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Find the electric field of a uniformly polarized sphere

  1. Jul 5, 2016 #1
    1. The problem statement, all variables and given/known data
    We want to calculate the field of a uniformly polarized sphere of radius=R

    2. Relevant equations
    [tex]
    V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \oint_{S} \frac{\sigma_{b}}{r} da' + \int_{V} \frac{\rho_{b}}{r} d\tau'
    [/tex]
    3. The attempt at a solution
    i)I know that
    [tex]\sigma_{b} = P \cos\theta[/tex]. I know that [tex]\rho_{b}=0[/tex]
    Now using the cosine law I find that
    [tex]
    \tilde{r}
    = \sqrt{R^2 + r^2 - 2Rr\cos\theta}[/tex]
    and using spherical coordinates I find that
    [tex]da' = R^2 \sin\theta d\theta d\phi[/tex]
    So here comes the problem :When I try to use the voltage equation I can't find the right answer.I've searched everywhere on this matter and I can't understand why this [tex]
    V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}} \cos\theta' R^2 \sin\theta' d\theta' d\phi[/tex] is correct and not this [tex]
    V(\vec{r}) = \frac{1}{4 \pi\epsilon_{0}} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{P \cos\theta}{\sqrt{R^2 + r^2 - 2Rr\cos\theta'}}R^2 \sin\theta' d\theta' d\phi[/tex] and where does this extra [tex]\cos\theta'[/tex] come from and why?
    ii) Secondly I don't understand why when I integrate from 0 to pi we don't have to multiply the whole integral with the number 2.Do we perceive it only as the half sphere and if yes why?
    Thanks In advance and please help..this is baffling me for waay too long now o_O
     
    Last edited: Jul 5, 2016
  2. jcsd
  3. Jul 5, 2016 #2

    Charles Link

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    The solution to this problem can be found in most E&M textbooks in the form of a dielectric sphere in a uniform electric field. The polarization for this case turns out to be uniform. The solution involves Legendre polynomials to find the solution for the potential, both inside and outside the dielectric sphere. The case of uniform polarization in the absence of an applied field could be determined with a minor modification or two of these solutions ... Solving the problem with a Coulombs law evaluation of the integrals to get the electric field ## E ## and/or computing a Coulomb-type potential ## V ## I believe is quite difficult. The one exception to this is determining the electric field at the center of the sphere. It turns out for this problem the electric field inside the sphere is actually uniform and has the value that is found at the center of the sphere. Outside of the sphere, the potential function ## V ## isn't real complicated (from which the electric field can be computed), but I'd need to consult an E&M textbook to look up the solution involving the field outside the sphere. Again, it involves Legendre polynomials as solutions of what I believe is called the Laplace equation.
     
    Last edited: Jul 5, 2016
  4. Jul 5, 2016 #3
    Thank you for your time Sir.Although I would like to know if this can be solved with my method of integration and specifically I would really appreciate it If my first question was answered.My kind regards
     
  5. Jul 5, 2016 #4

    Charles Link

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    To answer you first question, I think the cosine term in the denominator should be ## cos(\theta '-\theta) ## , (comes from law of cosines=the angle in the law of cosines is ## (\theta'-\theta ##)),and there should be a single ## cos(\theta ' ) ## term in the numerator along with the ## sin(\theta ') ## term. The ## \theta ' ## angle is a polar angle. In spherical coordinates, it only needs to go from ## 0 ## to ##\pi ## cover the whole sphere. I don't see any simple transformation that gets the first answer above.(but I need to look this over more carefully because a ## cos(\theta) ## term that can come out of the integral may in fact be correct.) Meanwhile you need a "prime" on the cosine in the numerator and as I mentioned, the denominator term is a difference of angles. (Unprimed ## \theta ## is the location of the observation point, vector ## \vec{r} ## of ## V(\vec{r}) =V(r,\theta, \phi) ##.
     
  6. Jul 5, 2016 #5

    Charles Link

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    A follow-on: My law of cosines in post #4 with ## cos(\theta'-\theta) ## is incorrect. The diagram needs to be drawn in 3 dimensions and this cosine factor (in the law of cosines) is much more complex. I'm still working it to see if somehow a transformation might simplify things...editing... the solution from the Legendre polynomials does in fact contain a ## cos(\theta) ## term for the potential both inside and outside the polarized sphere in the final result, but I question whether your first expression above is correct. It might be, but getting it to that point would be difficult and further evaluation would also be difficult. This one was taught to us in advanced E&M courses using the Legendre polynomial solution. And I did work through the Legendre solutions for this case just now. It's probably the simplest complete solution to this problem... editing.. and just a comment on the Legendre solutions: You have a whole family of solutions to work with but you can make an educated guess and pick one of the simpler forms. There usually are one or two constants that you need to solve for and the solution involves showing that you satisfy any necessary boundary solutions and/or other conditions with the solutions for the potential ## V ##. If you can show the solution you picked meets all of the requirements, then in fact it is the correct solution... In this case (for the problem at hand) the electric field that gets computed is in spherical coordinates, but you can show that the electric field is uniform throughout the inside of the sphere and points in the "minus z" direction (and you also get a value for its amplitude).
     
    Last edited: Jul 5, 2016
  7. Jul 6, 2016 #6
    About the [tex]
    \cos\theta'[/tex] I have found from another post the following explanation:It states that it is used because we care about the Z components only ,but I don't quite get it.Thanks a lot ! :)
     
  8. Jul 6, 2016 #7

    Charles Link

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    The integral result (presuming they did the necessary algebra properly) is correct, but may not be able to be evaluated by any simple means. In any case, since the Legendre solutions are correct, they must be the solution to the integral. Let me provide you with them. They are ## V_{in}(\vec{r})=(P_o/(3 \epsilon_o))r \cos(\theta) ## and ## V_{out}(\vec{r})=(P_o/(3 \epsilon_o)) R^3 cos(\theta)/r^2 ##. (## V_{in} ## means inside the sphere, i.e. ## r<R ## and ## V_{out} ## is outside the sphere, i.e. for ## r>R ##). ## E=-\nabla V ## and you need the spherical coordinate form of the gradient. The result for ## E_{in} ## is ## E_{in}=-(P_o/(3 \epsilon_o))(cos(\theta)\hat{a_r}-sin(\theta)\hat{a_{\theta}})=-(P_o/(3 \epsilon_o))\hat{k} ## independent of ## \vec{r} ##. ## E_{out}(\vec{r}) ## meanwhile falls off as ## 1/r^3 ##(edited=I previously had ## 1/r^2 ##). In picking the Legendre solution, it helps to have a "clue" what the solution might be by evaluating the E field at the center of the sphere. (Suggest you do the surface integral using ## \sigma_p=P_o \cos(\theta) ## and computing ## E ## at the center.) That integral is a simple one and is ## E=-(P_o/(3 \epsilon_o)) \hat{k} ##. There is no guarantee that ## E ## is uniform throughout the inside of the sphere until you demonstrate that the Legendre potential is of the form ## V_{in}=(P_o/(3 \epsilon_o))r \cos(\theta) ##. (Notice for a uniform ## E ## in the minus z direction, the ## r \cos(\theta) ## is the form the potential will have.) With the Legendre method, you make an educated guess and show your answer works. (additional editing: for this case, amongst other things, you need to show Gauss's law holds at the surface of the sphere, i.e. that ## \int E \cdot dA =\sigma_p A /\epsilon_o ## etc.)
     
    Last edited: Jul 6, 2016
  9. Jul 6, 2016 #8

    Charles Link

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    I edited post # 7. Be sure and see the latest version.
     
  10. Jul 6, 2016 #9

    Charles Link

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    It is worthwhile to note that the solution we found for this problem can be used to solve the case of a dielectric sphere with susceptibility ## \chi ## in a uniform applied electric field ## \vec{E_a}= E_o \hat{k} ##. We can write polarization ## P=\epsilon_o \chi E_i ## and ## E_i=E_o-P/(3 \epsilon_o) ##. (This last equation follows from our solution above.) The result is ## E_i=E_o/(1+\chi/3) ##(in the ## \hat{k} ## direction.) Solving for ## P=\epsilon_o \chi E_o/(1+\chi/3) \hat{k} ##, the electric field outside the sphere can also be computed by adding the electric field from the polarized sphere ( with ##P=\epsilon_o \chi E_o/(1+\chi/3) ##) to ## \vec{E_a}=E_o \hat{k} ##.
     
    Last edited: Jul 6, 2016
  11. Jul 6, 2016 #10

    Charles Link

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    One additional comment or two. The problem of post #1 of the electric field for uniform (spontaneous/fixed) polarization is a good way to learn some of the concepts before advancing to the slightly more difficult problem with a polarization that is induced by an external electric field. The latter problem is more difficult because the applied electric field will induce a polarization so that additional sources of electric field (the polarization surface charge density ## \sigma_p ##) arise that serve to reduce the original applied electric field. The self-consistent algebra (of post #9) is somewhat straightforward, but much easier to follow after first having worked the slightly simpler problem of post #1. (For the problem of post #1 the polarization P was assumed to be unaffected by the electric field that arose from the polarization.)
     
    Last edited: Jul 6, 2016
  12. Jul 6, 2016 #11

    Charles Link

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    I have not been able to verify whether the first formula you give is in fact a correct expression for the potential V. (The expression I have does not yet have a ## cos(\theta) ## (unprimed in the numerator), and the denominator has a complicated couple of terms from the dot product of the R and r vectors. It involves 2 or 3 terms and is not simply a ## cos(\theta')##). The potential V is a scalar function, and it would be incorrect to arbitrarily multiply it by ## cos(\theta ') ## to try to filter out a "z-component".
     
  13. Jul 7, 2016 #12

    Charles Link

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    A couple of follow-on comments regarding posts #9 and #10 along with the original problem. The answer to the problem of the electric field ## \vec{E_p}_{in}=-(1/3) \vec{P}/\epsilon_o ## for a uniformly polarized sphere is a rather specialized result and also quite useful. The "-1/3" is a geometric factor that arises for a spherical shape. There are only a couple of geometric shapes that give a uniform reverse ## E_p ## for a uniform ## P ##. One other shape where this occurs is a flat disc or slab, and in that case the ## E_p=-P/\epsilon_o ##. This can easily be demonstrated with the equation ## -\nabla \cdot P=\rho_p ## a;ong with Gauss'law. The uniform E in the material from uniform P is important because that says if we put the object in a uniform field ## E_o ##, the resulting ## E_i ## and ## P ## will also be uniform. Post #9 shows how to compute the ## E_i ## and ## P ## for a dielectric sphere. It is a similar calculation that can be used to compute the results for a flat slab or disc. On other shape that is found to generate this uniform E from a uniform P is a cylinder where the polarization is in the x-direction (with the cylinder axis along the z-direction.) In this case ## E_p=-(1/2)\vec{P}/\epsilon_o ##. It is a little extra info from the original question of the OP, but very useful in solving various E&M electrostatic problems. One other item is I made an effort to try to solve the complicated integral of post # 1 in closed form. I have to believe a closed form solution exists, by first performing the ## \phi' ## integral and then subsequently doing the ## \theta' ## integral, but I have been unsuccessful at solving it or finding the answer in a table of integrals.
     
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