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Electric potential on concentric spheres

  1. Jan 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A metallic sphere at ##\phi = 0 ## is surrounded by a spherical layer with charge density ##\rho ##. Find the potential at every point.
    https://photos-6.dropbox.com/t/2/AACuekw5jWy2n0tNfMqPRKoakoEdSj24JOvheq4J6XNWRQ/12/28182931/png/1024x768/3/1421928000/0/2/spheres.png/CJOTuA0gAyACIAEoAigB/g4S2DgYrMDWxMbOXqlP8kHX8klBkIjXPbCRua9GoiiQ [Broken]
    2. Relevant equations


    3. The attempt at a solution
    The central zone is just 0.

    The potential in the yellow zone could be calculated assuming there is no charge in the grey zone, since that distribution would give the same boundary condition in the interior surface of the yellow zone, that is, ##\phi = 0 ##. However I don't know the potential in the outer surface, so I'm not sure if I can do that.

    Is there anyway to find the potential at the outer surface?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 22, 2015 #2

    BvU

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    To answer your last question: Yes there is. Find a relevant equation and work it out ;) . (sorry, this is PF culture -- and rule)

    But to make a step back: ##\phi=0## does not mean charge = 0 !

    If there were no grey sphere, would you be able to work out ##\phi## ? How ?
     
  4. Jan 22, 2015 #3
    I would use Gauss' Law to find the E normal to the sphere and then derive to get the potential. But if I have the grey sphere, whose charge I don't know, I don't know the internal total charge, and therefore Gauss is useless. Any method I can't think of requires that I know the charge on the grey zone...

    The only thing I can think of is as follows: the charge on the yellow grey zone is all the charge there is outside the yellow sphere, so with Gauss'Law I can get the potential that would be in the interior sphere if it way empty. Then the charge distribution that creates the opposite potential is the charge in the grey zone. Does that make any sense?
     
    Last edited: Jan 22, 2015
  5. Jan 22, 2015 #4

    BvU

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    you mean integrate.

    Yes. Very much so. Small correction: "all the charge there is outside the yellow sphere," -- you mean grey sphere.

    Remember the characteristics of a conductor: ##\phi## = constant (so that E = 0. if it weren't then the mobile charges in the conductor would move until it is). On the other hand, they can't move off the conductor, so at the boundary you can have a surface charge.
    This means that a simple charged spherical conductor has all the charge evenly distributed on the surface.

    Go to work and see what comes out ! Good luck!
     
  6. Jan 23, 2015 #5
    Ok, I've found mathematically that a charge in the center of the grey sphere equal to the total charge on the yellow zone would produce a null potential in the interior zone, which in retrospect seems obvious. But that means, according to Gauss' Law, that ## \phi(r>b) = 0##, since the total charge in ##r<b## is zero.

    So, if both boundary surfaces of the yellow zone have ##\phi = 0## and the potential function has no local maxima or minima, then the ##\phi_{yellow} = 0##, is that possible?
     
  7. Jan 23, 2015 #6

    BvU

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    Whoa, going the wrong way. may be my fault. ##\phi \;( r<a )= 0## but the yellow sphere attracts opposite charge that sits on the surface of the conducting sphere. And the charge in the yellow zone does generate a field ! a= 0 is here
     
  8. Jan 23, 2015 #7
    But if there is charge in the surface on the conducting sphere that's the opposite of the charge in the yellow sphere then the total charge inside the whole sphere (##r < b##) is zero, which means ##\phi(r>b) = 0##. Am I correct so far?
     
  9. Jan 23, 2015 #8

    BvU

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    Yes. I am clearly painting myself into a corner now. I want ##\phi = 0## at ##r\rightarrow \infty## and also for ##0<r<a##. The former I get from ##\rho \ne 0## in ##a<r<b##. So in case there is no conducting sphere
    $$ E = 0 \qquad r<a
    $$ $$ E = {\rho \over\epsilon_0} \; {{4\over 3}\pi (r^3-a^3) \over 4\pi r^2} \qquad a<r<b
    $$ $$ E = {\rho\over \epsilon_0} \; {{4\over 3}\pi (b^3-a^3) \over 4\pi r^2} = {Q\over 4\pi \epsilon_0 r^2}\qquad b<r$$
    [edit]: Must have been seeping already. I took b to stand for the outer radius of the whole thing instead of the thickness.
    So where it says b read a+b
    (Our Original poster seems inclined to make the same mistake..)


    But this gives me ##\phi \ne 0## for ##0<r<a##

    Now what with the conducting sphere to get ##\phi = 0## for ##0<r<a## ?

    Too late on the clock to clear this up; bedtime -- sorry, perhaps tomorrow. Perhaps someone else ?
     
    Last edited: Jan 24, 2015
  10. Jan 24, 2015 #9

    ehild

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    Assume some charge q on the metal sphere and find the electric field in terms of it both for a<r<(a+b) and r > a+b. Determine the potential in the layer and outside it and match them at the outer surface. You can find q from the continuity of the potential.
     
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