1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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.

    ehild
     
    Last edited: Aug 7, 2011
  4. Aug 7, 2011 #3
    Got it all sorted now! Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook