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Superposition principle for potential?

  1. Jan 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A solid, conducting sphere of radius "a" and charge -Q is concentric with a spherical conducting shell of inner radius "b" and outer radius "c". The net charge on the shell is +3Q. Take the zero of electric potential to be at some point at infinity.
    a.) Use Gauss's law to find the charge on the inner and outer surface of the shell.
    b.) Use the superposition principle for potential to find the potential at all points in space.
    c.) Use the results of part b) to find the electric field at all points in space.

    2. Relevant equations
    E= -dv/dx

    3. The attempt at a solution
    For part a) I took a gaussian surface between the radius b and c, so the E field is zero. Therefore Qinner-Q=0, so Qinner=Q. So outer charge must be 2Q then.
    I am completely stuck on part b). What I think I am supposed to do is get the net charge, 2Q, and multiply that by Ke, and divide it by the radius. So basically the answer will be 2QK/r, where r is the radius of the point in space I am trying to find, but that seems really wrong, especially when I am trying to find the E fields from that, as taking the negative derivative will just give me the same result. How should I do this?
     
  2. jcsd
  3. Jan 30, 2015 #2

    mfb

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    That is correct outside the outer shell, it is not correct everywhere.
    What do you mean?
     
  4. Jan 30, 2015 #3
    Well what my professor said to me as an example find the potential between the outer radius c and the inner radius b. He said that the answer to that would be the net charge of the whole system, 2Q, divided by the radius r, from the center to a point inbetween the inner and outer radius multiplied by Ke. So if I am going based off that then I will always have 2KQ over whatever the radius is, which seems wrong to me. It also makes it so if I try to get the E-field between, say, the inner and outer radius by the formula E= -dv/dx, I will not get zero, which the E-field should be.. I think I am just really not understanding how to use the superposition principle.
     
  5. Jan 30, 2015 #4

    mfb

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    For a specific radius, this happens to be right, but this is a weird approach.

    Potentials are continuous - they don't make jumps. You know the potential for all points outside the outer surface, so you know the potential at the surface.
    Now you have two alternative ways to proceed, both will lead to the potential between b and c:
    a) What is the electric field in the conducting shell? What does that tell you about the potential?
    b) Find a more general expression for the potential of a charged sphere where the potential at infinity can be some other value. Try to apply it to the region between b and c.
     
  6. Jan 30, 2015 #5
    The electric field in the conducting shell would be zero, so V would be the same as the V on the surface correct? So 2QK/c?
    I'm not sure how to go about doing it the way you describe in b) though.
     
  7. Jan 30, 2015 #6

    mfb

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    Right.
    This will be needed for the region between b and a.

    Potentials describe the same physics even if you add a constant to it because the potentials itself are not physical reality, only potential differences are real. You can use this constant to find a solution for the range between a and b. It has to follow the law for the field around a charge, but it also has to match the potential at a radius of b.
     
  8. Jan 30, 2015 #7
    I'm really not sure how to do that at all. How would I get this constant?
     
  9. Jan 30, 2015 #8

    mfb

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    It comes from the part where you get the potential at r=b right (=continuous).
     
  10. Jan 30, 2015 #9
    What do you mean with "right"? And the potential at r=b, that would be 2QK/b right?
     
  11. Jan 30, 2015 #10

    mfb

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    "right" as in "not wrong".
    Yes, and that potential should not make a "jump" at r=b if you go to the formula for the next region.
     
  12. Jan 30, 2015 #11
    So then when r=a the potential would be 2QK/a which is the same when r<a. If that is right all I have left is a<r<b which seems to be the trickiest one. I calculated the electric field using a gaussian surface and got the e-field to be -QK/r^2. So the potential between a and b should be -QK/r right? Now how to find that using the superposition principle I still don't get. And what do you mean by the formula for the next region?
     
  13. Jan 31, 2015 #12

    mfb

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    No. How did you get that result?
    You'll need the potential between b and a to find the potential for r<a.

    That does not work, the potential at r=b would not be continuous. -QK/r reproduces all potential differences, but the potential has to be something else.
    Exactly the point you are working on.
     
  14. Jan 31, 2015 #13
    For potential between a and b, would it work if I made the problem into 3 separate spheres, so one sphere of radius c, one of radius b, and one of radius a? so when r<c the potential would be 2QK/c, when r<b the potential would be QK/b, and when r>a the potential would be -QK/r? So the adding all those together the potential between a and b would be QK(2/c + 1/b - 1/r)
     
  15. Jan 31, 2015 #14

    mfb

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    You already calculated that there is an electric field - the potential cannot be constant. Also, your solution would be not continuous.

    Adding them would give a constant potential everywhere, which is certainly not true either.

    You found the potential for ##r\geq c## and for ##c \geq r \geq b##. You found the derivative of the potential for ##b \geq r \geq a##. If you know the potential at one point (b=r) and its derivative, how can you find the potential (between a and b)?
     
  16. Jan 31, 2015 #15
    Would Vb-Va= -∫-QK/r^2 work? Starting to feel even more lost.
    Also earlier when I said that the potential at r=b is 2QK/b, are you certain that is right? Because if I do Vc-Vb= -∫Edr, the right side will equal zero, so Vc=Vb which would mean Vb=2QK/c
     
  17. Jan 31, 2015 #16

    mfb

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    Integrating the electric field is a good idea, yes.

    Oops, that should be c in the denominator, as you wrote in post 5.
     
  18. Jan 31, 2015 #17
    OK for the integral the limits are 'a' to 'b' so -∫-QK/r^2 = QK/a - QK/b. But then wouldn't I still need the derivative of that to be equal to -QK/r^2? Since both of those are constants that wouldn't work.
     
  19. Jan 31, 2015 #18

    mfb

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    You found the potential at "a" by setting one border to "a" - but you can choose a different integration range.
     
  20. Jan 31, 2015 #19

    TSny

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    The question asks you to use superposition to get V in all the regions of your problem. The charge in your system is located on three separate spherical surfaces. Start by considering V due to a charge q uniformly spread over a single spherical surface of radius R. What are the expressions for V outside and inside the surface? Once you have that, it should be easy to use superposition to get the answer for V in the different regions of your problem.
     
  21. Jan 31, 2015 #20
    Well if I took it from r' to b, then I would end up with QK/r' - QK/b. The negative derivative of that would not give me the correct answer for the electric field. If I set it from a to r' it would work, but that doesn't seem right either does it?
     
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