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Gauss's Law (I think!)

  1. May 12, 2006 #1
    At present its more the wording of the question I don't understand, so could you see if you think it means what I think it does.

    Charges +q and -q are placed on two conducting spherical shells of radii a and b (a<b). Find the electric field between the shells, and show that it has a maximum value given by,

    E(max) = Vb/a(b-a)

    Where V is the potential difference between the spheres.

    I think its set up like this:
    [​IMG]

    Does this mean I can find the field like I would with oppositely charged parallel plates?
     
  2. jcsd
  3. May 12, 2006 #2

    Tide

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    Not exactly - the geometry is quite different. What you need to do is set up a Gaussian sphere concentric to the charged spheres with radius intermediate to them. You know how much charge is enclosed by that surface and electric field follows from Gauss' Law and symmetry. You can use it to calculate the potential.
     
  4. May 12, 2006 #3
    So I set up a surface inbetween them?

    So the charge enclosed will be just +q?

    The radius of this new surface I don't know do I? But I don't need to know it do I?
     
  5. May 12, 2006 #4
    The radius of the sphere can be taken as 'r' to apply Gauss' law.Now you can find the max value for E in terms of a,b and V (E will be maximum on the surface of the smaller sphere).

    Arun
     
  6. May 12, 2006 #5

    nrqed

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    that's right
    that's correct
    you need to specify it. Give it a name, say "r", with the understanding that [itex] r_a < r < r_b [/itex]. That's the radius of your gaussian surface so when you calculate the electric flux through that surface, it will contain "r". When you will solve for the magnitude of the E field, it *will* contain r. Now, you r is arbitrary, within the restriction [itex] r_a < r < r_b [/itex]. Si you will have found the E field at *any* point between the two shells.
     
  7. May 12, 2006 #6
    I've got the answer out to be:

    [tex] E = \frac{q}{\epsilon_{0}4\pi r}[/tex]

    I know the final bit should be simple, putting the Gaussian surface on the smaller sphere (so that r=a), so why doesn't that just give me E=V/a?
     
  8. May 12, 2006 #7

    nrqed

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    This is not quite right. When you calculate the flux through your gaussian surface, you ended up multiplying the magnitude of the E field by the area of your gaussian surface, right? But that is [itex] 4 \pi r^2[/itex]! So you should have a r^2 there, not r.
     
  9. May 12, 2006 #8
    I did get [tex]4 \pi r^2[\tex], I just copied it down wrong.

    The r came from working out E=V/r (because V=kq/r right?)
     
    Last edited: May 13, 2006
  10. May 12, 2006 #9

    nrqed

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    I think I misunderstood your post the first time around so I deleted my initial response.

    If you do set r=a in your expression, then of course you get [itex] {q \over 4 \pi \esilon_0 a^2} [/itex] and this *is* potential at r=a divided by a so it does work. Do you agree?
     
  11. May 13, 2006 #10
    It was posted above that the maximum value for the electric field will be when the electric field is on the smaller sphere (i.e. when r=a), but the question says to show that the maximum value is :
    [tex]E_{max} = \frac{Vb}{a(b-a)}[/tex]

    which I don't see how it comes about. Surely if r=a when its maximum you end up with E=V/a (where V is the potential difference)?
     
  12. May 13, 2006 #11
    Brewer, potential difference is different from potential.
    Try to find the pd between the two spheres.
    You can start by treating the shells as point charges as you might have learnt from Gauss' law, find the potentials at the outer and inner shells and subtract to get the pd=V in the question.
    Believe me the electric field is greatest at the inner sphere.
     
  13. May 13, 2006 #12

    nrqed

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    Arunbg said it correctly.

    Here V is the potential *difference* between the shells. (a better symbol would b [itex] \Delta V [/itex]). Using [itex] k = { 1 \over 4 \pi \epsilon_0} [/itex], you have that the potential on the inner shell is [itex] {k q \over a} [/itex] and on the outer shell is is [itex] {k q \over b} [/itex]. And V is the difference between the two. The maximum E field is [itex] {k q \over a^2} [/itex]. If you reexpress the max E field in terms of V, you will get the expression they give.
     
  14. May 13, 2006 #13
    Ok guys I'm with you now (I think!). So as I read it I find the potential at b and the potential at a, and subtract them to find the pd which is v?
     
  15. May 13, 2006 #14
    Thanks for your help by the way
     
  16. May 13, 2006 #15

    nrqed

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    Yes. and then express the maximum electric field in terms of this difference of potential. You will get the expression you quoted in your first post.
    (the reason they express it in terms of a *difference* of potential rather than, say, the potential on one of the spheres is that the potential difference is easy to measure. Just connect the two terminals of a voltmenetr to the two spheres and the value you will read will be the potential difference). But, as I said, it is a bit misleading (although very common) to use the symbol "V" for a difference of potential.

    And you are welcome.
     
  17. May 14, 2006 #16
    nrqed, are you sure those are the exp. for potentials at the two shells?

    To bring a unit charge to the outer shell, the inner shell does equal and opposite work as compared to the outer shell. So the potential will be 0.
    To bring a unit charge to the inner shell however, once the charge is past the outer shell, only the inner shell can do work (as you might have learnt from Gauss' law).
    So the total work done to bring the charge to the inner shell (potential)becomes
    [tex] {k q \over a} + {k (-q) \over b} [/tex]

    Therefore potential difference [itex] \Delta V = P_{a} - P_{b}[/itex]
    [tex] {k q \over a} + {k (-q) \over b} - 0 = {k q \over a} - {k q \over b}[/tex].
    Now you can answer your problem Brewer.

    And you are welcome.
     
  18. May 14, 2006 #17
    Yep thats what I got in the end (after trying to do b-a for a while!), but I have the correct answer now.
     
  19. May 14, 2006 #18

    nrqed

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    The potential at any point is totally arbitrary. I was using a certain choice of ground and you are using a different one (I made a choice which gives a potential equivalent to the potential produced by a point charge, I should probably have specified my choice of ground). So we are both correct, within our choice of ground. There is no way to "prove" that the potential is zero at a point, it is completely arbitrary. If you calculate the work done in moving a charge from one point to another, onlt the difference of potential between the initial and final points appear, so it is impossible to use the work done in moving a charge to fix the potential at a point.
    we get the same result for the *difference* of potential, which is *not* arbitrary (it is independent on the choice of ground).

    Regards

    Patrick
     
  20. May 15, 2006 #19
    Well Patrick, I guess it was a case of misunderstanding.
    When you gave the exp. for potential of the spheres without specifying the ground I got a bit doubtful and certainly the OP must have felt the same.
    So for Brewer, Patrick chose his 0 potential as -kq/b when potential is measured from infinity. So the potential difference is not affected on the whole.

    Arun
     
  21. May 15, 2006 #20

    nrqed

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    It is understood that whenever the potential is given at a point, a choice of ground has been made. If I caused confusion to the OP then I apologize.

    The reason for my choice of ground was to have the same expression for the potential between the two spheres as for the potential of a point charge q located at the origin (namely, kq/r). Since the electric field in that region is the same as for the electric field of a point charge located at the origin, it is intuitively clear that the electric potential *may* be chosen to have the same expression as for the potential of a point charge located at the origin. To me this seemed the most natural expression.

    Just a quick comment to make things very clear to the OP . You wrote in a previous post:
    (Here you were talking about moving a charge between two points oustide of both spheres, I assume).
    The correct statement would have been to say that the net work is zero, therefore the potential difference between two points on the outside of both spheres is zero, therefore the potential must be constant outside of the two spheres. The value is has there is totally arbitrary.

    I am sure you know that, I just want to make things clear to the OP.

    To be exact, itis kq/b. So, with my choice of ground, the potential is kq/r at all points between the two spheres, dropping from kq/a to kq/b as one moves from the inner sphere to the outer sphere and then it remains constant at kq/b outside of the two spheres.

    With you choice of ground, the potential is zero outside of both spheres, and it has a slightly different expression between the two spheres. I chose my ground so that it would be simply kq/r between the two spheres.


    Regards

    Patrick
     
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