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Homework Help: Magtnitude and direction of e&m

  1. Sep 14, 2008 #1

    A thin nonconducting uniformly charged spherical shell of radius R has total positive charge of Q. A small circular plus is remoced from the surface. What are the magtnitude and direction of E at the center of the hole?

    So this shell has a hole now. Can I just treat this as a warped sheet and do E*dA = Q/2*epsilon? Then I can do E*dA = Q/2*epsilon. And then I will have E = Q/A*2*epsilon =

    Q/8*pi*R^2? Somehow this seems wrong to me.

  2. jcsd
  3. Sep 14, 2008 #2
    Re: E&m

    instead of removing a positive charge maybe you should think of it as adding a negative charge.
  4. Sep 15, 2008 #3
    Re: E&m

    So if it's a negative charge, then wouldn't the charge there still be E*dA = Q/e, where Q now is the positive charge plus (or in this case minus) the negative charge? But I don't know what the negative charge is, so if it is q, would I still do.

    E*dA = Q/e. Since dA = 4*pi*r*2, so E is (1/4*pi*e)(Q -q/r^2)?

    Not sure if this is the right answer.
  5. Sep 15, 2008 #4


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    Homework Helper

    Re: E&m

    What's a "small circular plus"? And what do you mean by "warped sheet"? I'll think in terms of symmetry and consider the fact that since a small spherical cap is removed, there is nothing to cancel out the electric field due to the spherical cap of equal shape and size on the opposite end of the spherical shell. So just find that and you're done.
  6. Sep 15, 2008 #5
    Re: E&m


    Thanks, but if I'm not given a r of the small spherical gap, how can I come up with the exact charge? Because wouldn't I be finding the portion towards the total charge of the sphere this small gap contributes? And wouldn't that depend on the area of the small gap?
  7. Sep 15, 2008 #6
    Re: E&m

    my guess is that youre not given r because it doesnt matter. my guess is that the field there is zero. no that cant be right. the charge there is zero but the field cant be zero.

    calculate the field without the missing plug. then calculate the field dues solely to theimaginary negative plug. then add them. the first one is easy.
  8. Sep 15, 2008 #7
    Re: E&m

    yes. thats the whole point.
  9. Sep 15, 2008 #8
    Re: E&m

    cant you just treat it as an infinite flat plate. then all you need is the charge density.
  10. Sep 15, 2008 #9
    Re: E&m

    So this means that I can treat it as an infinite plate? But can you tell me why I can treat it as an infinite plate? THanks.
  11. Sep 15, 2008 #10
    Re: E&m

    because you are finding the field precisely on the surface of the shere. from the point of view of an infinitesimal point an infinitesimal distance from the surface the surface is an infinite plane.

    I'm assuming that when they say 'small' plug that they mean so small that its curvature can be neglected.
  12. Sep 15, 2008 #11
    Re: E&m

    Got it. Thanks.
  13. Sep 15, 2008 #12
    Re: E&m

    So I can do E*dA = ro/2epsilon. And here the Area would still be 4*pi*r^2? where r is the r of the sphere?

  14. Sep 15, 2008 #13
    Re: E&m

    forget equations for a moment. just picture the flux lines. isnt it obvious that it will be 1/2 the field strength everywhere else on the sphere? thats how it looks to me.

    the field lines for th e sphere only go in one direction. the field lines for an infinite plane go in 2 directions. hence 1/2

    am I wrong?
  15. Sep 15, 2008 #14
    Re: E&m

    Okay I see. So this is because a plate the field lines can go through the front and then outwards from the back? And because E is uniform throughout the sphere and all the field lines go out of the sphere, and so you can just do 1/2 the field strength of the sphere. I guess I got confused between flux and E because for some reason I was thinking that E would be different throughout the sphere and that the greater the area the greater E would be. But that's not correct. Thanks!
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