Is there such thing as an evenly-charged sphere?

Let's begin with a very typical model seen in almost every electromagnetics textbook: a solid, evenly-charged sphere.
First it cannot be a conducting sphere because charge will automatically distribute itself upon the surface of a conductor. So this sphere must be made of something dielectric. Suppose its relative dielectric constant is $\epsilon_r$ which should be greater than 1.
Now suppose the sphere has a volume charge density $\hat{\rho}$ (the "hat" above $\rho$ indicates that it is not the density of free charge), and has a radius $R$. Then it is quite easy to give the electricity field in the space (and of course nothing else in the surroundings): $$\vec{E}(\vec{r})=\cases{{\frac{\hat{\rho}}{3\epsilon_0}\vec{r}}&r\in[0,R)\\{\frac{\hat{\rho}R^3}{3\epsilon_0 r^3}\vec{r}}&r\in[R,+\infty)}$$ where $\vec{r}$ is the radius vector from the center of the sphere.
Clearly, $\vec{E}(\vec{r})$ as a vector function is continuous on the points where $|\vec{r}|=R$, say, on the surface of this sphere. But today I was reading about theories concerning the dielectric medium. One of them just caught me eye, which is called "the boundary relationship" for $\vec{D}$ and $\vec{E}$ on the interface between two different dielectric media. It tells that if the $\vec{E}$ line passes throuh the interface then it must satisfy that $$\frac{{E_{1n}}}{{E_{2n}}}=\frac{\epsilon_2}{\epsilon_1}$$ where, extremely close to the interface, ${E_{1n}}$ and ${E_{2n}}$ (of the same field line) are respectively the normal component of field strengths in medium 1 and medium 2, whose dielectric constants are respectively $\epsilon_1$ and $\epsilon_2$.

Now that the sphere is dielectric (media 1) whose relative dielectric constant is $\epsilon_r$ and it is surrounded by vacuum (media 2) whose relative dielectric constant is 1, therefore, from the above theory (which is rather properly proved in the textbook but not to be detailed in this post), we can conclude that near the interface, the exterior field strength should be stronger than the interior field strength. But this would obviously contradict the fact that the field strength function is continuous on every point on the interface, which is to say, near the interface, the interior field strength should always be equal to the exterior field strength.
To me neither of the two facts can be proved wrong, so here is my doubt : can such an "evenly-charged" dielectric sphere exist at all?

• Nitpicking: Technically in proper science terminology, that's a ball, not a sphere. E.g. conductors have surface (not bulk) charges. Jan 9 '15 at 18:57
• @Qmechanic sorry that I don't quite follow you. Do you mean that I should leave out the "surface area" of the dielectric sphere (or ball)? But since it is evenly charged everywhere, I guess even removing the surface area from our consideration will not make a difference.
– Vim
Jan 10 '15 at 2:28
• @Qmechanic Or maybe it's a misuse of English language? Well, honestly it's not my mother tongue so I'm perhaps ignorant of some subtle things in it like the difference between a "sphere" and a "ball".
– Vim
Jan 10 '15 at 2:33

For this simple example, you're assuming that the polarization of the material is zero, and therefore, it's dielectric constant is $1$, just like for vacuum. It's true that real materials initially with charge density $\rho$ would self-polarize, and would pick up a surface charge density.

The self-polarization would mess up the simple linear relationship of the field to the radius and make the whole thing a differential equation to solve (since the field at $r$ would depend on the field generated by the charge interior to the shell at radius $r$), and this would be too complicated for simple examples to demonstrate Gauss's law, so it's left out in introduction classes.

• Thank you but I don't think the polarization of the material is zero. In my textbook there is a relation between polarization strength $\vec{P}$ and electric field strength $\vec{E}$ which is $$\vec{P}=\epsilon_0 \chi\vec{E}$$, since $\vec{E}$ is not zero except in the center then $\vec{P}$ won't be zero either, I think.<br/> and do you mean that actually the ball will pick up extra surface charge induced by self-polarization, which breaks the even distribution? So can I tell that such an "even-ly charged" ball doesn't exist at all, in real life?
– Vim
Jan 10 '15 at 2:38
• @Vim I'm saying, that yes, for real materials, $\chi$ will be nonzero, but for the simple Gauss's law example you described, it is set to zero for simplicity. But yes, if the polarization is nonzero, then you will break the even distribution, because, at minimum, you'll pick up a surface charge. But you can still conceptually think of a charged distribution with $\chi=0$, and this will satisfy the simple linear equation you have in your question statement. Jan 10 '15 at 4:11
• Another way of seeing the discontinuity is writing the solution as: $$\vec{E}(\vec{r})=\cases{{\frac{\hat{\rho}}{3\chi\epsilon_0}\vec{r}}&r\in[0,R)\\{\frac{\hat{\rho}R^3}{3\epsilon_0 r^3}\vec{r}}&r\in[R,+\infty)}$$ Where the discontinuity shows up because vacuum has zero polarization, and you no longer have the fields equal at the boundary. Jan 10 '15 at 4:13
• Thanks for your solution, but I'm still confused about your words that the polarization should be zero in my conceptual model. There seems to be an inevitable contradiction: if the polarization within the ball must be set to zero, then where does the net charge come from?
– Vim
Jan 10 '15 at 5:55
• Oh I see, it can not self- induce the net charge. That's a horrible mistake I've made in my model. you're right. The net charge should come from something else than this ball.
– Vim
Jan 10 '15 at 13:19

Gauss' law for dielectric materials is (takes into account charge displacement):

$$\int_{Volume} \rho_{free} dV = \int_{Area} \epsilon_0 \epsilon \vec{E} d\vec{S}$$

Taking a sphere concentric with the ball and with radius $r < R$, we find the electric field at $r$ from the centre:

$$\frac{4}{3}\pi r^3 \rho_{free} = 4 \pi r^2 \epsilon \epsilon_0 E \Rightarrow$$ $$E = \frac{\rho_{free} r}{3 \epsilon_0 \epsilon}$$

However, taking the "true" Gauss' law, where taking charge displacement into account is to be done manually, we find (the result from the question):

$$E = \frac{(\rho_{free} + \rho_{avg.disp})r}{3\epsilon_0}$$

Equating last two formulas give us:

$$\frac{\rho_{free}}{\epsilon} = \rho_{free} + \rho_{avg.disp} \Rightarrow \rho_{avg.disp} = const$$

As $\rho_{avg.disp}$ does not vary with $r$, the displacement charge inside the sphere is really homogeneous ($\rho_{avg.disp} = \rho_{disp}$ everywhere inside). The induced "bump"-like surface charge can directly be compensated by adding exactly the opposite amount of charge to the surface.

• I'm afraid, however, that in the LHS of your first equation, $\rho_{given}$ should be the volume density of the free charge excluding the polarized charge enclosed by the Gaussian surface.
– Vim
Jan 10 '15 at 2:52
• @Vim: Thanks for pointing out. I meant that each layer of $d$ would be given an additional amount of charge $\rho_{given} S$. In this case, $\rho_{given}$ would really have coincided with the definition of free charge. However reading the text again showed that I said something else. I changed $\rho_{given}$ to $\rho_{free}$ now. Jan 10 '15 at 11:21