Boundary conditions (E and D) for a dielectric sphere

[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.

 

[happyparticle]

Yes=posts 44 and 49.

Even 17
Ah! and $E_{in} = -\frac{\vec{P}\hat{z}}{3 \epsilon_0}$ ?

 

[Charles Link]

@EpselonZero Please read the last couple sentences of post 63 by @kuruman where the boundary conditions are summarized for the case where $q_{free}=0$.

With the first condition $D_{out \, perpendicular}=D_{in \, perpendicular}$,
we get, (note $D_{in \, perpendicular}=\vec{D}_{in} \cdot \hat{n}$, etc., and $\vec{D}_{in}=\epsilon_o \vec{E}_{in}+\vec{P}_{in}$, etc.)

$\epsilon_o E_{out \, perpendicular}=\epsilon_o E_{in \, perpendicular}+\vec{P} \cdot \hat{n}$,

because $\vec{P}_{out}=0$.
Note $\vec{P} \cdot \hat{n}=\sigma_p$.

This is consistent with everything else that we have calculated.

Working the problem with $D's$ rather than $E's$ really doesn't give us anything extra.

 

[happyparticle]

Ah, In this case
Since, $\vec{E}^{out}_\perp - \vec{E}^{in}_{\perp} = \frac{\sigma_p}{\epsilon_0}$
And we know that $\vec{E}_{in} = -\frac{P \cdot \hat{z}}{3 \epsilon_0}$

Then, $-\frac{P}{3 \epsilon_0} + \frac{\sigma_p}{\epsilon_0} = \frac{2P}{3\epsilon_0}$

Earlier, I found $\vec{E}_{out} = \frac{2\vec{P} \cdot \hat{r} R^3}{\epsilon_0 r^3 3}$

Since we evaluate $E_{out}$ really close to the sphere where r --> R

I have
$\vec{E}_{out} = \frac{2\vec{P} \cdot \hat{r}}{3 \epsilon_0}$
It sounds fine, but I have to deal with those vectors now.

 

[Charles Link]

@EpselonZero Looks like you are making progress. Please see post 61 with the "links" there. I think you also saw the calculations I added to post 72, but please take another look. In any case, I do think you are making progress. :)

See also again https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5513730 especially posts 2 and 7. You should be able to start to follow these discussions in detail. You are really in about the same place these other students were a couple of years ago.

 

[kuruman]

Ah, In this case
Since, $\vec{E}^{out}_\perp - \vec{E}^{in}_{\perp} = \frac{\sigma_p}{\epsilon_0}$
$\dots$

For this equation (and the others that follow) you must write the left-hand side as a scalar quantity, not as a vector. Pick a point on the sphere at angle $\theta$ from the z-axis and relate the "in" and "out" normal field components (which are scalars) to the surface charge density at that angle. Same with the tangential components except that the right-hand side is zero.

 

[Charles Link]

One thing that is very important in this problem of the polarized sphere is the result that the electric field when the sphere is polarized to the right is $E_p=-P/(3 \epsilon_o)$, i.e. it is uniform inside the sphere and points to the left with a factor of $1/3$. (This results from the surface polarization charge density of $\sigma_p=\vec{P} \cdot \hat{n}=P \cos{\theta}$).

This result can even be used to solve for the electric field when there is an applied field on a dielectric sphere of $E_o$. . The Legendre method is important, but this simple result that comes out of the Legendre method is a very important one.

It may be worthwhile to also look at the electric field inside a flat slab that is polarized, (e.g. spontaneous polarization). This does not need Legendre methods to calculate, and the result with $\sigma_p=\vec{P} \cdot \hat{n}=\pm P$ is $E_p=-P/\epsilon_o$. Note that the $E_p$ is the same from the $P$, even when the $P$ is the result of an applied electric field $E_o$. We can write $E_i=E_o+E_p$, and $P=\chi \epsilon_o E_i$ for linear materials. The result with a little algebra is $(1+\chi)E_i=E_o$, so that $E_i=E_o/\epsilon_r$ for a dielectric slab with an applied $E_o$. Note that $\epsilon_r=1+\chi$, where $\epsilon=\epsilon_o \epsilon_r$.

A similar calculation can be done for the dielectric sphere with an applied electric field using the $1/3$ factor result.

 

[Charles Link]

Besides the two results mentioned in post 76, (the sphere and the dielectric slab), a third geometry where the axis of the cylinder is perpendicular to the polarization vector $\vec{P}$ is also important. For this case, the electric field $\vec{E}_p$ inside the cylinder from the surface polarization charge that forms is $\vec{E}_p=-\vec{P}/(2 \epsilon_o )$.

I'm mentioning these results here and trying to emphasize them, because with the Legendre method it is so easy to get lost in the mathematics. It is important to take notice of these results that come out of the Legendre method=i.e.for the sphere and the cylinder. Hopefully the OP @EpselonZero sees this post and the previous one, and takes note of these important cases.

 

[Charles Link]

I would like to add something about the original statement of the problem, and the boundary conditions. Normally for this type of problem, I think the boundary conditions are to have $V_{in}=V_{out}$ everywhere on the boundary, and that $E_{out \, perpendicular}-E_{in \, perpendicular}=\sigma_{total \, boundary}/\epsilon_o$. (The electric fields are computed from derivatives of the potential $V$). Note $\sigma_{total}=\sigma_{free}+\sigma_p$.

Using $D's$ instead of $E's$, we get $D_{out \, perpendicular}-D_{in \, perpendicular}=\sigma_{free \, boundary}$.

The tangential components for $E$ will be the same if $V_{in}=V_{out}$ everywhere on the boundary, and this last condition is very necessary. Perhaps @kuruman can add to this.

 

[kuruman]

I would like to add something about the original statement of the problem, and the boundary conditions. Normally for this type of problem, I think the boundary conditions are to have $V_{in}=V_{out}$ everywhere on the boundary, and that $E_{out \, perpendicular}-E_{in \, perpendicular}=\sigma_{total \, boundary}/\epsilon_o$. (The electric fields are computed from derivatives of the potential $V$). Note $\sigma_{total}=\sigma_{free}+\sigma_p$.

Using $D's$ instead of $E's$, we get $D_{out \, perpendicular}-D_{in \, perpendicular}=\sigma_{free \, boundary}$.

The tangential components for $E$ will be the same if $V_{in}=V_{out}$ everywhere on the boundary, and this last condition is very necessary. Perhaps @kuruman can add to this.

I agree 100%. The only addition I have is that the continuity of the potential across the boundary is preferred when one crafts solutions by first finding the potential (e.g. the Legendre polynomial method) whilst the equivalent continuity of the tangential component of E is preferred when one has to deal with time-varying EM fields at interfaces (e.g. EM radiation).