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Homework Help: Gauss's Law on a Charged Sphere

  1. Aug 7, 2011 #1
    Hi I'm sorry if this is posted in the wrong section or it's laid out wrong but I have a question that I need a bit of help with.

    1. The problem statement, all variables and given/known data
    I'm given: A spherically symmetrical charge distribution is contained within a sphere of radius a with no charge outside. At a distance r (r [itex]\leq[/itex] a) from the centre of the sphere the potential may be written as:

    V = -Ar[itex]^{3}[/itex] + B

    Then I'm asked to calculate a bunch of things which hopefully I've got right.

    (a) Calculate the electric field intensity for r < a.

    (b) What is the electric field intensity for r = a.

    (c) Using Gauss's law calculate the total charge in the sphere.

    (d) What is the electric field for r > a?

    (e) What is the potential for r > a if V[itex]_{\infty}[/itex] = 0?

    (f) Show that the difference in potential between r = a/2 and r = 2a is given by (19/8)Aa[itex]^{3}[/itex].

    It's the last one I'm stuck on.

    2. Relevant equations
    Gauss's Law.

    E = -[itex]\nabla[/itex] V

    3. The attempt at a solution

    Ok. So for (a) I differentiated and got E = 3Ar[itex]^{2}[/itex]

    (b) I just subbed r=a and got E = 3Aa[itex]^{2}[/itex]

    (c) I got that E for r=a with Gauss's Law: E = [itex]\frac{Q}{4\pi\epsilon_{0}a^{2}}[/itex]. Then equated that with answer from (b) to get Q = 12[itex]\pi\epsilon_{0}[/itex]Aa[itex]^{4}[/itex]

    (d) I used Gauss's Law again to obtain: E = [itex]\frac{Q}{4\pi\epsilon_{0}r^{2}}[/itex].

    Then I subbed in Q from part (c) to get: E = 3A[itex]\frac{a^{4}}{r^{2}}[/itex].

    (e) I integrated -E with boundaries r and 0 and got: V = 3A[itex]\frac{a^{4}}{r}[/itex].

    (f) Now I subbed r=a/2 into the equation for V given in the question. And r=2a into the equation I just got in part (e). But cannot seem to get (19/8)Aa[itex]^{3}[/itex] at all.

    Thanks for any help.
  2. jcsd
  3. Aug 7, 2011 #2


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    Homework Helper

    For e, you need to integrate between r and infinity as the potential is zero at infinity.
    For f, you need to match B so as V=-Ar3+B be the same at r=a as the outer potential.

    Last edited: Aug 7, 2011
  4. Aug 7, 2011 #3
    Got it all sorted now! Thanks!
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