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Electric field of a charged dielectric sphere

  1. Oct 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A dielectric sphere of radius R with uniform dielectric constant ε has an azimuthally symmetric density charge σ = σ0 cos θ placed on the surface. Outside the sphere is vacuum. (a) Obtain the electrostatic potential inside the sphere, φin. (b) Obtain the electrostatic potential outside the sphere, φout. (c) What is the electric field inside the sphere?

    2. Relevant equations
    ##\int \vec{D} \cdot \vec{dA}=Q_{free,enc.}##

    3. The attempt at a solution
    I'm not sure if my approach was correct. I assume not because part c) says to find the electric field, yet I found it without needing to find the potential. I'm sure I went wrong somewhere.

    **My attempt:**

    I draw a Gaussian sphere around the given sphere. Then I find ##\vec{D}##, the electric displacement.

    ##\int \vec{D} \cdot \vec{dA}=Q_{free,enc.}##
    ##D=\frac{\sigma_0 \cos \theta R^2}{r^2} \quad ; \quad r>R##

    Using this, I can find the electric field by ##\vec{D}=\epsilon \vec{E}##, but this doesn't feel right. Normally, with these types of problems there's a reason to solve the steps in order.
     
  2. jcsd
  3. Oct 23, 2016 #2

    Charles Link

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    The surface charge that is applied to the sphere I believe is a free surface charge, and the distribution is such that it has a well-known solution that ## E_o=-\sigma_o /(3 \epsilon_o) \hat{z} ## everywhere inside the sphere. This (uniform)applied electric field to the dielectric sphere also has a well-known solution that involves a self-consistent mathematics. ## \\ ## This one is solved by Legendre polynomial methods, but the best method I know for this one involves making an educated guess and showing the solution is the correct one. What I am showing you should be within the Physics Forum rules because for this one, you almost need a good part of the answer in order to make the correct "Ansatz" solution. ## \\ ## The problem of a dielectric sphere in a uniform electric field, which is what we need to work from here, has a simple solution that can be arrived at in a couple of stages. To begin with, we can take a guess that the uniform electric field causes a uniform polarization ## P ## which will cause surface charge density ## \sigma_p =P \cdot \hat{n} ## everywhere on the surface. This causes a field ## E_p=-P/(3 \epsilon_o) \hat{z} ## everywhere inside the sphere. (A simple Coulomb's law calculation will show the field at the center of the sphere is ## E_p=-P/(3 \epsilon_o) ##. A detailed calculation with Legendre Polynomials shows the electric field is indeed uniform everywhere inside the sphere.) Next is some self-consistent mathematics: ## E_i=E_o+E_p ## where ## E_o ## (i.e. ## E_o \hat{z} ##) is the applied field and ## E_i ## is the internal field. Also ## P=\epsilon_o \chi E_i ##. (## \epsilon=\epsilon_o (1+\chi) ##). A little algebra allows for a solution: ## E_i=E_o/(1+\chi/3) ##. The solution for ## P ## is ## P=\epsilon_o \chi E_i=\epsilon_o \chi E_o/(1+\chi/3) \hat{z} ##. You can take this solution and convert it to Legendre polynomial solutions for the ## \phi_{in} ## and you can show it is indeed correct. ## \\ ## One item to note is the manner in which the surface charge was applied puts the applied electric field on the dielectric in the "minus z " direction. ## \\ ## The final item for this is to compute the potentials from the electric field ## \phi_{in} ## and also come up with a ## \phi_{out} ## that satisfies all the constraints. It helps in making an educated guess for the correct ## \phi_{in} ## by showing you the calculations which I presented. You need to look up Legendre solutions in spherical coordinates, and it should be clear the terms that you need to use to match the potential functions inside and out. You will need to solve for a couple of the coefficients. Once you have the expression for the potential (inside and out), you should be able to find the electric field and the surface charge, etc. to verify your solution is correct. ## \\ ## If you have additional questions about this problem, I'd be happy to try to answer them.
     
    Last edited: Oct 23, 2016
  4. Oct 23, 2016 #3

    I'm going to have a lot of dumb question, hope you're prepared :)

    First, what do you mean by "a dielectric sphere in a uniform electric field"? The problem does not state that there is an external electric field.
     
  5. Oct 23, 2016 #4

    Charles Link

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    By putting a surface charge on the dielectric sphere with the distribution that they chose made this the case. The result is well-known to those who have worked a number of these problems, but may not be at all obvious to someone kind of new to the subject. This really is not a simple problem to illustrate the concepts. They really might do better to give the student a couple introductory problems. In any case, it's quite solvable, and the surface charge distribution makes for a uniform applied electric field inside the sphere. (Outside the sphere this applied field will not be uniform.) The problem of a dielectric sphere in a uniform electric field in the absence of free charges on the sphere is a very standard electrostatics problem that gets solved using Legendre polynomials. (It is recommended you carefully study the problem of a dielectric sphere in a uniform field to see how to apply the Legendre polynomials.) ## \\ ## In the case which they gave you, the free charge which they apply to the sphere will create a uniform electric field ## E_o ## which will create a uniform polarization inside the sphere and also causes additional surface polarization charges to appear on the surface of the sphere. The polarization charge will have an electric field ## E_p ##(which also happens to be uniform for this case because ## \sigma_p=P \cdot \hat{n} ## = again the same surface charge distribution which gives a well known-result=notice this polarization charge has a similar (geometrical) distribution as the free charge that was initially applied. It only cancels a fraction of it.) which serves to reduce the electric field=thereby the self-consistent algebra of the previous post to solve for ## E_i ## and ## P ##.## \\ ## Additional note: Without the simple result of the uniform field and polarization inside the sphere, this problem would be next to impossible. The fact that it gives a simple and uniform solution inside the sphere makes this problem workable. ## \\ ## Hopefully you can work through the algebra of my first post. To show explicitly as a vector equation, ## \ \vec{E_i}=\vec{E_o}+\vec{E_p} ## . I am hoping it was recognizable that the vectors were all pointing in the z or minus z direction, etc.
     
    Last edited: Oct 23, 2016
  6. Oct 23, 2016 #5

    Charles Link

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    A follow-on: Try also this problem that was previously posted on PF. https://www.physicsforums.com/threa...field-of-a-uniformly-polarized-sphere.877891/ One other post that is a good way to learn about the Legendre type solution is: https://www.physicsforums.com/threads/potential-of-sphere-given-potential-of-surface.887477/ Both of these I think you could find quite useful at solving your homework problem. Especially if you are starting with limited experience with these dielectric spheres, your homework assignment is no easy task, but the solution is actually a little simpler than you might think.
     
    Last edited: Oct 23, 2016
  7. Oct 23, 2016 #6
    Ok so we have a dielectric sphere. Then we put a surface charge on it. This creates a uniform electric field. Now this uniform electric field causes uniform polarization, which means that the individual atoms on the dielectric sphere are now lined up parallel to the electric field. But this next part confuses me - the polarization we just created causes a surface charge density? You mean in addition to the one we already had? What does this part mean?
     
  8. Oct 23, 2016 #7

    Charles Link

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    The answer is yes, in addition to the surface charge distribution that we already had. These is a surface polarization charge that comes from ## - \nabla \cdot P=\rho_p ##. For uniform polarization ## P ## there are no polarization charges, except at the surface boundary where ## \sigma_p=P \cdot \hat{n} ##. This last equation is readily derived using gauss's law at the surface. ## P=0 ## outside the surface. Polarization charges create electric fields just like free charges. Now the polarization ## P ## is not fixed, but it is determined by a combination of ## E_o ## and ## E_p ##: ## E_i=E_o+E_p ##. ## P=\epsilon_o \chi E_i ##. We see that any ## P ## that we create will cause a ## E_p ## that serves to reduce ## E_o ##. The mathematics is really kind of simple once you practice with it a little. It is a self-consistent type that involves solving for ## E_i ## and ## P ##. (2 equations/two unknowns). The one result you need is ## E_p=-(P/(3 \epsilon_o) )\hat{z} ## for a sphere. This (-1/3) factor only occurs for spherical geometries. For most other geometries, ## E_p ## would not be uniform inside the solid. ## \\ ## It really should be a supplemental problem before beginning the problem you were given is to show/prove that for fixed polarization ## P ## (with no external fields) that ## E_p=-(P/(3 \epsilon_o))\hat{z} ## everywhere inside the sphere. (Notice that ## \sigma_p=P \cdot \hat{n}=P cos(\theta) ##.)This result comes from the Legendre solutions and is the problem of one of the two links which I gave above. This also tells you immediately that if ## \sigma_{free}=\sigma_o cos(\theta) ## that ## E_o=-(\sigma_o/(3 \epsilon_o))\hat{z} ## (inside the sphere). If you've seen these types of problems before, you would likely see this right away. The professor gave you a problem that was really easy for him, but not easy for someone first seeing the problem of a dielectric sphere.
     
    Last edited: Oct 23, 2016
  9. Oct 23, 2016 #8
    The electric field that is caused by the original surface charge - am I safe in reasoning that it points in the radial direction (outwards)? If so, then by what I've quoted above, ##E_p## reduces ##E_0## because it is directed radially inwards, right? I think in order for one field to "reduce" another, then they must be pointing in opposite directions.
     
  10. Oct 23, 2016 #9

    Charles Link

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    The ## E_o ## points in the minus z-direction and the ## E_p ## for this problem will point in the plus z-direction. (This is only for inside the sphere). (The P in the minus z direction will have a postive ## \sigma_p ## on the left (minus z) side of the sphere.) (Additional note: ## E_o ## will in general be larger in amplitude than ## E_p ##.) They are not radially outward or inward. It is in the z-direction. ## \\ ## See also the answer provided in the "post #7 of the first link I provided above" for the direction and amplitude of ## E_p ## that a uniform ## P ## inside a sphere will give. The Legendre solution for the potential in spherical coordinates is shown in that post, and you can compute ## E_i=E_p=-\nabla \phi_{in} ##. The answer is in spherical coordinates, it takes a step or two of algebra to show that indeed ## E_i=E_p=-(P/(3 \epsilon_o) )\hat{z} ##. (Note ## E_o=0 ## for the problem of the link since there is no external field).
     
    Last edited: Oct 23, 2016
  11. Oct 23, 2016 #10

    Charles Link

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    @1v1Dota2RightMeow I edited the last post=made an addition to it, so be sure and read the second paragraph in the above post.
     
  12. Oct 23, 2016 #11
    Yup, I noticed. I'm reading everything you've said, including what you've linked. I want to fully understand this problem.
     
  13. Oct 23, 2016 #12

    Charles Link

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    Additional item: In post #7 of the first link I worked out the potential functions for a surface charge distribution ## \sigma_p=P cos(\theta) ##. It should be a simple matter to write out the potential for ## \sigma_{total}=\sigma_{free}+\sigma_p ## since they both are of the form surface charge density ##\sigma= A cos(\theta) ##. ## \\ ## Note that in post #2 above, we solved for ## P ## and thereby ## \sigma_p=Pcos(\theta) ##.
     
    Last edited: Oct 23, 2016
  14. Oct 23, 2016 #13

    Charles Link

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    Additional note: They are asking for the answer for the potential ## \phi ## in terms of ## \sigma_o ## and ## \epsilon ##, so that you will need to convert ## \chi ## to ## \epsilon ## and the ## E_o ## needs to be written in terms of ## \sigma_o ##, but that was all computed/shown in post #2 above. Hopefully the solution is starting to come together.
     
  15. Oct 24, 2016 #14

    Charles Link

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    And one more additional note: The "free charge" distribution that is put on the sphere is not free to move around and stays fixed throughout the problem. Meanwhile polarization charge arises from the field of these charges, but also has a field of its own which will affect how much polarization charge arises. It is a self-consistent algebra/mathematics that will determine the polarization ## P ## that occurs.(see post #2 above for the solution of ## P ## and ## E_i ##.) ## \\ ## Also note: The algebra of post #2 is a simplified solution that allows you to make an educated guess. It assumes that ## E_o ##, ## P ##, ## E_p ##, and ## E_i ## are all uniform and point in the z-direction and that the ## -(1/3) ## factor is the correct number for a uniform surface charge. These assumptions are used to make an "Ansatz" (German word for assumption) type solution for the potential ## \phi ## (which is comprised of Legendre polynomials) which is shown to be correct with a couple of subsequent calculations. ## \\ ## The formal Legendre polynomial solution is the complete solution to this problem, but the method I often use to quickly get the answer (for the electric field inside the sphere) is that of post #2 above.
     
    Last edited: Oct 24, 2016
  16. Oct 24, 2016 #15
    In one of the links, you noted ##V_{in}## and then took the gradient, but your result ##E_{in}=-P_0/3 \epsilon_0 (cos \theta \hat a_{r} - sin \theta \hat a_{\theta})## is different from what I got. How did you group the trigonometric terms like that?

    I got ##E_{in}=-P_0/3\epsilon _0 cos \theta \hat r + sin \theta \hat \theta##
     
  17. Oct 24, 2016 #16
    Oh wait, I made a small mistake. Never mind.
     
  18. Oct 24, 2016 #17

    Charles Link

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    One additional comment=I think very few people ever memorize the complete mechanics of the Legendre method. For this problem of the sphere, the result is very useful and once the (-1/3) factor is established, it can be used without going through the Legendre operation each time. The (-1/3) factor even comes up in the pole method of magnetostatics for a sphere=for a magnetic material (a sphere of magnetic material) in a uniform magnetic field. In this application, the D=-1/3 is known as the "demagnetizing factor".
     
  19. Oct 24, 2016 #18
    Well I'm only an undergrad, so I need to practice the Legendre method. Plus I don't think my professor would accept me just knowing this established result without showing it.

    I'm trying to work through the method now.
     
  20. Oct 24, 2016 #19

    Charles Link

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    Very good. I worked through it also many years ago as a sophomore undergraduate in an advanced E&M course at the University of Illinois at Urbana.. It took a lot of effort to learn it, and it was only later(=much later) that I started to see a couple of shortcuts.
     
  21. Oct 24, 2016 #20
    So if I'm right, I think the way to solve by method of Legendre polynomials is to solve

    ## V(r, \theta) = \sum_{l=0}^{\infty} (A_l r^l + \frac{B_l}{r^{l+1}})P_l(cos \theta) ##

    for ##V_{in}## and ##V_{out}##.

    Therefore, ##V_{in}=\sum_{l=0}^{\infty} A_l r^lP_l(cos \theta)## and ##V_{out}=\sum_{l=0}^{\infty} \frac{B_l}{r^{l+1}}P_l(cos \theta)##.

    But I need boundary conditions, don't I? I can only think of 1: that the potential at the center must be finite (that's how I got rid of the ##B## term in the first equation. Can anything be said of the potential at the surface of the sphere?

    Originally, I wanted to say that ##V=0## at the surface, but then I realized that this might not be the case due to the polarization of the dielectric.
     
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