1. Not finding help here? Sign up for a free 30min 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!

Jackson's problem 4.10, dielectric between 2 spheres

  1. Feb 20, 2013 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Two concentric conducting spheres of inner and outer radii a and b, respectively, carry charges +/-Q. The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ##\varepsilon / \varepsilon _0## ).
    a)Find the electric field between the spheres.
    b)Calculate the surface-charge density distribution on the inner sphere.
    c)Calculate the polarization-charge density induced on the surface of the dielectric at r=a.


    2. Relevant equations
    Laplace equation, ##\triangle \Phi =0## for the region between the 2 spheres.


    3. The attempt at a solution
    Working on a).
    To get ##\vec E## between the 2 spheres I'll go for the potential first and then use the relation ##\vec E =- \vec \nabla \Phi##.
    I use spherical coordinates ##(r, \theta, \phi)##. I choose my z-axis such that the half-filled (with the dielectric) hemisphere is for ##z \leq 0##. This way I get that the potential has no dependency on the coordinate phi. In other words I get azimuthal symmetry which simplifies greatly the expression of the solution to Laplace equation.
    Solving the Laplace equation in spherical coordinates leads to ##\Phi (r, \theta ) = \sum _{l=0}^ \infty [A_l r^l + B_l r^{-(l+1)}] P_l (\cos \theta )## (eq. 3.33 in Jackson's book).
    Now I'm looking for the boundary conditions. The fact that the 2 spheres are conductors means that the potential over both surfaces is constant. I'll call ##V_a## the potential ##\Phi (r=a)## and ##V_b## the potential ##\Phi (r=b)##.
    The boundary conditions are then ##\begin{array} \Phi (a, \theta ) = \sum _{l=0}^ \infty [A_l a^l + B_l a^{-(l+1)}] P_l (\cos \theta ) = V_a \\ \Phi (b, \theta ) = \sum _{l=0}^ \infty [A_l b^l + B_l b^{-(l+1)}] P_l (\cos \theta ) =V_b \end{array} ##.
    (I've no idea why the latex doesn't "work" here, it should read Phi(a, theta) at the start of the first line).
    I notice that ##V_a = V_a P_0 (\cos \theta )##. So that the infinite series collapses to only 1 term, the one with ##l=0##.
    This yields, if I made no mistake, ##B_0=\frac{V_a-V_b}{\left ( \frac{1}{a} - \frac{1}{b} \right ) }## and ##A_0 = V_a -\frac{(V_a-V_b)}{\left ( 1 - \frac{a}{b} \right ) }##.
    This is where I'm stuck. Previously, I was always given the potential on the surface of the conductor(s). But here I have ##V_a## and ##V_b## which are also unknowns. I must get them in terms of Q, a and b I guess. I just don't know how to proceed.
    I've been thinking about it and I don't even have the slighest idea on how to get those constant potentials. Any idea would be welcome.
     
  2. jcsd
  3. Feb 20, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    See if you can use your result for the potential to express the electric field in terms of Bo. Then you should be able to use E to find the surface charge density on the inner sphere in terms of Bo for z > 0 and z < 0. You can then determine Bo in terms of Q by requiring the total surface charge to equal Q. You won't need to determine the potentials Va and Vb.
     
  4. Feb 20, 2013 #3

    fluidistic

    User Avatar
    Gold Member

    Thank you very much!
    So my potential is ##\Phi (r)=K+\frac{B_0}{r}## where K is a constant that depends on ##B_0## and ##V_a##.
    Taking the gradient in spherical coordinates gave me ##\vec E = - \frac{B_0}{r^2} \hat r##. So that I see that E only depends on r (and not on theta), hmm okay so far.
    Then to get ##\sigma##, I used the relationship between the gradient of the potential across the surface of the conductor and the charge density over it. Namely ##\sigma = - \frac{\partial \Phi }{\partial r} \big | _{r=a}##. This gave me ##\sigma =\frac{\varepsilon B_0 }{\varepsilon _0 a^2}## for ##z<0## and ##\sigma = \frac{\varepsilon _0 B_0}{a^2}## for ##z>0##.
    Integrating the charge density over the inner sphere must give me Q. So that ##2\pi \varepsilon _0 B_0+2 \pi \frac{\varepsilon }{\varepsilon _0 } B_0=Q##.
    After a very little algebra I got ##B_0= \frac {\varepsilon _0 Q}{2\pi (\varepsilon + \varepsilon _0 ^2)}##.
    So now I have the expression for the potential: ##\Phi (r ) =K+\frac{Q \varepsilon _0}{2\pi (\varepsilon + \varepsilon _0 ^2)r}##. I don't really know how to deal with K. Can I simply "ignore" it and "set it" to 0 because it's a constant?!
     
  5. Feb 20, 2013 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Check the sign for E. Remember, E is the negative gradient of the potential.
    I think you're missing a dielectric constant in this equation.
    I get a different result for z < 0.
    Right. K has no physical significance. You could choose it to give any desired potential value at the surface of one of the spheres. You do not need to worry about it to answer the questions.
     
  6. Feb 20, 2013 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Couldn't Gauss make life a lot simpler here?
    BTW the"spheres" are really spherical shells, especially the outer one, right?
    Also, the problem doesn't state whether the dielectric is over the outer or inner section of the space between the shells.
     
  7. Feb 20, 2013 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, rude man.

    The setup is as shown.
     

    Attached Files:

  8. Feb 20, 2013 #7

    fluidistic

    User Avatar
    Gold Member

    Oh right, I made a sign mistake for the E field.
    Ok for the charge distribution part, but I still reach the same for z<0.
    I get ##\sigma = - \frac{\varepsilon }{\varepsilon _0 } \frac{\partial \Phi}{\partial r } \big | _{r=a}=\frac{\varepsilon B_0 }{\varepsilon _0 a^2}## for when ##z<0##. I really don't see my mistake here.
    Thanks for the K value.
    P.S.:Sorry rude man if I posted no picture. :(
     
  9. Feb 20, 2013 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Does the ##\varepsilon _0## belong here?
     
  10. Feb 20, 2013 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    "No harm, no foul"!
    Yeah, a bit tougher than I had envisaged. So much for Gaussian symetry ... should've known you guys were frying bigger fish than that ...
     
  11. Feb 20, 2013 #10

    fluidistic

    User Avatar
    Gold Member

    Ok I've been a bit more careful. Hopefully I've got the answer by now. The epsilon_0 does not belong there.
    I get
    a) ##\vec E (r)=-\frac{Q}{2\pi (\varepsilon + \varepsilon _0)r^2} \hat r##.
    b)##\sigma (r)= \begin{array}{ll} \frac{\varepsilon _0 Q}{2\pi a^2 (\varepsilon + \varepsilon _0)} \text{for z>0} \\\frac{\varepsilon Q}{2\pi a^2 (\varepsilon + \varepsilon _0)} \text{for z<0} \end{array}##.
    c)##\sigma _{\text{pol}}=\frac{(\varepsilon _0 - \varepsilon ) Q}{2 \pi a (\varepsilon + \varepsilon _0)}##. ( I used the relationship ##\sigma _{\text{pol}} = - \vec \nabla \cdot \vec P \big | _{r=a}## where ##\vec P = (\varepsilon _0 - \varepsilon ) \vec E##.)
    Edit... Wait it means that the polarization charge density does not depend on anything (not on the region! Unlike the charge density), it's just a constant. If that is true, it would be really hard for me to construct some intuition on this. :bugeye:
     
  12. Feb 20, 2013 #11

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Those results look good except for dropping the square on the ##a## in ##\sigma_{pol}##.

    I don't understand your last comment. Why do you claim that your result for ##\sigma_{pol}## at ##r = a## holds in both regions?
     
  13. Feb 20, 2013 #12

    fluidistic

    User Avatar
    Gold Member

    Whoops, yes indeed.
    What I mean is, the charge polarization is the same over the whole surface of the inner sphere. It does not depend whether vacuum or the dielectric is "touching" it.
    Does this make sense?
     
  14. Feb 20, 2013 #13

    TSny

    User Avatar
    Homework Helper
    Gold Member

    No, it doesn't make any sense. There can't be any polarization charge in the region z > 0 where there is no dielectric. That's why I was curious about why you claim there is polarization charge in that region. ##\vec{P}## is certainly zero in that region, so how can ##\vec{\nabla}\cdot\vec{P}## be nonzero there?
     
  15. Feb 20, 2013 #14

    fluidistic

    User Avatar
    Gold Member

    Aww I just realize my mistake.
    My algebra seems right but indeed, this result is only valid for the region where the dielectric is, namely z<0.
    I totally forgot to do the same algebra in the upper z region where indeed P is 0 there.
    Now all makes sense! Phew.

    I have 1 question though, it ocurred to me that almost all I did is wrong. Because the problem statement says that there's a charge Q and -Q over the spheres while when I integrated the charge density over the inner sphere and equated that to Q, I did not take into account the charge due to polarization. I only considered that the total free charge on the inner sphere was worth Q, not the free charge+polarization charge=Q.
    Did I therefore do something wrong?
     
  16. Feb 20, 2013 #15

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Good.
    I think Q is meant to be interpreted as just the free charge that was placed on the sphere. So, I think everything is ok.

    For fun, what do you get for the total charge density on the inner sphere for each region?
     
  17. Feb 20, 2013 #16

    fluidistic

    User Avatar
    Gold Member

    Ok, I was starting to get worried, because I'm reading that the polarization creates an electric field which goes against the E field I calculated (with magnitude of P/epsilon _0).

    Wow, I get ##\sigma _{\text{total}}=\frac{\varepsilon _0 Q}{(\varepsilon + \varepsilon _0 )2\pi a^2}##, this time, for all the space between the 2 spheres/shells. That's really strange, though I guess it should not appear strange to the accustomed physicist.
    If I multiply this by the surface area of a sphere I don't get a total charge of Q but ##Q_{\text{total}}=\frac{2Q\varepsilon _0 }{\varepsilon + \varepsilon _0 }##. In fact it would be worth Q if there would be no dielectric (epsilon = epsilon_0), if I'm not wrong.
    Thanks for all the help so far, you've been extremely helpful to me.
     
  18. Feb 21, 2013 #17

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Looks good. Note that since E at the surface of the inner sphere is uniform over the surface, then you can use Gauss' law to check that the total charge at the surface of the inner sphere agrees with what you got by adding together the free and polarization charge.
     
  19. Feb 21, 2013 #18

    fluidistic

    User Avatar
    Gold Member

    I had no time to work in this problem today.
    But something worries me a lot, it seems like I never used the fact that the outer sphere/shell had a total charge of -Q. Not even for the calculation of the E field.
    I could have put any charge on that shell, it would not have modified the expression of my E field. Intuitively I'm totally at a loss at understanding this.
    How is that possible?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Jackson's problem 4.10, dielectric between 2 spheres
  1. Griffiths problem 4.10 (Replies: 2)

  2. Jackson problem (Replies: 3)

Loading...