are equivalent. At first glance I would say they are not. So I probably made a mistake somewhere.
     
  6. Nov 13, 2013 #5

    rude man

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    I've never encountered "s" in any cylindrical coordinate system. The cylindrical coordinates are usually denoted r, theta, z or r, phi, z.

    Also, your ## expressions are not being translated, at least not on my computer.
     
    Last edited: Nov 13, 2013
  7. Nov 13, 2013 #6
    I get that a lot. It's something, I think, that is inherited from Griffiths since his books are popular on campus.

    We normally do ##(s,\phi, z)##
     
  8. Nov 13, 2013 #7

    rude man

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    OK, that's fine. Griffith is very popular so maybe things have changed since my time ...

    And I apologize for my comment about your ## expressions. I was looking at the "go advanced" window ...

    Gotta think a bit.

    Meanwhile, you might re-post this in the math section since it involves a dot-product in cylindrical coordinates. I would have to translate the components into cartesians before taking the dot product unless it's very simple.
     
    Last edited: Nov 13, 2013
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