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Electric Field due to Shell Hemisphere

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data

    Calculate the Electric Field generated by a hemisphere of radius ##R## (and uniform charge density ##\sigma##) at its center.

    2. Relevant equations

    ##\displaystyle E = \frac{Q}{4\pi \varepsilon_0 r^2}##

    3. The attempt at a solution

    I think I'm doing some mistakes with the math. When trying to find the field by spherical coordinates I get something different if I try by using rectangular coordinates.

    Attempt:

    ##\displaystyle \boxed{\mathrm{d}E = \frac{\mathrm{d} Q}{4\pi \varepsilon_0 R^2}}##

    - Spherical Coordinates:

    image.jpg

    ##\displaystyle \mathrm{d}Q = \sigma \mathrm{d}S##

    ##\displaystyle \mathrm{d}Q =\sigma R^2 \sin\phi \mathrm{d}\phi \mathrm{d}\theta##

    By symmetry ##\mathrm{d}E_y = \mathrm{d}E_z = 0## and ##\mathrm{d}E = \mathrm{d}E_x##, so

    ##\displaystyle E = \frac{\sigma}{4\pi \varepsilon_0}\int\limits_{\pi / 2}^{3\pi / 2}\int\limits_{0}^{\pi}\sin\phi \mathrm{d}\phi \mathrm{d}\theta = \frac{\sigma}{2\varepsilon_0}##

    - Rectangular Coordinates:

    image.jpg


    ##\displaystyle \mathrm{d}Q = \sigma\times \text{Circumference}\times \text{Width}##
    ##\displaystyle \mathrm{d}Q= 2\sigma\pi R^2\sin\alpha \mathrm{d}\alpha##

    Again we apply symmetry. Hence,

    ##\displaystyle \mathrm{d}E_x = \mathrm{d}E \cos\alpha##
    ##\displaystyle E = E_x = \frac{\sigma}{2 \varepsilon_0}\int\limits_{0}^{\pi /2}\sin\alpha\cos\alpha \mathrm{d}\alpha = \frac{\sigma}{4\varepsilon_0}
    ##

    There is a difference of a factor of 2 between the first and the second way. I don't know what I'm missing :/.

    Thanks!
     
    Last edited: Jun 16, 2015
  2. jcsd
  3. Jun 16, 2015 #2

    rude man

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    Hard to make out from your drawings but my vote is for the second result.
     
  4. Jun 16, 2015 #3

    Nathanael

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    The problem with your first integral is that it neglects the vectorial nature of the E-field. That is, it sums it up like a scalar instead of only summing up the x-component.
     
  5. Jun 16, 2015 #4
    Yep. That's the answer after verifying with the textbook, but I'm not sure what I'm doing wrong with the first result.

    It looks like ##\mathrm{d}E_x = \mathrm{d}E_x\cos\theta##, so the integral would be something like this:

    ##\displaystyle E = \frac{\sigma}{4\pi \varepsilon_0}\int\limits_{\pi / 2}^{3\pi / 2}\int\limits_{0}^{\pi}\sin\phi\cos\theta \mathrm{d}\phi \mathrm{d}\theta##

    By integrating it, I get ##-\displaystyle\frac{\sigma}{\pi \varepsilon_0}##, which it's not correct.

    The other thing I tried, and which gave me the right answer, is to choose the ##z## axis as the axis of symmetry instead of ##x##. In that case ##\mathrm{d}E_x = \mathrm{d}E\cos\phi##, so

    image.jpg

    ##\displaystyle E = \frac{\sigma}{4\pi \varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi / 2}\sin\phi\cos\phi \mathrm{d}\phi \mathrm{d}\theta = \frac{\sigma}{4\varepsilon_0}##

    But I still don't know how to set up the correct integral with the previous drawing. It doesn't seem that ##\mathrm{d}E_x = \mathrm{d}E\cos\phi##.
     
    Last edited: Jun 16, 2015
  6. Jun 16, 2015 #5

    Nathanael

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    Yes this is exactly what I was going to suggest: measure the angle Φ from the axis of symmetry. That makes things much simpler.

    As for how to solve it the way you originally set it up... I can tell you that the ratio of the x-component to the total force will depend on both Φ and θ.

    I haven't made sure, but my first idea is that, in your original coordinate system, it should be |dEx/dE| = |sinΦcosθ|
    (You can imagine the special cases when θ=pi/2 or θ=3pi/2 then the x-component should be zero for all Φ, and also when Φ=0 or Φ=pi it should be zero for all θ. Both these special cases are satisfied by dEx/dE=sinΦcosθ but not by your equation dEx/dE=cosΦ)

    Thus I think the integral should be:

    [itex]\displaystyle E = \frac{\sigma}{4\pi \varepsilon_0}\int\limits_{-\pi/2}^{\pi/2}\int\limits_{0}^{\pi}\sin^2\phi\cos\theta \mathrm{d}\phi \mathrm{d}\theta[/itex]

    (I changed the bounds on θ from [pi/2 to 3pi/2] to [-pi/2 to pi/2] simply so we don't get an extra negative sign from cosθ but as long as you know which direction the force is in, this should get you the same magnitude for either set of bounds)

    I haven't calculated it to see if it's correct or not but as far as I can tell it ought to be.

    Anyway, this is a good example of how choosing the right orientation of the coordinate system can make things simpler.
     
    Last edited: Jun 16, 2015
  7. Jun 17, 2015 #6
    Yep. That's the integral I got after being more careful with the projections of ##\mathrm{d}\boldsymbol{E}##. ##\mathrm{d}E\sin\phi## just gives the component of the field parallel to the ##xy##-plane. From this one the final component on the ##x## axis has to be found: ##\mathrm{d}E\sin\phi\cos\theta##.

    image.jpg

    This gives the integral you've indicated. I've evaluated it (with that regard in the bounds for ##\theta##). It's a bit more involved but the answer is the correct one anyway.

    It becomes clear the issue with ##\phi## if one tries to imagine a ring being swept around the plane perpendicular to the chosen axis of symmetry. In the first case (##x## as the axis of symmetry and the ring as in the picture) ##\phi## changes as one moves along the circle, while in the second case (ring parallel to the ##xy##-plane) it remains fixed.

    Pretty neat. Thanks!! (:
     
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