Boundary conditions (E and D) for a dielectric sphere

[happyparticle]

Homework Statement

Find the Boundary conditions for (E and D) for a dielectric sphere.
$\vec{P} = P\hat{z}$

Relevant Equations

$\sigma_b = \vec{P} \cdot \hat{n}$

Since there is no free charge $\int_S \vec{D} \cdot d\vec{a} = 0$ and
$\rho_f = 0$
$\sigma_f = 0$

$\vec{nabla} \cdot \vec{P} = 0$ since P is a constant
$\rho_b = - \vec{nabla} \cdot \vec{P} = 0$

For a simple surface we can find the boundary conditions for $\vec{E}$ using a Gauss' surface to verify $E^{\perp}_{above} - E^{\perp}_{below}$ and $E^{||}_{above} - E^{||}_{below}$

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

 

[kuruman]

For a simple surface we can find the boundary conditions for $\vec{E}$ using a Gauss' surface to verify $E^{\perp}_{above} - E^{\perp}_{below}$ and $E^{||}_{above} - E^{||}_{below}$

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

Boundary conditions are relations of field components between two different regions in space which are written as equations. I see no equations here. I think what you are asked to do here is find $\vec D$ and $\vec E$ inside and outside the sphere without using the boundary conditions and then see whether their normal and tangential components match as prescribed by the known boundary conditions.

 

[happyparticle]

If there's no free charge $\vec{D}$ is 0 everywhere. How can I see if his normal and tangential component match as prescribed? I mean, I don't know how to check their normal and tangential components for a sphere.

 

[kuruman]

If there's no free charge $\vec{D}$ is 0 everywhere.

How do you figure? Only the flux of $\vec D$ through a closed surface is zero. You can use this to find $\vec D$ and $\vec E$ inside. What closed surface do you think you should use?

 

[happyparticle]

How do you figure? Only the flux of $\vec D$ through a closed surface is zero. You can use this to find $\vec D$ and $\vec E$ inside. What closed surface do you think you should use?

Ah, you are right.
Here's my difficulty. I'm never sure which Gauss' surface to use and since we only did examples using planes or cubic volume, I think I should use a sphere here. I think this is the best way to get $\vec{D}$ and $\vec{E}$ inside and outside, but in this case I'm not sure how to find the normal and tangential components.

 

[kuruman]

Ah, you are right.
Here's my difficulty. I'm never sure which Gauss' surface to use and since we only did examples using planes or cubic volume, I think I should use a sphere here. I think this is the best way to get $\vec{D}$ and $\vec{E}$ inside and outside, but in this case I'm not sure how to find the normal and tangential components.

If you are not sure how to find the normal and tangential components, then you are considering the wrong Gaussian surface. The criteria for choosing a Gaussian surface (check the examples you are familiar with to see if this is true) are (a) the field must be constant or zero on the surface; (b) the field must be perpendicular to the surface if it isn't zero.

Start in the region inside the sphere and write the equation(s) relating $\vec D$, $\vec E$ and $\vec P$. Can you imagine what an appropriate $S$ would be? Your goal is to find $\vec E$ on the surface and you already know that $\int_S \vec D\cdot\hat n ~dA=0.$

 

[happyparticle]

I thought the surface should have the same shape as the volume.
I think a cylinder could work, but how can I find the electric field if the sphere has no charge?

 

[kuruman]

Start in the region inside the sphere and write the equation(s) relating $\vec D$, $\vec E$ and $\vec P$.

 

[happyparticle]

They are all 0
Since $\int_S = \vec{D} \cdot d\vec{a} = 0$ and $\vec{E} =\frac{\vec{D}}{\epsilon}=0$
$\vec{P} = \epsilon_0 \chi_e \vec{E} = 0$

 

[kuruman]

You have gone around in a circle using $\epsilon$ and $\chi_e$ and concluded that $\vec P =0$ when the problem clearly states that $\vec P = P~\hat z$ for the question to make sense. Also, you made the mistake that you were warned against making. One more time, just because the integral is zero does not mean that D is zero.

The equation, which I always use and trust in problems like this, is $\vec D=\epsilon_0\vec E+\vec P$. When the left hand side is zero, the right hand side is not (excepting the trivial case).

 

[happyparticle]

Does it means that $\vec{E}$ and $\vec{P}$ always points in the opposite direction when $\vec{D} = 0$ and $\vec{P} \propto \vec{E}$

 

[kuruman]

What do you think? Read and interpret the equation.

 

[happyparticle]

That's what I think, but I'm never sure of what I think. It's like I always forget something. I don't trust myself.

 

[kuruman]

Not trusting yourself could be a problem because you will be living 100% of the rest of your life with yourself. If you think that when $\vec D = 0$, $\vec E$ and $\vec P$ must point in opposite directions but are not sure, try to prove yourself correct by looking at the problem in terms of what you know in general.

For example, the equation you have is of the form of adding two vectors to get zero
$\vec A+\vec B=0$. What must be true about the angle between them?
If you still are not sure, write out the sum of the two vectors in component form, add them and then see what must be true about their components for their sum to be zero. I will get you started, but you have to finish and draw your conclusions.

$\vec A=A_x~\hat x+A_y~\hat y+A_z~\hat z$
$\vec B=B_x~\hat x+B_y~\hat y+B_z~\hat z$
$0=\vec A+\vec B=\dots$

Learn to rely on your own abilities by convincing yourself that yes, you can do it without anyone's help or stamp of approval. When you're right, you're right.

 

[happyparticle]

Not trusting yourself could be a problem because you will be living 100% of the rest of your life with yourself.

I know, but I'm always wrong. Sometime I know I'm wrong, but I just can't say what's wrong.

I thought if a volume wasn't charged the electric field = 0. How can I trust myself.

I drew my sphere with the electric field lines. I think they move from top to bottom.
However, I was using a sphere as a Gauss' surface, but this is still my initial question, I don't see how to check the boundary conditions.

Should I use a different Gauss' surface for boundary conditions?

 

[kuruman]

Should I use a different Gauss' surface for boundary conditions?

Yes. Use the criteria I gave you. First you have to find the fields inside.

 

[happyparticle]

sphere top to bottom.

I have to check the boundary conditions for all 3 faces of the cylinder.
For the top face.

since if $\vec{D}$ is parallel to the surface $\vec{D}^{above}_{||} - \vec{D}^{below}_{||} = 0$
$\int_S \vec{D} \cdot d\vec{a} = \vec{D}^{above}_{\perp}- \vec{D}^{below}_{\perp} = Q_f = 0$
$\int_S \vec{E} \cdot d\vec{a} = -\vec{D}^{above}_{\perp}+ \vec{D}^{below}_{\perp} = \frac{\sigma}{\epsilon_0} = \frac{\sigma_b}{\epsilon_0}$

 

[kuruman]

I don't understand this. Please describe your cylindrical Gaussian surface in more detail or perhaps post a drawing of it and the dielectric sphere.

 

[happyparticle]

 

[kuruman]

That will not do much for you. I don't want to lead you astray by offering misleading advice. 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?

If "no" I need to know exactly what was given to you. If "yes" you need to review your notes or textbook where the boundary conditions were found for a flat surface and make sure you fully understand the derivation. It requires drawing a Gaussian surface for the normal component of D and a closed loop for the tangential component of E. The goal is to adapt the derivation to the case of a dielectric sphere. I can guide you through that but not if you do not fully understand the derivation; it will be a waste of your and my time.

Let us know when you're ready.

 

[happyparticle]

I'm using the Griffiths as textbook and there is only 3 pages about the boundary conditions (88-90). I really don't see what I'm doing wrong.

In this example, I have a curved plane and we have $\int_S \vec{E} \cdot d\vec{a} = \frac{\sigma A}{\epsilon_0}$
He uses a pillbox as the Gauss'surface and since the sides of the pillbox contribute nothing to the flux, in the limit as the thickness $\epsilon$ goes to zero and $\vec{E}^{||}_{above} - \vec{E}^{||}_{below} = 0$ (parallel to the surface). We are left with $\vec{E}^{\perp}_{above} - \vec{E}^{\perp}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}$

 

[kuruman]

I'm using the Griffiths as textbook and there is only 3 pages about the boundary conditions (88-90). I really don't see what I'm doing wrong.

In this example, I have a curved plane and we have $\int_S \vec{E} \cdot d\vec{a} = \frac{\sigma A}{\epsilon_0}$
He uses a pillbox as the Gauss'surface and since the sides of the pillbox contribute nothing to the flux, in the limit as the thickness $\epsilon$ goes to zero and $\vec{E}^{||}_{above} - \vec{E}^{||}_{below} = 0$ (parallel to the surface). We are left with $\vec{E}^{\perp}_{above} - \vec{E}^{\perp}_{below} = \frac{\sigma}{\epsilon_0} \hat{n}$

What you are doing is not really wrong but it won't take you anywhere either because these are the boundary conditions for a conductor. Here you have a dielectric. You need to
(a) Write down the boundary conditions for a dielectric-vacuum interface.
(b) Find the D and E fields inside the dielectric sphere and in the vacuum. Griffiths does that for you and you have to find it.
(c) Use the results of (b) in the equations of (1) and see whether the left-hand side of each equation matches the right-hand side. That's the verification part.

 

[happyparticle]

I think I know (b), but I'm really not sure about (a), since there is not example, I'm trying to guess and as you see I'm really bad.

However, I was thinking about the example with the curved plane and the pillbox as the Gauss' surface.
Maybe I'll sound stupid one again, but the top of the sphere and the curved plane seems the same can I use a pillbox for the Gauss' surface and I can pick A to be really small and the sides goes to 0. Thus, I have $\vec{E}^{\perp}_{above} - \vec{E}^{\perp}_{below} = \frac{\sigma}{\epsilon_0} \hat{n} - \frac{\sigma}{\epsilon} \hat{n}$However, $\vec{E} = \frac{\vec{P}}{-\epsilon} = -\frac{\sigma_b}{\epsilon}$

 

[kuruman]

I think I know (b), but I'm really not sure about (a), since there is not example, I'm trying to guess and as you see I'm really bad.

However, I was thinking about the example with the curved plane and the pillbox as the Gauss' surface.
Maybe I'll sound stupid one again, but the top of the sphere and the curved plane seems the same can I use a pillbox for the Gauss' surface and I can pick A to be really small and the sides goes to 0. Thus, I have $\vec{E}^{\perp}_{above} - \vec{E}^{\perp}_{below} = \frac{\sigma}{\epsilon_0} \hat{n} - \frac{\sigma}{\epsilon} \hat{n}$However, $\vec{E} = \frac{\vec{P}}{-\epsilon} = -\frac{\sigma_b}{\epsilon}$

Look in Griffiths chapter 4 for (a). Look carefully. What else did you see?
You can use a very small pillbox for a Gaussian surface, so small that the surface of the sphere that it encloses is flat, just the floor of the room you' re in which is on the surface of the spherical Earth. Put it at some arbitrary angle $\theta$ relative to the z-axis and definitely not along the z-axis.

 

[happyparticle]

The only thing I can see is that $\vec{E} = \frac{\sigma}{\epsilon}$ inside the sphere and $\vec{E} = \frac{\sigma}{\epsilon_0}$ outside. I have to say that I thought I was close in post #23.

 

[kuruman]

The only thing I can see is that $\vec{E} = \frac{\sigma}{\epsilon}$ inside the sphere and $\vec{E} = \frac{\sigma}{\epsilon_0}$ outside. I have to say that I thought I was close in post #23.

You couldn't have seen that. The left-hand sides of these equations are vectors and the right-hand sides are scalars. This book dos not have typos like that.

Look more closely. Use the index in the back of the book. What is the title of this thread? Maybe there is something about it in the book. Maybe you will find something about dielectric spheres in the book or if you visit the web.

 

[happyparticle]

I don't know if you have the book, but page 185, you can see $E_{above}$ bold for a vector and on the same page $E_{above}$ not bold. I'm confuse about that aswell.

I don't know what I'm looking for. Which part is wrong? does all I did is wrong?

 

[kuruman]

I don't know if you have the book, but page 185, you can see $E_{above}$ bold for a vector and on the same page $E_{above}$ not bold. I'm confuse about that aswell.

You are confused because you did not interpret correctly what you saw. Griffiths does exactly have "$E_{above}$ bold for a vector and on the same page $E_{above}$ not bold." He has $\mathbf{E}_{above}^{\parallel}$ bold for a vector and on the same page $E_{above}^{\perp}$ not bold. That is consistent with vectors being boldface. At a given point on a surface, the tangent can be in many directions and needs a vector to be specified whereas there is only outward normal direction to the surface.

 

[happyparticle]

It makes sense, but I didn't think about that. Now I know. There a lot of thing that the author left without explanation and I guess in my case I need explanation for every detail. I mean it's perfectly logic, but I didn't see that. I don't blame the author.

 

[happyparticle]

$\vec{E} = -\frac{\sigma}{\epsilon} \hat{r}$ inside the sphere
$\vec{E} = -\frac{\sigma}{\epsilon_0} \hat{r}$ outside the sphere

I don't even know if I'm looking for the right thing, but I guess this is correct.

 

[Charles Link]

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.

 

[happyparticle]

Does it means that $\epsilon \vec{E} + \vec{P} =\vec{D} = 0$
$\vec{E} = - \frac{\vec{P}}{\epsilon}$
$\sigma_b = \vec{P} \cdot {\hat{n}}$ is wrong?

 

[Charles Link]

$\sigma_p=\vec{P} \cdot \hat{n}$.
The solution inside the sphere is simple and really very well-known $\vec{E}=-\vec{P}/(3 \epsilon_o)$.
For this problem of spontaneous $P$, there is no $\epsilon$ that is of any use.

Meanwhile $\vec{D}=\epsilon_o \vec{E}+ \vec{P}$.

 

[happyparticle]

I really don't understand.
Why is $\vec{E} = - \frac{\vec{P}}{\epsilon0}$ doesn't work?
That what is used in my textbook to find $\vec{E}$
Why is it $\epsilon_0$ inside the sphere even if the sphere is a dielectric ($\epsilon$)

I'm also really confuse with boundary now.

 

[Charles Link]

For linear materials, $\vec{P}=\epsilon_o \chi \vec{E}$, so that $\vec{D}=\epsilon_o \vec{E}+\vec{P}=\epsilon_o (1+\chi) \vec{E}=\epsilon \vec{E}$.

For the case at hand, the $\vec{P}$ occurs spontaneously, independent of any electric field. The case of linear materials with an $\epsilon$ does not appply here.

 

[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.