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Charged sphere and charged conducting shell

  1. Jul 27, 2017 #1
    1. The problem statement, all variables and given/known data
    A + q = 5 pC charge is uniformly distributed on a non-conducting sphere of radius a= 5 cm , which is placed in the center of a spherical conducting shell of inner radius b = 10 cm and outer radius c = 12 cm. The outer conducting shell is charged with a -q charge. Determine:
    1) the charges on the inner and outer surfaces of the shell;
    2) the electric field (module, direction) everywhere;
    3) the electrostatic potential on the external surface of the conducting shell (r = c), on the internal surface of the shell (r = b) and on the outer surface of the internal sphere of radius a (r = a).
    Suppose now to replace the inner sphere with a spherical conductor of radius a charged with the same + q charge:
    4) Which of the previous answers will change and how?
    5) Determine the potential everywhere;
    6) Determine the capacitance of the spherical capacitor formed from the internal conducting sphere of radius a and the outer conducting shell;
    7) If a proton (m = 1.67·10-27 Kg) starts from rest from the spherical conductor of radius a, which will be its speed when it hit the inner surface of the outer spherical shell?


    2. Relevant equations
    /

    3. The attempt at a solution

    img043.jpg

    So, I would like to know if my attempt to solve the problem is correct, and how can I continue it in the parts I didn't manage to solve...
    Thank you!
     
  2. jcsd
  3. Jul 27, 2017 #2

    haruspex

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    No.
    Please explain how you arrived at your answer to 1)
     
  4. Jul 27, 2017 #3
    For the first point my reasoning is: since on the inner sphere we have a + q charge, because of the induction phenomenon, the inner surface of the outer shell has to "balance" that charge, and thus has a charge -q. Since the problem statement tells us that -q is the total shell charge, we know that there is no charge on its outer surface...
     
  5. Jul 27, 2017 #4

    rude man

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    Seems to me his answer to (1) is correct.
    Placing gaussian sphere at b < r < c forces qb = -q.
    Placing gaussian sphere at r > c forces qc = 0.
     
  6. Jul 27, 2017 #5

    rude man

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    You can just integrate the E field using your data from (2).
     
  7. Jul 27, 2017 #6

    haruspex

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    My mistake - I read it as +5q instead of +q = 5pC.

    I don't understand your attempt at 5). Looks more like an attempt at 6).
     
    Last edited: Jul 27, 2017
  8. Jul 27, 2017 #7

    haruspex

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    Yes, but we're not told where to set the zero potential. The usual convention is to set that at infinity.
     
  9. Jul 27, 2017 #8

    rude man

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    Yes, and that is where he should and can integrate the E field from. What's the problem?

    The custom is to set potential to zero at infinity unless that is impossible - as with a charged wire of infinite length.
     
  10. Jul 27, 2017 #9

    haruspex

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    No problem, just adding a detail to your guidance.
     
  11. Jul 27, 2017 #10

    rude man

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    10-4 ( means "OK", from an old US television show called "Highway Patrol" from which hopefully you were spared! )
     
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