Actual electrostatics question

Discussion in 'Physics & Math' started by §outh§tar, Jan 16, 2007.

  1. §outh§tar is feeling caustic Registered Senior Member

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    If 13 stationary source charges are placed at the vertices of a 13-sided polygon (a triskaidecagon), what is the total electric field provided by them at the geometric center of the triskaidecagon?

    I have a numeric/symbolic answer but I'd like to see someone solve it to see if our results are the same, especially since explaining my way would involve appeals to geometric ideas I can't convey in ascii.
     
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  3. Pete It's not rocket surgery Registered Senior Member

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    Are the charges identical, or arbitrary?
    Is the polygon a regular polygon?
     
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  5. CANGAS Registered Senior Member

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    I understand the thread.

    Pete: in order, your questions answered:
    Yes
    Yes

    Pete; your solution?
     
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  7. Pete It's not rocket surgery Registered Senior Member

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    Thanks CANGAS.
    It is indeed natural to assume that the answers are as you suggest, but since the question is posed by SouthStar I'd rather he clarified it.
     
  8. §outh§tar is feeling caustic Registered Senior Member

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    Cangas is right. Yes to both. I'd like also to see if anyone has the same ideas I do on generalizing to an n-gon (n>2).
     
  9. Pete It's not rocket surgery Registered Senior Member

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    By symmetry, the field at the center must be zero.

    Did you mean the total electrostatic potential?
     
  10. §outh§tar is feeling caustic Registered Senior Member

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    Ah yep. You're right. I had an analytic way of doing it though and Mathematica came up with ~3^-15, which is a poor excuse for 0.

    I'm having trouble with this one though. You have a spherical shell of charge, radius a and surface density sigma from which a SMALL circular piece of radius b << a is removed. What is the direction and magnitude of the field at the center of the hole in the shell (ignoring the removed circular piece).

    I got a field of 2*pi*sigma. I'd like to see if anyone can replicate.
     
  11. Pete It's not rocket surgery Registered Senior Member

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    Too much work for me, I think.

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