Boundary conditions (E and D) for a dielectric sphere

[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).