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Electric field of a charged nucleus (sphere) Gauss Law

  1. Sep 11, 2009 #1
    ok guys.. this question is regarding a spherical nucleus with radius R related to Gauss's Law... so where the electric field given as the following:

    E(r) = Q/4*pi*epsilon-nought(r^2+R^2)

    1. the questions asks "why this model/behaviour for r>>R might be "reasonable"
    2. what q(r) is needed for Gauss's law to produce the above E(r)
    3. what rho(r) in turn is needed to produce this q(r)?
     
  2. jcsd
  3. Sep 11, 2009 #2

    Hootenanny

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    Welcome to Physics Forums.

    According to our guidelines (see https://www.physicsforums.com/showthread.php?t=5374"), you are expected to show an attempted solution when asking for help, or at the very least detail your thoughts. This is why we have a homework template, please use it.
     
    Last edited by a moderator: Apr 24, 2017
  4. Sep 11, 2009 #3
    hmm ok.. so where the question is concerned...

    1. could it be that because the model assumes that where Gauss's law is used, 2 Gaussian surfaces are separately used and calculated with total charge enclosed over the 2 surfaces still being Q, and hence the (r^2+R^2) instead of just r^2? if that makes any sense. i know that outside the original sphere, for distances r >> R, the relationship between E & r is such that E inversely proportional to r^2. could it be that due to the possibility of that the charge is not uniformly distributed also?

    2. As for no 2, using gauss's law, integral of E.dA = Qin/epsilon-nought. not too sure how to get about after though. is it just rearranging every term in terms of Qin?

    3. i know its related to no 2 in a way that Qin = rho x Vsphere and via differentiation and integration.

    Some thoughts and help here much appreciated.
     
  5. Sep 11, 2009 #4

    Hootenanny

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    You're on the right lines. In fact, you hit the nail on the head here:
    In the limit r >> R, the expression reduces to the expression for a point charge. In other words, the electric field at a large distance from the nucleus is virtually identical to that of a point charge, as it should be. Do you follow?
    Indeed, that is it. You will also need to evaluate the integral of course.
    The differential form of Gauss' law would be helpful here.
     
  6. Sep 11, 2009 #5
    how do you use the differential form of gauss's law.
    So integral of E.da=qin/empslom
    if we differentiate both sides we get E= dq/da*(1/epslom).... so we have determined dq/da, which is equal to E*epslom, but how does that allow you to determine the charge density... because wouldnt charge density equal dq/dv.
     
  7. Sep 13, 2009 #6

    Hootenanny

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    http://en.wikipedia.org/wiki/Gauss's_law#Differential_form
     
  8. Sep 14, 2009 #7
    Oh yh. So the charge density is simply equal to epslom multiplied by the electric field?? So is this the only way to approach this qs.
     
  9. Sep 14, 2009 #8

    Hootenanny

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    No, that it not Gauss' law, is it? The charge density is equation to the permittivity of free space (epsilon zero) multiplied by the divergence of the electric field.

    P.S. Are you Boon28?
     
  10. Sep 15, 2009 #9
    I don't really get how to do the 'divergence' thing.
    Can u explain it in a simple way?

    P.S I'm not warrior_1 :/
     
  11. Sep 17, 2009 #10
    hmm so for no. 2:
    integral of E.dA = Q/epsilon-zero

    solving the integral and rearranging, I got q(r) = Qr^2/(r^2+R^2)..

    for no. 3: charge density i was told i can get by finding the derivative of q(r) with respect to r.

    and so i got rho(r) = dq/dr = 2QrR^2/(r^2+R^2)^2

    Do those seem reasonable?
     
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