Dielectric constant limit at infinity I have a question involving an electrically neutral dielectric cylinder of radius a with $\epsilon = \epsilon_r \epsilon_0$ placed in uniform background electric field $E_0\hat x$. Ive found in the limit $\epsilon \to \infty$ of the electric potential and field are:
$\phi(r, \varphi) =\left\{
     \begin{array}{lr}
       0& ,   r \le a\\
       E_0r\cos(\varphi)(\frac{a^2}{r^2}-1) & ,   r> a
     \end{array}
   \right.\\$
$E(r, \varphi) =\left\{
     \begin{array}{lr}
       0& ,   r \le a\\
       E_0[(1+2a^2+\frac{a^2}{r^2} )\cos(\varphi)\hat r -(1-\frac{a^2}{r^2})\sin(\varphi)\hat \varphi] & ,   r> a
     \end{array}
   \right.\\$
But I want to know what this means. I think it makes sense that inside the dielectric the potential and field go to zero since the dielectric is getting stronger? I'm not quite sure what is physically happening in the dielectric and I'm not quite sure what it means outside the cylinder. 
 A: First of all: the comment by @GeeJay is absolutely correct. I'd like to give some additional details though. Sure you know this page. So looking at the field for $\epsilon=2.5$ and $\epsilon=100000$, i.e. "almost infinity" you get

One interesting thing to notice is that the field inside is still homogeneous while it is highly deformed outside. This is due to the following fact. A homogeneous polarization in body with a surface that can be described by a  quadratic form (i.e. $\vec x  \mathbf A\vec x+\vec b \vec x+c=0$) results in a homogeneous field. So a cylinder, an elliptic cylinder, an ellipsoid, or even a hyperbolic paraboloid has this property. The superpositions of two homogeneous fields is again homogeneous. In case of the infinite cylinder symmetry makes everything much easier , as it is rather clear in which direction the polarization is pointing. From that it also clear that the surface charge must follow a $\cos \varphi$ law. With a little research you will figure out that this produces a dipole field (2D here!) outside and a homogeneous field inside. As a hint: have a look at multipole expansion. Finally to the $\epsilon \rightarrow \infty$. This means that, from the charging point of view, the material behaves like a perfect (not grounded) conductor (actually in case of an external homogeneous field there is no difference between grounded and non-grounded). In a perfect conductor the charge arranges such that no field penetrates the conductor and the field lines are perpendicular on its surface. This is exactly what happens here.
Some additional note to the dipolar field outside the cylinder. Sure you know the calculus of a charge outside a conducting sphere and how to solve this by image charges (if not you should have a look). The homogeneous field outside can be consider as the field of two large point charges $\pm Q$ at $\pm X_0$, with $X_0\rightarrow \infty$ and $Q \rightarrow \infty$. Then you see that you'll get two according image charges that eventually form a dipole. Same could be done for the infinite cylinder with infinite line charges. You find information here, and what I mean can be summarized like

In the upper left you have the mirror charge solution for a single line charge. It is not valid inside the cylinder, of course. The upper right shows the case of two opposite charges on opposite sides ($\pm 2.1$). increasing their distance and charge density with $1/\epsilon$ gives a more and more homogeneous field at the cylinder position ($X_0=\pm 5$, lower left and $X_0=\pm 500$, lower right). The mirror charge also increases with $1/\epsilon$ while the distance between the two mirror charges decreases with $\epsilon$. Charge times distance is, hence, constant and in the limit $\epsilon\rightarrow\infty$ you get a 2D dipole. The outside field is the same as above (only with opposite sign here, as I placed the positive line charge right).
