Electric Field Inside a Dielectric-Filled Spherical Capacitor

[BOAS]

Homework Statement

A spherical capacitor consists of two concentric spherical conductors of radii $R_{1}$ and $R_2, (R_2 > R_1)$. The space between the two conductors is filled with a linear inhomogeneous dielectric whose relative permittivity varies with the distance $r$ from the centre of the spheres as $ε_r(r) = (c + r)/r$, with $c$ a constant. The inner sphere carries a total charge $q$ and the outer conductor is grounded.

Using Gauss’s law in dielectrics, compute the electric field (direction and magnitude) at a distance $R_1 < r < R_2$ from the centre of the spheres.

Homework Equations

The Attempt at a Solution

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I think the charge on the inner sphere $q$, can be considered the free charge.

Gauss' law in dielectrics;
$\oint \vec D . d\vec a = q$

I don't know the polarisation vector.
$\vec D = \frac{q}{4 \pi R_1^2} \vec r = \epsilon_0 \vec E + \vec P$$\vec D = \epsilon_r \epsilon_0 \vec E$

where $\epsilon_r = \frac{\epsilon}{\epsilon_0}$

$\vec E = \frac{\vec D}{\epsilon_r \epsilon_0}$

I think that this expression gives me the electric field inside the dielectric, but I am concerned that I have not considered the effect of the grounded outer shell.

Do I need to compute the electric displacement inside the outer shell due to the induced charge on it, and the field inside is the linear super position of the two?

 

[rude man]

What does Gauss's law really say? Does it care about anything ouside the Gaussian surface?

 

[BOAS]

What does Gauss's law really say? Does it care about anything ouside the Gaussian surface?

Ah, I didn't think of that.

Gauss's law in a dielectric says that the flux through a closed surface is equal to the enclosed free charge. My argument seems reasonable in light of this. Although I notice my choice of gaussian surface (a sphere of radius $R_1$) should really be a sphere of radius $r , R_1 < r < R_2$

 

[rude man]

Ah, I didn't think of that.

Gauss's law in a dielectric says that the flux through a closed surface is equal to the enclosed free charge. My argument seems reasonable in light of this. Although I notice my choice of gaussian surface (a sphere of radius $R_1$) should really be a sphere of radius $r , R_1 < r < R_2$

Good! Finish and show your work if you care to.

 

[BOAS]

Good! Finish and show your work if you care to.

$\oint \vec D . d\vec a = Q_{fencl}$

I choose a spherical gaussian surface of radius $r, (R_{1} < r < R_{2})$.

$D(4 \pi r^2) = Q_{fencl}$

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

*this step seems fishy to me* $Q_{fencl} = q$ Am I justified in saying that this is the case?

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

Since I do not know the polarisation vector $\vec D = \vec E \epsilon_0 + \vec P$, I can rearrange this to get $\vec D = \epsilon \vec E$ where $\epsilon = \epsilon_r \epsilon_0$

The electric field is therefore $\vec E = \frac{\vec D}{\epsilon_0 \epsilon_r}$

$\epsilon_r \epsilon_0 = \frac{\epsilon_0 (c + r)}{r}$

$\vec E = \frac{q}{4 \pi r^2} \frac{r}{\epsilon_0 (c + r)} \hat r = \frac{q}{4 \pi \epsilon_0 (cr + r^2)} \hat r$ for $r, (R_{1} < r < R_{2})$

 

[rude man]

$\oint \vec D . d\vec a = Q_{fencl}$

I choose a spherical gaussian surface of radius $r, (R_{1} < r < R_{2})$.
$D(4 \pi r^2) = Q_{fencl}$
$\vec D = \frac{Q_{fencl}}{4 \pi r^{2}} \hat r$
*this step seems fishy to me* $Q_{fencl} = q$ Am I justified in saying that this is the case?

Why fishy? q is the total free charge within the gaussian surface, is it not?
The rest looks fine! You might, strictly for the sake of elegance, have moved the r out of the denominator parenthesis.

 

[BOAS]

Why fishy? q is the total free charge within the gaussian surface, is it not?
The rest looks fine! You might, strictly for the sake of elegance, have moved the r out of the denominator parenthesis.

The question states that the conductor carries a total charge of q, which I'm imagining as having been added to an electrically neutral conductor. Why is it ok to ignore the electrons in the conduction band of the metal? Is it a case of free charge residing on the surface, and these electrons clearly don't meet that criteria?

Great, thank you.

 

[rude man]

The question states that the conductor carries a total charge of q, which I'm imagining as having been added to an electrically neutral conductor. Why is it ok to ignore the electrons in the conduction band of the metal? Is it a case of free charge residing on the surface, and these electrons clearly don't meet that criteria?

Because the conduction-band electrons are equal in number to the positive-charge "holes" left behind, leaving a net charge of q only inside the gaussian surface. It's the net charge that counts.

BTW the concept of "holes" is a matter for quantum mechnanics. Holes are not merely positively charged ions: https://en.wikipedia.org/wiki/Electron_hole

 

[BOAS]

Because the conduction-band electrons are equal in number to the positive-charge "holes" left behind, leaving a net charge of q only inside the gaussian surface. It's the net charge that counts.

BTW the concept of "holes" is a matter for quantum mechnanics. Holes are not merely positively charged ions: https://en.wikipedia.org/wiki/Electron_hole

Ok, thanks.

I have met holes in the various solid state physics experiments I've done, but haven't taken a QM course yet.