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Gauss' law for concentric circles

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has total charge -2q and the outer shell has charge +4q.
    a) calculate the electric field (magnitude and direction) in termsof q and the distance r from the common center of the 2 shells fori) r<a; ii) a<r<b; iii) b<r<c; iv) c<r<d; v)r>d.
    b) what is the total charge on the i) inner surface of the smallshell; ii) outer surface of the small shell; iii) inner surface of the large shell; iv) outer surface of the large shell.



    2. Relevant equations
    E.4πr^2=q/ε0

    3. The attempt at a solution
    concentric.png

    This is drawn in my textbook
    I have following doubts
    1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.
    2)what is the difference between the positions p3 and p4 because my textbook states at P3 electric field would have some magnitude but at P4 electric field would be zero because it lies inside the conductor then why P3 is not considered to be inside the conductor?
     
    Last edited: Sep 19, 2015
  2. jcsd
  3. Sep 19, 2015 #2

    blue_leaf77

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    In the question it's stated that
    But the picture shows otherwise, which is the correct one?
     
  4. Sep 19, 2015 #3
    I have fixed that.
     
  5. Sep 19, 2015 #4

    Zondrina

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    Let me start by saying the innermost cavity with radius ##a## is going to have zero enclosed charged, and therefore the electric field is zero inside of the cavity.

    Now, what about between ##a## and ##b##? Remember the charge is distributing itself uniformly across the outer surface of the inner conductor.

    Then the diagram indicates a ##-2 q## charge on the outside of the innermost shell and a ##+2 q## charge on the inside of the outermost shell. This indicates an electric field that points from the inside of the outermost shell to the outside of the innermost shell. You may use Gauss' law to find ##\vec E##.

    Now what is the electric field between ##c## and ##d##? Hint: Think about the field between ##a## and ##b##.

    Finally, the outermost shell has a ##+2 q## charge on its surface, and therefore the electric field points radially outward from the sphere. Use Gauss' law to find the electric field once again.

    Well for one, the two points are at different locations. The electric field at ##P3## is due to the opposing charges on the inside of the outer conductor and outside of the inner conductor. There must be an electric field at ##P3## as a result, whereas at ##P4## the electric field should be zero because you are on the conductor itself. ##P3## is not inside a conductor because it is outside the inner conductor which has a charge on its surface and ##P3## is inside the outer conductor which also has a charge on its inner surface.
     
  6. Sep 19, 2015 #5
    P1 is also not inside the conductor,right?
     
  7. Sep 19, 2015 #6
    This question is still remaining.
     
  8. Sep 19, 2015 #7

    blue_leaf77

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    Such charge redistribution for the larger sphere is required to satisfy the requirement that the electric field inside a conductor must be zero. In order to have ##E=0## for ##c<r<d##, the net charge enclosed by a spherical surface with radius ##r## with ##c<r<d## must be zero. Since you already have ##-2q## total charge from the small sphere, you need ##+2q## to be placed on the inner surface of the large sphere. You are left with the remaining ##+2q## in the large sphere and these charges must reside on the outer surface for the same reason as above, ##E=0## inside any conductor.
     
  9. Sep 19, 2015 #8

    Doc Al

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    Imagine a Gaussian surface inside (within the material) the outer conductor (intersecting P4). What's the field over that Gaussian surface? Apply Gauss' law.
     
  10. Sep 19, 2015 #9
    zero.electric field inside the conductor (here gaussian surface) is zero.
     
  11. Sep 19, 2015 #10

    Doc Al

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    Good. So what must the total charge enclosed by that surface be?
     
  12. Sep 19, 2015 #11
    zero.
     
  13. Sep 19, 2015 #12

    Doc Al

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    Good. So what's the total charge on the inner conductor? Given that, what must be the charge on the inner surface of the outer conductor to cancel it?
     
  14. Sep 19, 2015 #13
    Got your points.Thanks @Doc Al & @blue_leaf77
     
  15. Sep 19, 2015 #14
    The same question asks me to make a graph of E as a function of r.Could you please help me in that too?
     
  16. Sep 19, 2015 #15

    blue_leaf77

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    Have you calculated the electric field in each region?
     
  17. Sep 19, 2015 #16
    Yes.
     
  18. Sep 19, 2015 #17

    blue_leaf77

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    Then, making a graph of E(r) should not be a problem.
     
  19. Sep 19, 2015 #18
    Hmm...let's see.
     
  20. Sep 19, 2015 #19
    I want to know how the blue lines are figured out?
    graph.png
     
  21. Sep 19, 2015 #20

    blue_leaf77

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    I think the question asks you to calculate the magnitude of the field, and hence should be positive everywhere.
    Figure out the blue lines, you mean the values on the vertical axis? You have figured it out, haven't you?
     
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