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Flux through a Sphere

  1. Nov 13, 2013 #1
    I was told it might be better to post this here.
    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.

    Also, this is the work I've done:

    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.
     
  2. jcsd
  3. Nov 13, 2013 #2
    Now supposing you got E right you need [itex] d{\bf A} = r^2 \sin \theta d \theta d\phi \widehat{r}[/itex],
    and that product is the inner product.
     
  4. Nov 13, 2013 #3
    Okay,

    ##\int_0^{2\pi}\int_0^{\pi}r^2sin^2(\theta)(sin(\theta)\hat{r}+cos( \theta)\hat{\theta})r^2sin(\theta)d\theta d\phi##
    =##2\pi r^4\int_0^{\pi}sin^4(\theta)\hat{r}+sin^2(\theta)cos(\theta)\hat{\theta}##

    second integral is zero. first integral evaluates to ##\frac{3\pi}{8}##

    so finally the flux is: ##\frac{3}{4}\pi^2r^4##

    edit:corrected power on sine
     
    Last edited: Nov 13, 2013
  5. Nov 13, 2013 #4
    I'm not sure what you are doing.

    [itex] {\bf E}=(rsin(\theta))^2(sin(\theta)\hat{r}+cos(\theta)\hat{\theta}) [/itex]
    and
    [itex] d{\bf A} = r^2 \sin \theta d \theta d\phi \widehat{r}[/itex],
    tell you
    [itex] {\bf E} \cdot d{\bf A} = r^4 sin^4(\theta) d\theta d\phi [/itex].
     
  6. Nov 13, 2013 #5
    I did ##sin^4(\theta)## but put 3 up there instead. What we have then is the same since the theta component goes to zero in the integral.
     
  7. Nov 13, 2013 #6
    that's what i'm worried about. the right answer for the wrong reason. do you see what's different
    between what i'm saying and what you are?
     
  8. Nov 13, 2013 #7
    I see that you are only doing the ##\hat{r}## direction but I don't know why you would ignore the ##\hat{\theta}##.
     
  9. Nov 13, 2013 #8

    Dick

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    Because as qbert has been trying to tell you, ##{\bf E} \cdot d{\bf A}## isn't a vector. It's a scalar. It's the inner product of the vector ##{\bf E}## and the vector ##d{\bf A}##.
     
  10. Nov 13, 2013 #9
    There is no [itex]\widehat{\theta}[/itex] direction, it is wiped out when you take the inner product in [itex] {\bf E} \cdot d{\bf A}[/itex], and that is my point. You have to do the inner product when you're doing flux integrals.
    It's only the piece of electric field which pierces the surface element which contributes to flux.
     
  11. Nov 13, 2013 #10
    I'm trying to figure out what you guys mean by inner product. I haven't heard this term before.
     
  12. Nov 14, 2013 #11

    Dick

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    Your relevant equation has a dot in it that means 'dot product', same thing as 'inner product'. Do you know what dot product means? Also sometimes called scalar product. It's an operation between two vectors that give you a scalar (i.e. a number).
     
  13. Nov 14, 2013 #12
    I know dot product. I've just never heard it called inner product. However, I don't understand, physically, why we would ignore one direction of the field. It's one thing to say if the direction of the field is only in ##\hat{r}## then that is all fine and understandable (if I did the integral in cylindrical coordinates this would work wonderfully). But, if I had 3 components to the field direction would I still ignore the other two in favor of r? I know they are all perpendicular but I don't grasp why they go away physically.
     
  14. Nov 14, 2013 #13

    Dick

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    The point is that ##\hat r \cdot \hat \theta=0##. That's why you can ignore it. dA only has an ##\hat r##. Physically the intuition is that the ##\hat \theta## component of the field isn't going into or out of the sphere, it's only going around. It doesn't matter for flux.
     
    Last edited: Nov 14, 2013
  15. Nov 14, 2013 #14
    I get it, sort of. I understand the math but when I think of flux I think of everything, no matter direction, going through a surface. I want to count all of it, not just what is perpendicular to the surface.

    I've finished the problem and it matches perfectly with the divergence of E and Gauss's theorem. Thanks a lot for the help guys!
     
  16. Nov 14, 2013 #15

    Dick

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    Great! You can count all of it, but the part of the field that's parallel to the surface is not going to contribute to the flux. Only the perpendicular part will. That's what the 'dot product' part says. Keep thinking about it.
     
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