Boundary conditions (E and D) for a dielectric sphere

[Charles Link]

a follow-on: The boundary for $\vec{E}$ is basically Gauss' law for the given $\sigma_p$.

 

[happyparticle]

For a non linear material $\vec{P} = ?$
I had no idea about that. I thought I was right about the electric field inside the sphere.

 

[Charles Link]

Perhaps I misread the problem in the OP, but it wasn't real clear. If it is a dielectric, that might imply linear materials. If the $\vec{P}=P_o \hat{z}$ is given though, and that's how it appeared to me, it is a spontaneous polarization, and not a dielectric. Hopefully, I'm not making unnecessary confusion.

The spontaneous and uniform $P$ is a fairly simple problem with a simple solution.
In any case, the boundary conditions for a specified $P$, uniform or otherwise, are the same in both cases.

 

[happyparticle]

Sorry if the problem wasn't clear, but it is a dielectric sphere. Basically, I have to find the boundary condition for $\vec{E}$ and $\vec{D}$.

 

[Charles Link]

Sorry if the problem wasn't clear, but it is a dielectric sphere. Basically, I have to find the boundary condition for $\vec{E}$ and $\vec{D}$.

Please read the line that I added to the end of the previous post.

 

[Charles Link]

If you know $\sigma_p=\vec{P} \cdot \hat{n}$, what can you say about $E_{out \, perpendicular}-E_{in \, perpendicular}$?

also using $\nabla \cdot \vec{D}=0$ and Gauss' law, what can you say about the $\vec{D}$ in a similar manner?

 

[happyparticle]

I thought that $E^{out}_\perp = \frac{\sigma_p}{\epsilon_0}$, $\epsilon_0$ because outside we are in vacuum and $E^{in}_\perp = \frac{\sigma_p}{\epsilon}$. However, it seems like I'm wrong.

 

[Charles Link]

The only unbalanced electrical charges in the problem is the polarization charge $\sigma_p$ at the surface of the sphere. The electric field $E$ around these charges behaves the same whether it is vacuum or material. We write $\int E \cdot dS=q_{total}/\epsilon_o$

Now make a small Gaussian pillbox, and let the charge $q_{total} = \sigma_p A$ where $A$ is a small area.

 

[happyparticle]

I'm trying to figure out what that means.
Does it mean that for a dielectric without $\sigma_f$, $E = \frac{\sigma_p}{\epsilon_0}$ on the surface?
I thought with the boundary conditions we were looking for the electric field inside the dielectric and outside and those field depends of $\epsilon$, but as I can see it only means the electric field on the surface, thus in my case, $E^{out}_\perp - E^{in}_\perp = \frac{\sigma_p}{\epsilon_0}$.

 

[Charles Link]

The second one is correct, not the first one.
For a boundary value problem, $E_{out}$ means for $r$ just slightly greater than radius $a$, and $E_{in}$ means for $r$ just slightly less than radius $a$, although these terms can also be used for arbitrary radius.

 

[happyparticle]

Sorry, what exactly is not correct?

 

[Charles Link]

The known solution is $\vec{E}_{in}=-P_o/(3 \epsilon_o) \hat{z}$ everywhere inside the sphere, including just inside the radius $a$. This is in disagreement with what you are stating in the first expression you wrote.

The electric field will be different on the other side of the surface charge. Your second expression is what we are looking for and is correct.

also note that $\sigma_p=P_o \cos{\theta}$ for uniform $P=P_o$.

 

[Charles Link]

The first expression $E_{out \, perpendicular}=\sigma_p/\epsilon_o$ only holds for a perfect conductor, where $E_{in}=0$.

 

[happyparticle]

My textbook confuses me.
So here I can't use $\vec{P} = \epsilon_0 \chi_e \vec{E}$ because it's non linear?
$E^{out}_\perp - E^{in}_\perp = \frac{\sigma_p}{\epsilon_0}$
is not on the surface of the sphere?

 

[kuruman]

If I may add something that might help steer the OP in the right direction. See posts 2 and 7 of
https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5513730
Writing out the boundary conditions should be fairly straightforward, but I think it might help the OP if he has some idea of what the solution is, of this somewhat well-known E&M problem.

It is a well-known problem. OP has a copy of Griffiths which treats it. In my edition it is example 4.2. Evidently OP didn't notice its presence. @EpselonZero please look it up and study it.

 

[happyparticle]

It is a well-known problem. OP has a copy of Griffiths which treats it. In my edition it is example 4.2. Evidently OP didn't notice its presence. @EpselonZero please look it up and study it.

I see that, but I don't understand why what you said in post #10 doesn't work. I mean, that's why I was and I'm so confuse. Both doesn't give me the same result, so I thought example 4.2 wasn't correct for my problem.

Furthermore, the problem is about the boundary conditions, maybe I'm wrong, but I don't see the link between E inside the sphere and the boundary conditions.

 

[kuruman]

I think we have reached the point where we need to agree on where this is headed. In post #20 I asked

Your statement of the problem is "Find the Boundary conditions for (E and D) for a dielectric sphere." Is that exactly how it was given to you?

which you did not answer. Also farther down (post #1) you ask

However, how to verify the boundary conditions for a sphere?

"Finding" the boundary conditions requires you to derive expressions relating the normal and tangential components on the surface of the permanently polarized dielectric sphere. A derivation that applies to all dielectric-vacuum interfaces can be found in any EM textbook. Just pattern your derivation to it.

"Verifying" the boundary conditions means that you know (or have already derived) them and that you also know (or have derived) expressions for the D and E fields inside and outside the dielectric. The verification task involves substituting the known expressions for the fields in the known boundary conditions and seeing whether the left-hand side of each equation is the same as the right-hand side. Of course, the two sides better be the same because that's built in the expressions for the D and E fields which were derived using the boundary conditions in the first place. This is like going around in a closed loop and making sure that the loop is closed by ending up where you started.

There may be other interpretations of what you are asked to do, but then we will need the problem statement as given to you. Maybe @Charles Link has additional suggestions.

 

[Charles Link]

The textbooks do a somewhat ok job of this problem, but they seem to leave off some details. Instead of just asking for the boundary conditions, they would do well to teach this problem with a series of steps, beginning with the case of spontaneous and uniform polarization $\vec{P}=P_o \hat{z}$, and compute the surface polarization charge density $\sigma_p=\vec{P} \cdot \hat{n}$, (comes from polarization charge density $\rho_p=-\nabla \cdot \vec{P}$ along with Gauss' law. Note with $\nabla \cdot \vec{P}=0$ for uniform polarization, the result is that polarization charge results only on the surface of the sphere). They should then have you compute the electric field at the center of the sphere. This would present some of the concepts before solving it as a boundary value problem.

Computing the electric field at the center of the sphere is a straightforward integral with Coulomb's law, and the result is $\vec{E}=-P_o/(3 \epsilon_o) \hat{z}$. (@EpselonZero Please try to do this computation). Upon getting this result, they can then work towards showing/proving that the electric field is the same everywhere inside the sphere for this case.

Upon doing that, they can then introduce the linear dielectric, and start to explore Legendre methods with the student, including the problem with the applied external electric field.

 

[happyparticle]

Using $V(\vec{r}) \int_V \frac{\vec{P} \hat{}r}{r^2} d\tau$
I found $\vec{E} = - \frac{P\hat{z}}{3\epsilon_0}$ r<R
$\vec{E} = \frac{\vec{P}\hat{r} R^3 2}{3\epsilon_0 r^3}$ r>R

If the boundary conditions gives me the electric field at the surface of the sphere. Then $\vec{E}$ found with boundary conditions should be the same as the electric field found inside the sphere, since there are only the surface charges.

 

[Charles Link]

The one calculation that is fairly simple is to compute the electric field at the center of the sphere from Coulomb's law. I'm not sure what you did for the above, because you didn't show your work.

 

[happyparticle]

The one calculation that is fairly simple is to compute the electric field at the center of the sphere from Coulomb's law. I'm not sure what you did for the above, because you didn't show what you did.

I only have the griffith as textbook and that's what he did in the example 4.2. The example Kuruman told me to look.

 

[Charles Link]

I only have the griffith as textbook and that's what he did in the example 4.2. The example Kuruman told me to look.

I don't have Griffith's text, but I think this same example from Griffith's text came up a couple years ago on the Physics Forums. Let me see if I can find that "link", where I remember doing some detailed calculations...

 

[happyparticle]

I don't have Griffith's text, but I think this same example from Griffith's text came up a couple years ago on the Physics Forums. Let me see if I can find that "link", where I remember doing some detailed calculations...

I edited my post with the example from the book

 

[Charles Link]

See https://www.physicsforums.com/threa...ormly-polarized-cylinder.941830/#post-5956930
The above is the problem from Griffith's that we worked a couple years ago on the Physics Forums. Griffith's uses a different method to solve this one, but I do recommend you get familiar with the Legendre method of solution. For the Legendre method, you are given a series of terms of powers of $r$ and $\cos{\theta}$, etc. , and you make an educated guess for a solution. If you can show your solution satisfies the boundary conditions, it then is indeed the correct one.

 

[happyparticle]

I don't know what it means that my solution satisfies the boundary conditions. Are they the same?
There's not even an example on how to find the boundary conditions for E and D, I'm not even sure of what I'm doing. I can't find answers to my questions. I didn't know I can't use $D = \epsilon E$, so I lost a lot of time.

Example 4.9
A sphere of radius R is filled with material of dielectric constant $\epsilon_r$ and uniform embedded free charge $\rho_f$
$\vec{E} = \frac{\rho_f \vec{r}}{3\epsilon_0 \epsilon_r}$ r<R
Why here the electric field had $\epsilon$ and not $\epsilon_0$ and here he's using $\vec{D}$ to find $\vec{E}$
Is it because if $\vec{D} = 0$ I can't use it to find $\vec{E}$

My guess is if there's no free charges I can't use $\vec{D} = \epsilon \vec{E}$, since all the examples using this equations to find the electric field have a volume with free charges.

 

[Charles Link]

Without any free charges or external electric fields, the polarization is of a spontaneous type, and it won't be of the linear variety.

One suggestion is that you study the examples carefully, and be as patient as you can. This material is somewhat difficult, and it's likely to take a fair amount of time and effort to really get a good handle on it.

See https://www.physicsforums.com/threads/electric-field-of-a-charged-dielectric-sphere.890319/
for another example of a problem with a dielectric sphere. See also post 5 of this "link" for a couple other examples.

 

[happyparticle]

finally, something I can understand. I understand, but It seems like I can't understand this textbook and, you know with all the courses, I only have few days to understand and right now I'm really stuck.

 

[kuruman]

See https://www.physicsforums.com/threa...ormly-polarized-cylinder.941830/#post-5956930
The above is the problem from Griffith's that we worked a couple years ago on the Physics Forums. Griffith's uses a different method to solve this one, but I do recommend you get familiar with the Legendre method of solution. For the Legendre method, you are given a series of terms of powers of $r$ and $\cos{\theta}$, etc. , and you make an educated guess for a solution. If you can show your solution satisfies the boundary conditions, it then is indeed the correct one.

In the previous chapter to the one with the dielectric sphere Griffiths has an example in which he considers arbitrary surface charge density $\sigma(\theta)$ "plastered" on a sphere. He writes the potentials inside and outside as a series of Legendre polynomials and uses the continuity of the potential at $r=R$ to find a relation between the coefficients of the potentials inside, $A_l$, and outside $B_l$. This allows him to write the two potential in terms of $A_l$ only. Then he uses the discontinuity of the normal component of E (or D) at the boundary $$\left . \left( \frac{\partial V_{out}}{\partial r}-\frac{\partial V_{in}}{\partial r}\right)\right |_{r=R}=-\frac{\sigma(\theta)}{\epsilon_0}.$$ Then he shows how to pick out the coefficients using what he loves to call "Fourier's trick":$$A_l=\frac{1}{2\epsilon_0 R^{l-1}}\int_0^{\pi}\sigma(\theta)P_l(\cos\theta)~{\sin}\theta~d\theta.$$In short, it's a full blown derivation using Legendre polynomials which I summarized here for OP's benefit.

At this point, Griffiths does a "for instance" in which he assumes a specific form for the surface charge distribution, $\sigma(\theta)=\sigma_0~\cos\theta$ and finds the potentials and fields inside and outside. This for instance is the starting point for the dielectric sphere in the next chapter where the surface charge density is $\sigma_P(\theta)=P~\cos\theta.$

I don't know what it means that my solution satisfies the boundary conditions.

See the paragraph that starts with "Verifying $~\dots$" in post #52. Also for your benefit, the boundary conditions at the interface between vacuum and dielectric are

1. The normal component of $\vec D$ is continuous across the boundary.
2. The tangential component of $\vec E$ is continuous across the boundary.

 

[Charles Link]

@EpselonZero I added a couple of things at the bottom of post 61, including a "link" where you will find a couple other "links" in post 5 of that "link". I think you might find the discussions of these problems that appeared previously on Physics Forums rather useful.

 

[Charles Link]

This for instance is the starting point for the dielectric sphere in the next chapter where the surface charge density is $\sigma_p=P_o \cos{\theta}$.

I think that is precisely the problem that is discussed in detail in the "link" of post 61, which I will repost here: https://www.physicsforums.com/threads/electric-field-of-a-charged-dielectric-sphere.890319/

 

[kuruman]

I think that is precisely the problem that is discussed in detail in the "link" of post 61,$~\dots$

It certainly looks that way.

 

[happyparticle]

After all I still don't know how to find the boundary conditions which was the initial question.
Still from my book.
$\int_s \vec{E} \cdot d\vec{a} = \frac{\sigma A}{\epsilon_0}$
$\vec{E} \cdot \vec{n} A = \frac{\sigma A}{\epsilon_0}$
$\vec{E} \cdot \vec{n} = \frac{\sigma_p}{\epsilon_0}$
$\vec{E} \cdot \vec{n} = \frac{\vec{P} \cdot \hat{n}}{\epsilon_0}$

And then since, $\int \vec{E} \cdot d\vec{l} = 0$ over a closed path and the tangential component is continuous so we have $\vec{E}_above - \vec{E}_below = \frac{\vec{P} \cdot \hat{n}}{\epsilon_0}$

We said that the electric field inside a dielectric sphere is $-\frac{\vec{P}\hat{z}}{3 \epsilon_0}$
which doesn't match with the electric field inside the sphere.
Left hand side and right and side A's are the same, just make sure.

 

[Charles Link]

$\int E \cdot da$ is over a closed surface. The normal on $da$ on the part of the pillbox in the material points opposite the normal to the sphere which we call $\hat{n}$. From this, you should get $E_{out \, perpendicular}-E_{in \, perpendicular}=\sigma/\epsilon_o$.

(From the boundary conditions, we don't have a clue yet what the two E's are, except in the case of a conductor, where $E_{in}=0$. The only thing we know is the difference in their perpendicular components).

 

[happyparticle]

$\int E \cdot da$ is over a closed surface. The normal on $da$ on the part of the pillbox in the material points opposite the normal to the sphere which we call $\hat{n}$. From this, you should get $E_{out \, perpendicular}-E_{in \, perpendicular}=\sigma/\epsilon_o$.

That was I found few posts ago, which was right?

 

[Charles Link]

That was I found few posts ago, which was right?

Yes=posts 44 and 49. See also what I added to post 68.