Electric field of a charged dielectric sphere

[1v1Dota2RightMeow]

Homework Statement

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?

Homework Equations

$\int \vec{D} \cdot \vec{dA}=Q_{free,enc.}$

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.

 

[Charles Link]

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.

 

[1v1Dota2RightMeow]

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.

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.

 

[Charles Link]

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.

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.

 

[Charles Link]

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.

 

[1v1Dota2RightMeow]

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.

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?

 

[Charles Link]

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?

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.

 

[1v1Dota2RightMeow]

We see that any $P$ that we create will cause a $E_p$ that serves to reduce $E_o$.

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.

 

[Charles Link]

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.

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).

 

[Charles Link]

@1v1Dota2RightMeow I edited the last post=made an addition to it, so be sure and read the second paragraph in the above post.

 

[1v1Dota2RightMeow]

@1v1Dota2RightMeow I edited the last post=made an addition to it, so be sure and read the second paragraph in the above post.

Yup, I noticed. I'm reading everything you've said, including what you've linked. I want to fully understand this problem.

 

[Charles Link]

Yup, I noticed. I'm reading everything you've said, including what you've linked. I want to fully understand this problem.

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 $\sigma_p=Pcos(\theta)$.

 

[Charles Link]

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.

 

[Charles Link]

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.

 

[1v1Dota2RightMeow]

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 $E_{in}=-P_0/3\epsilon _0 cos \theta \hat r + sin \theta \hat \theta$

 

[1v1Dota2RightMeow]

Oh wait, I made a small mistake. Never mind.

 

[Charles Link]

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".

 

[1v1Dota2RightMeow]

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.

 

[Charles Link]

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.

 

[1v1Dota2RightMeow]

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.

 

[Charles Link]

Once you have a prospective potential function (one inside and one outside) with coefficients to be determined, you compute the electric field $E$ and also the $D$ field. $\int D \cdot dA=Q_{free}$ and $\int E \cdot dA=Q_{total}/\epsilon_o$ where $Q_{total}=Q_{free}+Q_{polarization}$ are the requirements. For a gaussian pillbox around the surface, these give requirements are the surface charge density. (evaluated at r=R). The free surface charge density is basically your boundary condition.

 

[1v1Dota2RightMeow]

For $Q_{total}$, I get $Q_{total} = 4\pi R^2 \sigma_0 cos \theta -\int_V \nabla \cdot \vec{P} dV = 4\pi R^2 \sigma_0 cos \theta - \int_V \rho _p dV$.

Here, I claim that $\rho_p$ is a constant and can be pulled out of the integral. So,

$\int \vec{E} \cdot \vec{dA} = 4\pi R^2 (\frac{\sigma_0 cos \theta - R \rho_p /3}{\epsilon_0})$

and therefore,

$E =\frac{R^2}{r^2} (\frac{\sigma_0 cos \theta - R \rho_p /3}{\epsilon_0})$, depending on if either $r<R$ or $r>R$.

For $r<R$, this should be equal to $-\nabla V_{in}$, so

$-\nabla V_{in} = -(\sum A_l l r^{l-1}P_l (cos \theta))\hat r - \frac{1}{r} (\sum A_l r^l \frac{d}{d\theta} (P_l (cos \theta)))\hat \theta$

$= -(0+A_1 cos \theta + A_2 r (3cos^2 \theta -1) +...)\hat r - (0+A_1 + A_2(3cos \theta) + ... )\hat \theta$

Am I going the right way?

 

[1v1Dota2RightMeow]

I also tried something else: if there is a defined surface charge, $\sigma_ = \sigma_0 cos\theta$, then I can say that

$V_{in,R}=V_{out, R} = V(R,\theta)$

$= \frac{1}{4\pi \epsilon_0} \int_0^{\pi} \int_0^{2\pi} \frac{\sigma_0 cos \theta}{|R-r'|} r'^2 sin \theta d\theta d\phi$

 

[Charles Link]

I can give you a couple of hints here: I just finished working it the long way: $\\$ Let $V_{in}=A \, r \, cos(\theta)$ and $V_{out}=B \, cos(\theta)/r^2$. The reasons for this should be apparent. $\\$ Compute the $E$'s and also the $D$'s using $E=-\nabla V$ (spherical coordinates) and $D=\epsilon E$ . Only one boundary equation with the free surface charge (using $D$) is needed. The other boundary condition to solve for $A$ and $B$ is $V_{in}(R,\theta)=V_{out}(R,\theta)$.i.e. the potential functions must be equal at r=R. The Legendre solutions that I picked should be suitable ones for the given boundary surface charge distribution. (They do in fact work). $\\$ I can help you with the boundary equation for $D$ if you get stuck. It uses $\int D \cdot \hat{n} dA=Q_{free}$ (edited) at the surface so that $D$ just outside the sphere and $D$ just inside the sphere (both at r=R) are needed to compute the flux of $D$. $\\$ Additional note: The second link that I provided above(post #5) is a good reference for selecting a potential function. The OP did a good job of writing out the complete Legendre terms that you can select from. editing... For this part I gave you the correct choice above. It should be apparent that the terms that were chosen have a good chance of working.

 

[Charles Link]

For the part of computing the flux with $D$ (just above and below the the surface at r=R), on the inside of the sphere $\hat{n}=-\hat{a_r}$ and outside $\hat{n}=\hat{a_r}$. Additional item, $dA=R^2sin(\theta)d \theta d \phi$, but this will appear on both sides of the equation, including on the right side where you have $\sigma_o cos(\theta) dA$.Hopefully you can work out the terms from $D \cdot \hat{n} dA$ on the left side. Once you have this equation, it is a simple matter to solve for $A$ and $B$, using also the equation $V_{in}(R)=V_{out}(R)$.

 

[1v1Dota2RightMeow]

I can give you a couple of hints here: I just finished working it the long way: $\\$ Let $V_{in}=A \, r \, cos(\theta)$ and $V_{out}=B \, cos(\theta)/r^2$. The reasons for this should be apparent. $\\$ Compute the $E$'s and also the $D$'s using $E=-\nabla V$ (spherical coordinates) and $D=\epsilon E$ . Only one boundary equation with the free surface charge (using $D$) is needed. The other boundary condition to solve for $A$ and $B$ is $V_{in}(R,\theta)=V_{out}(R,\theta)$.i.e. the potential functions must be equal at r=R. The Legendre solutions that I picked should be suitable ones for the given boundary surface charge distribution.

I think you've nailed my weakness. When we covered this in class, I didn't really understand boundary conditions/equations. But let's work through this!

I see what you're saying about $V_{in,R}=V_{out,R}$. That makes intuitive sense. But can I also say that $E_{in,R}=E_{out,R}$? The subscript $R$ just means "evaluated at $R$".

 

[Charles Link]

I think you've nailed my weakness. When we covered this in class, I didn't really understand boundary conditions/equations. But let's work through this!

I see what you're saying about $V_{in}=V_{out}$. That makes intuitive sense. But can I also say that $E_{in}=E_{out}$?

Not for $E$ because you have a charge at the surface: Your small gaussian box will contain a charge. It is easier to work it with $D$ for this case though. I actually also worked it also with $E$ (its trickier to do because you need to add the polarization charge) and I got the same equation as I did with the $D$. I then saw I needed a second equation and that was $V_{in}(R)=V_{out}(R)$. This second one gives you immediately that $A=B/R^3$. $\\$ Compute $E$ please using $E=-\nabla V_{in}$ and also for $V_{out}$ . Evaluate at r=R and also compute the $D$'s. The $E$ inside gets multiplied by $\epsilon$ and the $E$ outside gets multiplied by $\epsilon_o$ to get the $D$. Next, compute the flux of $D$ for an element dA at position $(R, \theta)$ (It's independent of $\phi$ ). Again $\int D \cdot \hat{n} dA=Q_{free}$, and in this case you do it for an infinitesimal elemental box that has faces just inside and just outside the sphere. Notice the $Q_{free}=\sigma_o cos(\theta) dA$. $\\$ Additional comment: It's a slightly complicated boundary condition when they give you the distribution of $\sigma_{free}$ as the boundary condition, but it's still quite workable. Once you get your equation using what is called the electric displacement vector $D$, you can solve for $A$ and $B$ and your solution for $V$ is complete. $\\$
(Notice in your OP that the potential is called $\phi_{in}$ and $\phi_{out}$, but $V$ is perhaps a better choice of letter.)

 

[1v1Dota2RightMeow]

Not for $E$ because you have a charge at the surface: Your small gaussian box will contain a charge. It is easier to work it with $D$ for this case though. I actually also worked it also with $E$ (its trickier to do because you need to add the polarization charge) and I got the same equation as I did with the $D$. I then saw I needed a second equation and that was $V_{in}(R)=V_{out}(R)$. This second one gives you immediately that $A=B/R^3$. $\\$ Compute $E$ please using $E=-\nabla V_{in}$ and also for $V_{out}$ . Evaluate at r=R and also compute the $D$'s. The $E$ inside gets multiplied by $\epsilon$ and the $E$ outside gets multiplied by $\epsilon_o$ to get the $D$. Next, compute the flux of $D$ for an element dA at position $(R, \theta)$ (It's independent of $\phi$ ). Again $\int D \cdot \hat{n} dA=Q_{free}$, and in this case you do it for an infinitesimal elemental box. Notice the $Q_{free}=\sigma_o cos(\theta) dA$.

Ok so this is what I got:

$E_{in,R}=-\nabla V_{in,R}=-A(cos \theta \hat r + sin \theta \hat \theta)$

$E_{out,R}= - \nabla V_{out,R}=B(\frac{2 cos \theta \hat r}{R^3} + \frac{sin \theta}{R^2} \hat \theta)$

$D_{in,R}=-A\epsilon (cos \theta \hat r + sin \theta \hat \theta)$

$D_{out,R}=B\epsilon_0 (\frac{2 cos \theta \hat r}{R^3} + \frac{sin \theta}{R^2} \hat \theta)$

Inside: $\int \vec{D} \cdot (-\hat a_r) dA = Q_{free}$

$\int \int A\epsilon cos \theta R^2 sin\theta d\theta d\phi = \int \int \sigma R^2 sin \theta cos \theta d\theta d\phi$

$A\epsilon R^2 (2\pi) \int cos\theta sin \theta d\theta = \sigma R^2 (2\pi) \int cos \theta sin \theta d\theta$

But the integrals on both sides evaluate to zero.

 

[Charles Link]

You have a couple of minor errors in your first two terms: You should have a (1/R^3) on the $E_{out}$ $\hat{a_{\theta}}$ term and you missed a sign on the $E_{in}$ $\hat{a_{\theta}}$ term. The total charge enclosed by this sphere is zero. (That is why the integrals you did gave you zero.) You want to evaluate the flux of $D$ for a small box of size $dA$ : I'm going to give you what you should get=please try to verify it: $\\$ $\\$ $\epsilon A cos(\theta) (R^2 sin(\theta) d \theta d \phi )+ \epsilon_o (+2B/R^3)cos(\theta) (R^2 sin(\theta) d \theta d \phi )=\sigma_o cos(\theta) (R^2 sin(\theta) d \theta d \phi )$ $\\$ Notice $dA=R^2 sin(\theta) d \theta d \phi$ cancels across. (We could have left it as $dA$ everywhere.) Quite a number of the factors will cancel and you get an equation of $A$ and $B$. $\\$ $\\$ Note: $E_{in}=-\hat{a_r} A cos(\theta)+\hat{a_{\theta}} A sin(\theta)$ and $E_{out}=\hat{a_r} 2 Bcos(\theta)/r^3 +\hat{a_{\theta}} B sin(\theta)/r^3$. (At least that's what I computed). The $\hat{a}_{\theta}$ term in the formula for the gradient in spherical coordinates has a (1/r) in it.

 

[Charles Link]

@1v1Dota2RightMeow Question for you: Do you see how the flux of $D \cdot dA$ gets evaluated for what is called a small gaussian pillbox of area $dA$ on the two biggest faces? This gives us an equation with $A$ and $B$. $\\$ An additional hint: You need to take $D_{in} \cdot \hat{n}_{in} dA+D_{out} \cdot \hat{n}_{out} dA=\sigma_o cos(\theta) dA$ . (Evaluated at $(R, \theta)$ ). The left side is the flux of $D$, and the right side is the free charge enclosed by this small pillbox. Note: $\hat{n}_{in}=-\hat{a}_r$ and $\hat{n}_{out}=\hat{a}_r$. $\\$ One question you might have is how do you determine what kind of boundary equation you need to apply? This one was somewhat complex, but the difference between $D_{in}$ and $D_{out}$ (flux)at r=R needs to give the free surface charge present at every location on the surface of the sphere. If there were no free surface charge, we would have $D_{in}=D_{out}$ everywhere at r=R. $\\$ One additional question you might have is how do we know that $V_{in}$ must equal $V_{out}$ at r=R? The reason is as we approach a layer of surface charge density, the electric field is finite. Therefore over an infinitesimal distance in crossing through the layer of surface charge, the change in potential must be infinitesimal or zero.

 

[1v1Dota2RightMeow]

Note: $E_{in}=-\hat{a_r} A cos(\theta)+\hat{a_{\theta}} A sin(\theta)$ and $E_{out}=\hat{a_r} 2 Bcos(\theta)/r^3 +\hat{a_{\theta}} B sin(\theta)/r^3$. (At least that's what I computed). The $\hat{a}_{\theta}$ term in the formula for the gradient in spherical coordinates has a (1/r) in it.

Wait I got something different. $E_{in} = -A (cos \theta (-\hat a_r) +sin \theta \hat a_{\theta})$ because in finding $E$ you take the negative of the gradient of $V$.

 

[Charles Link]

Wait I got something different. $E_{in} = -A (cos \theta (-\hat a_r) +sin \theta \hat a_{\theta})$ because in finding $E$ you take the negative of the gradient of $V$.

$V_{in}=Arcos(\theta)$ . This gives $E_{in}= -\nabla V_{in}=-Acos(\theta) \hat{a}_r +A sin(\theta) \hat{a}_{\theta}$. I do believe I have it correct. $\\$ One minor item of interest, but something you might find of interest is you had asked about a boundary condition involving $E$. Since $V_{in}=V_{out}$ everywhere at r=R, we can move parallel to the surface and come to a new location where again $V_{in}=V_{out}$. This means that the $\hat{a}_{\theta}$ components of E (just inside and just outside the sphere) will necessarily be equal at r=R. (Only the $\hat{a}_{\theta}$ components. ) Employing this requirement gives us the same thing as $V_{in}=V_{out}$: $A=B/R^3$. It's a somewhat minor detail, but you might find it of interest...

 

[1v1Dota2RightMeow]

$V_{in}=Arcos(\theta)$ . This gives $E_{in}= -\nabla V_{in}=-Acos(\theta) \hat{a}_r +A sin(\theta) \hat{a}_{\theta}$. I do believe I have it correct. $\\$ One minor item of interest, but something you might find of interest is you had asked about a boundary condition involving $E$. Since $V_{in}=V_{out}$ everywhere at r=R, we can move parallel to the surface and come to a new location where again $V_{in}=V_{out}$. This means that the $\hat{a}_{\theta}$ components of E (just inside and just outside the sphere) will necessarily be equal at r=R. (Only the $\hat{a}_{\theta}$ components. ) Employing this requirement gives us the same thing as $V_{in}=V_{out}$: $A=B/R^3$. It's a somewhat minor detail, but you might find it of interest...

Ok I see my mistake.

So I finished it and this is my final answer:

$V_{in}(r, \theta) = -\frac{\sigma_0 rcos \theta}{(2\epsilon_0 + \epsilon)}$ and

$V_{out}(r, \theta) = -\frac{-r \sigma_0}{(2 \epsilon_0 + \epsilon)}$.

The electric field inside is then $E_{in}=\frac{\sigma_0 ( cos \theta (-\hat a_r) - sin \theta \hat a_{\theta})}{2\epsilon_0 + \epsilon}$

Is that what you got?

 

[Charles Link]

Ok I see my mistake.

So I finished it and this is my final answer:

$V_{in}(r, \theta) = -\frac{\sigma_0 rcos \theta}{(2\epsilon_0 + \epsilon)}$ and

$V_{out}(r, \theta) = -\frac{-r \sigma_0}{(2 \epsilon_0 + \epsilon)}$.

The electric field inside is then $E_{in}=\frac{\sigma_0 ( cos \theta (-\hat a_r) - sin \theta \hat a_{\theta})}{2\epsilon_0 + \epsilon}$

Is that what you got?

I did get the $2 \epsilon_o+\epsilon$ in the denominator. I think you have a couple of corrections to make though. I got $A=\sigma_o/(2 \epsilon_o+\epsilon)$ and $B=AR^3$. The form of the $V_{out} =Bcos(\theta)/r^2$ will have a $cos(\theta)$ in the numerator and an $r^2$ in the denominator. Also I got "+" signs on both potential terms. It looks like a couple algebraic corrections will give you the answer... Also, notice the $E_{in}$ is actually uniform and points in the (minus) z-direction. A little vector algebra will show this. $\\$ You need to get the correct sign on $V_{in}$. (Note for $V_{in}$, the positive part of the free charge distribution is in the forward +z direction so that the potential will be positive at r=R for $\theta=0$.) Possibly the wrong sign on $V_{in}$ came from the wrong sign on $E=-\nabla V_{in}$ that was used in the boundary equation calculation. $\\$ Additional item: We can also compare this result to the $E_{in}$ that I computed in post #2 above. Just one simple way to show you that you did get the correct (-1/3) factor for a sphere: If you let $\epsilon=\epsilon_o$, you do get that $E_{in}=-(\sigma_o/(3 \epsilon_o)) \hat{z}$. It also agrees completely with the result in post #2 for $\epsilon=\epsilon_o (1+\chi)$ as a little additional algebra will show. $\\$ Additional comment: The couple of these Legendre problems that have appeared on Physics Forums, including yours, have been very good practice problems for solving. With a little practice, this method is starting to get a little easier.

 

[Charles Link]

And a follow-on: To summarize the results of post #2 above:
$E_i=E_o/(1+\chi/3)$ $\\$ $E_o=-((\sigma_o)/(3 \epsilon_o)) \hat{z}$ $\\$ $\epsilon=\epsilon_o (1+\chi)$ $\\$ A little algebra gives $E_i=-((\sigma_o/(2 \epsilon_o+\epsilon)) \hat{z}$ in complete agreement with the Legendre result.