Calculating permittivity in a constant field

[Argelium]

Homework Statement

Having a conducting sphere with radius $R$ and charge $Q$, dielectric is put on it so that a spherical shell with inner radius $R$ and outer radius $3R$ is formed. Calculate:

1. Electric permitivitty $\epsilon$ such that $E(r), R<r<3R$ is constant and there's no polarisation charge on the surface of radius $3R$.

Homework Equations

$$\iint \vec{D} \cdot d\vec{S} = Q_{free}$$
$$\iint \vec{P} \cdot d\vec{S} = -Q_{bound}$$

$$\vec{D} = \epsilon_0 \vec{E} + \vec{P}$$

The Attempt at a Solution

So, if there's no polarisation charge we have that on $r=3R$:

$$\iint \vec{P} \cdot d\vec{S} = -Q_{bound} = 0$$

And then $\vec{P} = 0$ and:[/B]

$$D = \epsilon_0\frac{Q}{4\pi\epsilon (3R)^2} $$

Since we want it constant, then for $R<r<3R$:

$$D(r) = \frac{Q}{4\pi(3R)^2} = \epsilon_0\frac{Q}{4\pi\epsilon r^2}$$

And then

$$\epsilon (r) = \frac{\epsilon_0}{(3Rr)^2} $$

However I'm not sure, cause $P$ is not supposed to be 0 in that zone. But if it isn't, how do i calculate $P$?

 

[Charles Link]

This problem makes little sense. One question is in your first sentence, do you mean $9R$ or $3R$? $\\$ There is no way to make the electric field constant in the dielectric, as far as I know, when you have a spherically symmetric geometry, unless it is zero. And the only way to achieve zero field in the dielectric is to make it a conductor, and then the charge $Q$ will move to the surface at $r=3R$. $\\$Note: If the electric field is constant in the dielectric, that means $\nabla \cdot E=0$, so that there will be zero polarization charge density in the dielectric. The electric field in the dielectric will necessarily have the form $E=\frac{A}{r^2}$. $\\$ If the dielectric has infinite dielectric constant, the electric field will then be zero, the charge $Q$ will be canceled by $-Q$ coming from the conductive dielectric that was added, and a charge $Q$ then goes to the surface at $r=3R$. For this case of a dielectric with dielectric constant $\epsilon_r = +\infty$, it is then a matter of preference whether you call the charge $Q$ at $r=3R$ polarization charge or free charge. $\\$ And if you want to compute $P$ from zero electric field and infinite $\chi$, you can compute it at either surface by using $\sigma_p=\vec{P} \cdot \hat{n}$, (you will different answers at $r =R$ and $r=3R$), and can assume it drops off as an inverse square in the conductive dielectric, so that you have continuity and also have $-\nabla \cdot \vec{P}=\rho_p=0$.

 

[Argelium]

This problem makes little sense. One question is in your first sentence, do you mean $9R$ or $3R$? $\\$ There is no way to make the electric field constant in the dielectric, as far as I know, when you have a spherically symmetric geometry, unless it is zero. And the only way to achieve zero field in the dielectric is to make it a conductor, and then the charge $Q$ will move to the surface at $r=3R$. $\\$Note: If the electric field is constant in the dielectric, that means $\nabla \cdot E=0$, so that there will be zero polarization charge density in the dielectric. The electric field in the dielectric will necessarily have the form $E=\frac{A}{r^2}$. $\\$ If the dielectric has infinite dielectric constant, the electric field will then be zero, the charge $Q$ will be canceled by $-Q$ coming from the conductive dielectric that was added, and a charge $Q$ then goes to the surface at $r=3R$. For this case of a dielectric with dielectric constant $\epsilon_r = +\infty$, it is then a matter of preference whether you call the charge $Q$ at $r=3R$ polarization charge or free charge.

Why is this not possible? Is it due to the dipoles generated in the dielectric?

So, if it were possible for the field to be constant in the dielectric, it's redundant to ask for no polarisation charge?

 

[Charles Link]

Why is this not possible? Is it due to the dipoles generated in the dielectric?

So, if it were possible for the field to be constant in the dielectric, it's redundant to ask for no polarisation charge?

Again, the only way to get constant electric field in the dielectric for this spherically symmetric geometry is that it is zero. And the only case that occurs is for a "dielectric" that is basically a conductor.$\\$ If that's the answer they were looking for, then I think I got it right. I've got a feeling they had something else in mind, and that perhaps they made a mistake when they formulated the problem. It is also possible that I have erred somewhere in my logic, but I think I answered this one correctly.

 

[Argelium]

Again, the only way to get constant electric field in the dielectric for this spherically symmetric geometry is that it is zero. And the only case that occurs is for a "dielectric" that is basically a conductor.$\\$ If that's the answer they were looking for, then I think I got it right. I've got a feeling they had something else in mind, and that perhaps they made a mistake when they formulated the problem. It is also possible that I have erred somewhere in my logic, but I think I answered this one correctly.

Makes sense to me, I suppose. Thanks anyway for the help!

 

[TSny]

You have the general idea that you need to find $\epsilon (r)$ such that $E(r)$ is constant in the region of the dielectric. But I don't follow your work. You have

So, if there's no polarisation charge we have that on $r=3R$:

$$\iint \vec{P} \cdot d\vec{S} = -Q_{bound} = 0$$

OK, This says that the total surface charge is zero at $r = 3R$. (I don't think there should be a minus sign in front of $Q_{bound}$.)

And then $\vec{P} = 0$

You want $P = 0$ at $r = 3R$ so that there is no surface charge density on the outer surface of the dielectric. But $P$ is not going to be zero throughout the dielectric.

and:

$$D = \epsilon_0\frac{Q}{4\pi\epsilon (3R)^2}$$Since we want it constant, then for $R<r<3R$:

$$D(r) = \frac{Q}{4\pi(3R)^2} = \epsilon_0\frac{Q}{4\pi\epsilon r^2}$$

I'm not following this. It is not $D$ that you want to be constant inside the dielectric, it's $E$ that you want to be constant. (There is no way for $D$ to be constant in this region.)

And then

$$\epsilon (r) = \frac{\epsilon_0}{(3Rr)^2} $$

Shouldn't $\epsilon (r)$ have the same dimensions as $\epsilon_0$?

However I'm not sure, cause $P$ is not supposed to be 0 in that zone. But if it isn't, how do i calculate $P$?

Once you find the correct expression for $\epsilon (r)$ such that $E$ is constant inside the dielectric, you can find $P(r)$ from the basic relations between $P, D$, and $E$.

 

[Argelium]

You have the general idea that you need to find $\epsilon (r)$ such that $E(r)$ is constant in the region of the dielectric. But I don't follow your work. You have OK, This says that the total surface charge is zero at $r = 3R$. (I don't think there should be a minus sign in front of $Q_{bound}$.)

You want $P = 0$ at $r = 3R$ so that there is no surface charge density on the outer surface of the dielectric. But $P$ is not going to be zero throughout the dielectric.

I'm not following this. It is not $D$ that you want to be constant inside the dielectric, it's $E$ that you want to be constant. (There is no way for $D$ to be constant in this region.)

And then

Shouldn't $\epsilon (r)$ have the same dimensions as $\epsilon_0$?

Once you find the correct expression for $\epsilon (r)$ such that $E$ is constant inside the dielectric, you can find $P(r)$ from the basic relations between $P, D$, and $E$.

So, what about the other poster who said there could be no constant electric field?

Well, I thought $D$ would be constant since $D=\epsilon_0 E$ which is supposed to be constant. I suppose it doesn't hold in this case? Where can I start then?

 

[TSny]

Where can I start then?

What is $D$ as a function of r inside the dielectric? For a given $\epsilon (r)$ of the dielectric constant as a function of $r$, how can you get $E(r)$ inside the dielectric from $D(r)$?

 

[Charles Link]

@Argelium It's kind of a clumsily worded problem in that I think they are wanting you to conclude/show that the dielectric must in fact be a conductor to get the result they are asking for, in which case the electric field inside the material is not only constant, it is also equal to zero.

 

[TSny]

@Argelium It's kind of a clumsily worded problem in that I think they are wanting you to conclude/show that the dielectric must in fact be a conductor to get the result they are asking for, in which case the electric field inside the material is not only constant, it is also equal to zero.

I think there's a solution for a dielectric material (not a conductor). Hope I'm not missing something. If I am missing something, then it will be revealed as I try to guide the OP.

 

[Charles Link]

@TSny : The OP @Argelium posted a homework just this past week that involved very similar concepts, so it made it easier to solve this exercise. See: https://www.physicsforums.com/threa...t-voltage-and-dielectric.943991/#post-5973268

 

[TSny]

@TSny : The OP @Argelium posted a homework just this past week that involved very similar concepts, so it made it easier to solve this exercise. See: https://www.physicsforums.com/threa...t-voltage-and-dielectric.943991/#post-5973268

In that problem, the permittivity $\epsilon$ of the dielectric is assumed to be a constant. In this problem, I am assuming that you can allow $\epsilon$ to be a function of r. I agree that for a constant $\epsilon$ there is no solution.

 

[Charles Link]

In that problem, the permittivity $\epsilon$ of the dielectric is assumed to be a constant. In this problem, I am assuming that you can allow $\epsilon$ to be a function of r. I agree that for a constant $\epsilon$ there is no solution.

Perhaps mathematically that would work, but how would you ever create a material where you can control the dielectric constant as a function of radius? $\\$ I think they might be looking for the answer I came up with in post 2. The alternative is that they made a mistake when they came up with this problem.

 

[TSny]

The problem states that you want $E(r)$ to be a constant inside the dielectric. I interpret this to mean that the magnitude of $\vec E$ is constant inside the dielectric. But the E-field is nevertheless radial. So, the direction of $\vec E$ is not constant.

That is, $\vec E = E_0 \hat r$ for some constant $E_0$ (to be determined). So,$\vec \nabla \cdot \vec E \neq 0$ inside the dielectric (as you assumed in post #2).

I don't know how one would manufacture a dielectric with a specific spatially varying dielectric constant.
[Edit: Here's an example of an academic exercise with a non-constant permittivity: http://www.chegg.com/homework-help/questions-and-answers/dielectric-sphere-radius-b-spatially-varying-dielectric-constant-volume-charge-density-sho-q18264624]

 

[Charles Link]

I don't know how one would manufacture a dielectric with a specific spatially varying dielectric constant.

In general, the dielectric "constant" of a material is assumed to be constant in these electrostatic E&M problems.

 

[TSny]

"Spatially varying dielectric constant" does sound like an oxymoron . "Spatially varying permittivity" might be a better way to say it. Anyway, we agree that if the problem intends the permittivity to have the same value throughout the dielectric, then there is no solution. But, as you pointed out, you could replace the dielectric shell with a conducting shell with a small insulating gap between the shell and the inner sphere. Then you would need to place a nonzero net charge on the shell to satisfy the condition of no surface charge density on the outer surface of the shell.

 

[Argelium]

Wow, so I don't know, I'll ask my prof what he meant later in the day, but in the meanwhile

$\vec{D}=\epsilon_0\vec{E}+\vec{P}$

So in $R$ and $3R$, polarisation is zero, and we can solve for $\epsilon$ in both those points (boundary conditions)

Then we know that $\epsilon(r)r^2=c$ differentiating we obtain a differential equation

I have yet to solve it, but I'm open to suggestions

 

[Charles Link]

@Argelium My guess is he is looking for the solution that the dielectric is a conductor. And if that is what he wants for an answer, this one is really a much simpler problem than the previous one, where it took a fair amount of mathematics to arrive at the answer.

 

[TSny]

Wow, so I don't know, I'll ask my prof what he meant later in the day, but in the meanwhile

$\vec{D}=\epsilon_0\vec{E}+\vec{P}$

So in $R$ and $3R$, polarisation is zero, and we can solve for $\epsilon$ in both those points (boundary conditions)

$P$ will not be zero except at the outer surface (r = 3R).

Don't forget the basic relation $E(r) = \frac{1}{\epsilon (r)}D(r)$.

So, if you know $D(r)$ you can express $E(r)$ in terms of $\epsilon (r)$ and and $D(r)$.

$D(r)$ is easy to get from Gauss' law for D.

 

[Argelium]

$P$ will not be zero except at the outer surface (r = 3R).

Don't forget the basic relation $E(r) = \frac{1}{\epsilon (r)}D(r)$.

So, if you know $D(r)$ you can express $E(r)$ in terms of $\epsilon (r)$ and and $D(r)$.

$D(r)$ is easy to get from Gauss' law for D.

But at $r=R$ there's no bound charge is it?

Also, I thought the relation you just proposed is just valid in linear dielectrics which this clearly isn't, otherwise the oermittiperm would be constant

 

[Charles Link]

But at $r=R$ there's no bound charge is it?

You can treat a dielectric with $\epsilon_r \rightarrow +\infty$ , (which makes it essentially a conductor), mathematically as having bound polarization charge. That puts bound polarization charge $-Q$ at $r=R$, and also a charge $+Q$ at $r=3 R$. It seems to be a matter of choice whether you call the charge at $r=3 R$ free charge or bound polarization charge for the case $\epsilon_r \rightarrow +\infty$.

 

[TSny]

But at $r=R$ there's no bound charge is it?

There is a bound surface-charge density at r = R. Also, everywhere inside the dielectric, there is a nonzero, bound, volume-charge density that varies with r. But you don't need to worry about these bound charges in solving the problem.

Also, I thought the relation you just proposed is just valid in linear dielectrics which this clearly isn't, otherwise the oermittiperm would be constant

"Linear" means that at any point of the dielectric the polarization is proportional to the electric field at that point. This in turn implies that the D field at that point is proportional to the E field at that point. The proportionality constant between D and E is the dielectric constant (or permittivity) at that point. But the value of the dielectric constant can vary from one point to another.

 

[Charles Link]

For these problems, $D$ is really just a convenient mathematical construction that simplifies the mathematics in some cases. (The same is the case with the $H$ that gets introduced into magnetostatic calculations). $\\$ The actual physics really depends upon looking at the polarization function $\vec{P}$, where $-\nabla \cdot \vec{P}=\rho_p$, where $\rho_p$ is the polarization charge density that behaves just like free electric charges, in that it is a source for the electric field $\vec{E}$, so that $\nabla \cdot \vec{E}=\frac{\rho_{total}}{\epsilon_o}=\frac{\rho_{free}+\rho_p }{\epsilon_o}$, with $\rho_{free}$ and $\rho_p$ treated on equal footing. $\\$ (You will also have the result, (that comes from $-\nabla \cdot \vec{P}=\rho_p$), that the surface polarization charge density $\sigma_p=\vec{P} \cdot \hat{n}$ at an interface of the dielectric with the vacuum). $\\$ You also have the linear equation $\vec{P}= \epsilon_o \chi \vec{E}$, where $\chi$ is the polarizability of the material. $\\$ The above gives the result that, with $\vec{D}$ defined as $\vec{D}=\epsilon_o \vec{E}+\vec{P}$, that $\vec{D}=\epsilon \vec{E}=\epsilon_o \epsilon_r \vec{E}$ where $\epsilon_r=1 +\chi$, and also that $\nabla \cdot \vec{D}=\rho_{free}$. $\\$ The real physics though is contained in the equation $\vec{P}=\epsilon_o \chi \vec{E}$. The introduction of the electric displacement vector $\vec{D}$ and its associated equations is really just introducing a mathematical construction that simplifies the mathematics in some cases, but it doesn't change any of the physics.

 

[TSny]

The introduction of the electric displacement vector $\vec{D}$ and its associated equations is really just introducing a mathematical construction that simplifies the mathematics in some cases, but it doesn't change any of the physics.

Yes. Working with D in this problem is very helpful. It is easy to find D everywhere inside the dielectric. Then you can use D = εE to see how ε must vary with r so that E is constant inside the dielectric.

 

[Argelium]

Alright, so talked to the prof and used used @TSny hints and arrived at this solution:

The electric field at $r$ is as follows:

$$\vec{E}=\frac{Q}{4\pi\epsilon r^2} \hat{r}$$

Using Gauss' Law the displacement at $r$ is

$$\vec{D}=\frac{Q}{4\pi r^2}\hat{r}$$

Now, if $E$ is constant, we have that $\epsilon r = k$ and then:

$$\epsilon_r = \frac{k}{\epsilon_0 r^2} $$

Now to determine the value of $k$ consider, $\vec{P}(3R) = \epsilon_0 (\epsilon_r-1)\frac{Q}{4\pi\epsilon(3R)^2} = 0$ So the value of $k$ is:$$A = \epsilon_0(3R)^2$$

And then

$$\epsilon = \frac{\epsilon_0(3R)^2}{r^2}$$

So, as it turned out it wasn't really a conductor (I think).

 

[Charles Link]

And $\chi=\epsilon_r-1$ is non-zero at $r=3R$. This means with the non-zero $\vec{ E}$ that $\vec{P}$ must also be non-zero at $r=3R$, since $\vec{P}=\epsilon_o \chi \vec{E}$. This does not meet the specifications that $\sigma_p=\vec{P} \cdot \hat{n}=0$ of the original problem statement. $\\$ And the additional result is that we require a material that has a specified permittivity that depends on radius in a very precise manner in order to keep the electric field constant in a spherically symmetric geometry, and still have $\nabla \cdot \vec{D}=0$. (Materials with constant $\epsilon_r$ are relatively easy to manufacture. This material with its permittivity function is very hypothetical and could never be manufactured). $\\$ Perhaps an interesting mathematical exercise, but the requirements of the original problem statement are still not met.

 

[Argelium]

And $\chi=\epsilon_r-1$ is non-zero at $r=3R$. This means with the non-zero $\vec{ E}$ that $\vec{P}$ must also be non-zero at $r=3R$, since $\vec{P}=\epsilon_o \chi \vec{E}$. This does not meet the specifications that $\sigma_p=\vec{P} \cdot \hat{n}=0$ of the original problem statement. $\\$ And the additional result is that we require a material that has a specified permittivity that depends on radius in a very precise manner in order to keep the electric field constant in a spherically symmetric geometry, and still have $\nabla \cdot \vec{D}=0$. Perhaps an interesting mathematical exercise, but the original requirements of the problem are still not met.

About the first part, are you sure? The polarization as it is is calculated as:

$$P = \frac{Q}{4(3R)^2}(\epsilon_0(1-\frac{3R}{r}^2))$$

At $r=3R$ polarisation is 0.

 

[Charles Link]

About the first part, are you sure? The polarization as it is is calculated as:

$$P = \frac{Q}{4(3R)^2}(\epsilon_0(1-\frac{3R}{r}^2))$$

At $r=3R$ polarisation is 0.

Let me study that further. Perhaps it meets that requirement. $\\$ Editing: Looks like it is correct after all. I stand corrected. There is indeed one parameter in the permittivity function that makes it possible to have $\chi=0$ at $r=3R$.

 

[Charles Link]

@Argelium See the Editing: comments of post 28. Looks like you have a good solution. :) Even if it would be difficult to manufacture, it is interesting. :)

 

[Argelium]

@Argelium See the Editing: comments of post 28. Looks like you have a good solution. :) Even if it would be difficult to manufacture, it is interesting. :)

Thanks for all the help! :)