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Problem regarding induction and Gauss' Theorem

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data

    There are three concentric thin spherical shells A,B and C of radii a,b and c respectively. The shells A and C are given charges q and -q respectively and the shell B is earthed. Find the charges appearing on the surfaces of B and C.

    2. Relevant equations

    Gauss' Theorem and Induction charging.

    3. The attempt at a solution

    I know I will have to show an attempt. But here is the thing which hinders by attempt :

    If the shell B is earthed then the net potential at its surface should be zero. Then how does this question expect me to evaluate for the net charge at shell B ? I cannot understand. Please clear this up. Then I will be happy to show my effort to solve further.

    Thanks in advance.... :smile:
     
  2. jcsd
  3. Jun 18, 2013 #2

    ehild

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    You can assign zero potential at any place. It does not influence the charge.

    ehild
     
  4. Jun 19, 2013 #3
    I know its up to the user to choose an arbitrary point as zero potential. By convention and mere logic the point in vicinity of charge is defined to be zero potential.

    But the potential of earth is always zero. If a shell is earthed, its potential can only be zero when the charges on the shell are neutralized by conduction. I can not understand.. :confused:
     
  5. Jun 19, 2013 #4

    ehild

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    No, that is not true. Think of a point charge. We say, the potential is kQ/r at distance r from is. Where is that potential zero?

    A conducting plate or sphere connected to the earth is at zero potential, but that does not mean it is not charged. Think of a parallel plate capacitor. One plate earthed, the other is given some charge. The earthed plate becomes oppositely charged, as the electric field of the charged plate repels like charges from it into the ground and attracts opposite charges from the ground on it.


    ehild
     
  6. Jun 20, 2013 #5
    Seems like a replied hastily and thus misused the vocabulary by inserting the word "vicinity" here. :redface: Apologies on my part. I meant that a point, infinitely far from the influence of any other charge is defined to be the point of zero potential, by convention.

    Ok got the concept. But I am feeling troubled as to how to begin this problem. "A" shell being +q charged induces -q charge on inner surface of shell B and +q charge on outer surface of the same shell B. Correct till now ?
     
  7. Jun 20, 2013 #6

    ehild

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    For an isolated charged object the zero potential is in infinity, but you can put an earthed object close and then the potential is zero there.

    It is correct that the q charge of shell A induces -q charge on the inner surface of shell B. But B is grounded, and can suck up charges from the earth. You do not know what charge is on the outer surface of B. Let it be Q. Then -Q charge is on the inner surface of C - what is on the outer surface? And then, what is the potential of C? The potential has to be zero at infinity, and also on the shell B.

    ehild
     
  8. Jun 20, 2013 #7

    rude man

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    Sometimes the potential at infinity cannot be zero. For example, the potential of an infinitely long wire cannot be zero at infinity since it requires an infinite amount of work to bring a point charge from infinity to any point a finite distance from the wire.

    In this case you can still define zero potential at any finite distance from the wire and then speak of the potential at any other point a finite distance from the wire.
     
  9. Jun 22, 2013 #8
    Ok, so let the final charge on shell B be -Q on inner surface and +Q on outer surface. Inner shell also induces charge q, so that inner surface of B now have charge -(Q+q) and outer surface has a charge of Q+q. Now the shell B again induces charge of -(Q+q) on inner surface of shell C and Q+q on outer surface of shell C.

    Now since shell B is earthed, the net potential on it is 0. So, k(Q+q)/b -k(Q+q)/c=0

    :confused:
     
  10. Jun 22, 2013 #9

    ehild

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    Remember, the electric field emerge from charges and end in charges (or infinity).
    Assume a<b<c, shell A is inside, C is outside and B is in between.

    Shell A has charge q, that entirely appears on its surface. That means q/ε electric field lines between shells A and B, according to Gauss' Law. These end on the surface charges on the inner surface of B, which means -q surface charge there.

    As B is grounded, it can have any charge. Assume that the surface charge on the outer surface is Q. Gauss' Law says that that the number of the electric field lines between B and C is Q/ε. These must end on the negative charges on the inner surface of C: The surface charge on the inner surface is -Q. The total charge of C is -q, the charge is shared among the inner and outer surfaces, so there is Q-q surface charge outside.

    The potential is zero in infinity and on shell B. You need to write up the potential difference between C and infinity and between C and B , in terms of q and Q and the radii b and c. That gives two expressions for the potential of C. Equating them, you get the charge Q.

    ehild
     

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  11. Jun 24, 2013 #10
    Thanks for the explanation ehild !! :smile:

    I got the correct answer !!

    BTW, thanks to "rude man" too.
     
  12. Jun 24, 2013 #11

    ehild

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    Congratulation!:smile:

    ehild
     
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