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Electric field of a hollow sphere WITH surface thickness.

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A hollow spherical shell carries charge density [tex]\rho[/tex] = [tex]\frac{k}{r^{2}}[/tex]
    in the region a [tex]\leq[/tex] r [tex]\leq[/tex] b. Find the electric field in the three regions:

    i. r < a
    ii. a < r < b
    iii. r > b

    2. Relevant equations

    Surface Area of a sphere
    Coulombs Law
    Gauss's Law

    3. The attempt at a solution

    i. This is easy, centre of a hollow charged sphere has E = 0

    ii. ?

    iii. Just used the sphere as a point charge and using the surface area of the sphere's outer radius, multiplied by the charge density to get the charge, and then just used Coulomb's Law to get E.

    My problem is with part ii. The Electric Field within the thickness of the sphere. The question never specified that it was an insulator or a conductor, so I'm just going to assume that the majority of the charge lies on the outer radius surface, and that the inner radius surface should have little-to-no charge to account for the 0 E in the hollow centre, and the Field increases as the radius increases in the thickness: E [tex]\propto[/tex] r

    Would i be right in this assumption? How might I describe this mathematically?
     
  2. jcsd
  3. Aug 24, 2009 #2

    kuruman

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    Your assumption is incorrect. The charge distribution is given as a function of r, so there is charge everywhere inside the shell. Consider using Gauss' Law. Your Gaussian surface should be a concentric sphere with radius r, a < r < b. Find the charge enclosed by this surface, find the flux through the surface and put it together.
     
  4. Aug 24, 2009 #3
    Cheers for that kuruman, I thought my assumption might have been wrong, but I had a few different conflicting ideas on this.

    So for example, if i was to use the differential form of Gauss' law: [tex]\nabla[/tex].E = [tex]\frac{\rho}{\epsilon}[/tex] , how might I infer the Electric Field itself from this divergence? 2nd Year Calculus coming back to haunt me..

    I'll have a go now, using some "reverse-calculus" to get E by itself, but any more advice on this would be really appreciated.
     
  5. Aug 24, 2009 #4
    Actually.. I think i have it now:

    Using the Flux equations: [tex]\Phi[/tex] = E.A = E.4.pi.r^2 = Q/eps0
    and then using Q = rho x Area of sphere, sub into flux and solve for E.

    correct? incorrect?

    EDIT: Actually.. this is how i worked out the Electric field for r > b.. damn.. back where I started
     
  6. Aug 24, 2009 #5

    ideasrule

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    Doesn't charge density mean the charge per unit volume, and not the charge per unit area? Use integration to find Q, then use Gauss' law as you've already done.
     
  7. Aug 24, 2009 #6
    You're absolutely right! I'm getting myself confused here.. Instead of integrating the for the volume, could i perhaps use the shortcut method of V(sphere) = 4/3.pi.r^3 and just deduct the volume of the inner sphere from the larger sphere? would certainly simplify my calculations.
     
  8. Aug 24, 2009 #7

    kuruman

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    You can do that only if the density is uniform (equal charge in equal volume). Here the density depends on the radius. You have to integrate.
     
  9. Aug 24, 2009 #8
    Kuruman is right..the distribution is not uniform, it depends from r...look at the thread i posted few days before by the title ''Gauss Law'' by Engels and u will find the solution. I had almost the same problem as you and i solved it
     
  10. Aug 24, 2009 #9
    Ok, not sure if I'm doing it correctly here, but here's what I've done:

    using Gauss' law:
    [tex]\int_V\frac{\rho}{\epsilon_0} d\tau = \frac{Q}{\epsilon_0}[/tex],

    so i then do:

    [tex]\frac{Q}{\epsilon_0} = \int^{\pi}_{0}\int^{2\pi}_{0}\int^{r}_{a} \frac{k}{\epsilon_0 r^{2}} dr d\phi d\theta[/tex]
    as it's a sphere with radial limits a through r.. correct?

    Then I use coulomb's law to calculate E once I know Q.. correct?

    PS. my equation typing skills are improving =]
     
  11. Aug 25, 2009 #10

    gabbagabbahey

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    That's not Gauss' Law. The equation [itex]Q=\int_{\mathcal{V}} \rho d\tau[/itex] comes straight from the definition of charge density.

    Not quite, the volume element in spherical coordinates is [itex]d\tau=r^2\sin\theta dr d\theta d\phi[/itex]

    No, you use Gauss' Law.
     
  12. Aug 25, 2009 #11
    argh!~ This is annoying the heck out of me!

    Right, so re-evaluated:

    [tex]\frac{Q}{\epsilon_0} = \int_V \frac{k}{\epsilon_0} sin(\theta) dr d\theta d\phi[/tex]

    This is calculated down to a nice easy equation [tex]Q = 4k \pi (r-a)[/tex] giving charge enclosed between the shells radius r and a.. correct? This seems a little simple, but ahh well let's see where this goes.

    So now I have the charge [tex]Q[/tex], now what do I do? I can't for the life of me figure this out. It should be so simple, there's some missing piece of the puzzle here..

    I know i have to use the surface integral part of Gauss' law, but I can't find any worked examples on how to get E outta that damned integral.

    anyone care for a brief tutorial?
     
  13. Aug 25, 2009 #12

    gabbagabbahey

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    Yes, it is simple. Luckily, it's also correct:smile:

    Well, if you can find a Gaussian surface [itex]\mathcal{S}[/itex] where E will be constant over the surface and point normal to the surface, then you can say

    [tex]\int_{\mathcal{S}} \textbf{E}\cdot d\textbf{a}=||\textbf{E}||\int_{\mathcal{S}} ||d\textbf{a}||=EA[/tex]

    Right?
     
  14. Aug 25, 2009 #13
    wait wait wait wait wait! I think I have it!

    Was flipping through Griffiths and spotted a nice little trick for symmetry where E points radially outwards with the same magnitude at all points with the same radius:

    [tex]\int_S E.da = \int_S |E|da = E4 \pi r^{2} = \frac{Q}{\epsilon_0}[/tex]

    Sub my Q into that and solve for E.. does this sound right at all?

    EDIT: argh! gabbagabbahey you beat me to it!
     
  15. Aug 25, 2009 #14

    gabbagabbahey

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    Right, you've already calculated the charge enclosed by a spherical surface of radius a<r<b. So as long as you can argue that E will point radially outward with the same magnitude at all points with the same radius, then yes, that's all you need to do.
     
  16. Aug 25, 2009 #15
    WONDERFUL! Thank you so much for the help!

    ..and thanks to all who've assisted me with this irritating problem.

    I've now concluded that [tex]E = \frac{k(r-a)}{\epsilon_0 r^{2}}[/tex] for the region enclosing a < r < b.
     
  17. Aug 25, 2009 #16

    gabbagabbahey

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    Looks good to me!:approve:
     
  18. Aug 25, 2009 #17
    I are champion =]
     
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