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Homework Help: Electric Charges

  1. Sep 2, 2004 #1

    I need some guidance. This is the problem: Imagine that we have a half sphere (hemishpere) that is hollow with a charge of +q distributed evenly across the inside of the "bowl". What is the electric field due to this charge at point P? I have attached an image to help visualize this problem. A hint I was given was to thing of this hemisphere as many rings of charge, each getting smaller and smaller. I am not sure if I need to be using a limit or a sum, or integral for that matter. If anyone knows how to work the problem, a few tips to get me in the right direction would be apprecitated!


    http://home.comcast.net/~dennis_dwyer/sphere.JPG [Broken]
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Sep 2, 2004 #2
    use a gaussian surface and take advantage of its symmetry, but make sure your point P is on the GS or else there will be error.

    [tex] \Phi = \oint E \cdot dA [/tex]

    E will be constant, so it can be taken out of the integration, etc.
  4. Sep 2, 2004 #3
    another hint - the electrical field outside a spherical shell of charge with radius r and charge q is directed radially and has magnitude

    [tex] E = \frac{q}{4\pi \epsilon_{0}r^2} [/tex]
  5. Sep 2, 2004 #4
    Wouldn't it be easier to think of the hemisphere as many rings of charge? If I know [itex] E [/itex] of a whole sphere, how can I cut that sphere in half and only calculate [itex] E [/itex] for the hemisphere?

    Due to the symmetry, I know that the resultant electrical field will be in the [itex] \hat{i} [/itex] direction.

    And if I were to consider a GS for this problem, it wouldn't be a sphere....the shape seems irregular to me and am not sure thats the right way.
  6. Sep 3, 2004 #5

    Claude Bile

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    Couloumb's law with the proper integration should suffice.

    You are on the right track in considering the surface as many rings of charge. The integral you need to perform will be two dimensional, but simple if you exploit the symmetry of the problem.

    First integrate around the 'ring' (This should be easy since the distance, R, and thus E are constant), then integrate from radius r to radius 0.

    Also, because you have a charge distribution rather than a point charge, you need to replace q with something more appropriate.

    Good Luck.

  7. Sep 3, 2004 #6


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    There is no symmetry involved in the problem that will allow you to pull E out of the integral(for Gauss's Law).

    The best method is the one you mentioned SpatialVacancy, with the rings of charge.
  8. Sep 3, 2004 #7

    Ok, I have gotten this far...

    [tex] E_x = \dfrac{kQ cos \theta}{r^2} [/tex]

    Which equals

    [tex] \dfrac{kQx}{r^3} [/tex]

    Which equals

    [tex] \dfrac{kQx}{(x^2 + R^2)^{\frac{3}{2}}} [/tex]

    So from here I am not really sure what to do. I found the equation:

    [tex] E_x = \dfrac{k\lambda 2 \pi RQx}{(x^2 + R^2)^{\frac{3}{2}}} [/tex]

    But I'm not seeing it..... :frown:
  9. Sep 5, 2004 #8

    Claude Bile

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    You are sort of on the right track, though you have left out some important steps. Let me start you off:

    [tex] E_x = \dfrac{kQ cos \theta}{r^2} [/tex]

    Okay, this is Coulomb's law, but it only applies to a point charge. In order to utilize it, we need a more general form (i.e. one that applies to charge distributions).

    [tex] E_x = \iint \dfrac{k\sigma cos \theta}{r^2} dA [/tex]

    Essentially, we are now treating the charge distribution as a series of point charges. The total electric field will be the total sum of the electric fields due to these point charges where each point charge has magnitude sigma.dA where sigma is the surface charge density and dA is a small area element.

    Now you need to evaluate the integral, and this is where the original hint applies, that is, breaking the hemisphere into rings of charge.
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