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Electric Field in a Hemisphere

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A nonconducting hemispherical cup of inner Radius R has a total charge Q spread uniformly over its inner surface. Find the electric field at the center of curvature.

    2. Relevant equations

    [itex]\sigma[/itex] = [itex]\frac{Q}{2piR^{2}}[/itex]

    dq = [itex]\sigma[/itex] dA = [itex]\sigma[/itex] R[itex]^{2}[/itex]Sin[itex]\theta[/itex]d[itex]\theta[/itex]d[itex]\phi[/itex]
    k = [itex]\frac{1}{4\pi\epsilon_{}0}[/itex]

    3. The attempt at a solution

    dE = [itex]\frac{kdq}{R^{2}}[/itex]
    dE = [itex]\frac{k \sigma R^{2} Sin\theta d\theta d\phi}{R^{2}}[/itex] = [itex]k \sigma Sin \theta d \theta d \phi[/itex]

    E = [itex]k \sigma \int^{2 \pi}_{0} \int^{\frac{\pi}{2}}_{0} Sin \theta d \theta d \phi[/itex]
    E = k [itex]\sigma 2 \pi [/itex] = k [itex]\frac{Q}{R^{2}}[/itex]

    But for some reason, the answer is listed as k [itex]\frac{Q}{3R^{2}}[/itex]. I have no idea why my answer is factor of 3 larger. Any help would be appreciated.
     
  2. jcsd
  3. Sep 12, 2011 #2

    ehild

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    Do not forget that the electric field is a vector quantity, so is dE. You can add the contributions as vectors: Determine the x, y and z components of dE and integrate.

    ehild
     
  4. Sep 12, 2011 #3
    Alright, so the x and y components of the field will cancel by symmetry correct? Therefore, only the z-component dEz = dECos[itex]\theta[/itex] will remain.

    Ez = [itex]k \sigma \int^{2 \pi}_{0} \int^{\frac{\pi}{2}}_{0} Sin \theta Cos \theta d \theta d \phi[/itex]
    Ez = [itex]k \sigma \int^{2 \pi}_{0} \frac{1}{2} [Sin^{2} \theta]^{\frac{\pi}{2}}_{0} d \phi[/itex]
    Ez = [itex]k \sigma \int^{2 \pi}_{0} \frac{1}{2} d \phi[/itex]
    Ez = [itex]k \sigma \pi[/itex] = k [itex]\frac{Q}{2 R^{2}}[/itex]

    This still seems to be off from the correct answer unless I did this wrong.
     
  5. Sep 12, 2011 #4

    BruceW

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    It looks correct to me. I don't know why the answer is listed differently...
     
  6. Sep 12, 2011 #5
    So the correct answer is k [itex]\frac{Q}{2 R^{2}}[/itex] ?
     
  7. Sep 12, 2011 #6

    BruceW

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    Yes, that's what I get the answer to be.
     
  8. Sep 12, 2011 #7
    My definition of the center of curvature is correct right? It would be the center of the bottom of the hemisphere? If so, I guess the book's answer is wrong.
     
  9. Sep 12, 2011 #8

    BruceW

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    Yes, I would have thought the centre of curvature was the centre of the bottom of the hemisphere. (since this point is equidistant from any point on the curve).

    It seems to me that the book is wrong. This does happen sometimes. (Rarely, though).
     
  10. Oct 28, 2013 #9
    If the cup is closed (with a cover), it adds /pi r^2 to the surface and the book's answer is correct.
     
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