# Finding an equivalent system of an infinitely long cylinder with polarization vector $\vec P = P\hat y$

I am given an infinitely long cylinder, with the axis of symmetry on the $z$ axis, $y$ is right to left, and $x$ towards us.

It is polarized with $\vec P = P\hat y$.

My solution manual states that an equivalent system is two cylinders, one with density $\sigma_B=P$ and one $\sigma_B=-P$.

I don't understand how exactly they arranged the two cylinders.

One on top of the other? Or are they next to each other? And what is that based on?

And if they are next to each other, which one do I assign the positive or negative charge density

The polarization will cause a displacement of all the positive charge relative to the negative charge in the y-direction, so that a top view of the cylinders will look like this:

Since the polarization vector points in the positive y direction, the electric dipole moment vector will point the same way, which tells you that the negatively charges cylinder is to the left and the positively charged one to the right ($\vec{p}$ points from negative to positive charge)

Now, all you have to do is find the potential using the Legendre polynomials (at least that is how Griffiths solves it), then take the gradient to find the electric field both outside and inside the cylinder.

Side note:

However, there is a neat trick that you can use to find the electric field due to the cylinder (I find it is relevant to your question, and I want to put it out there) Note, however that this can only be used when the polarization is constant in one direction, as in your case. You can exploit this constant polarization that effectively displaces all negative and positive charges by the same distance relative to each other. Then, you treat the situation as that of two separate cylinders with a constant volume charge density ($\rho_0$) Let the radius of the cylinder be R

Inside the cylinder

$\vec{E}$=$\frac{\rho_0 r}{2\epsilon_0}$ (General formula for E-field inside uniformly charged cylinder)

So, from the two cylinders,(refer to figure)

$\vec{E_{net}}$=$\frac{\rho_0 \vec{r}}{2\epsilon_0}-\frac{\rho_0 (\vec{r}-\delta \vec{r})}{2\epsilon_0} =\frac{\rho_0 \vec{\delta r}}{2\epsilon_0}=-\frac{\vec{P}}{2\epsilon_0}$

Outside the cylinder

$\vec{E}$=$\frac{\rho_0 R^2}{2\epsilon_0 r}=\frac{Q}{2 \pi \epsilon_0 hr}$ (General formula for E-field outside uniformly charged cylinder)

So, from the two cylinders,(refer to figure)

$\vec{E_{net}}$=$\frac{1}{2 \pi \epsilon_0 h}(\frac{Q}{r}-\frac{Q}{r-\delta r})\hat{r}=\frac{1}{2 \pi \epsilon_0 h}\frac{\vec{p}}{r^2}\hat{r}$, where $\vec{p}$ is the net dipole moment in the volume $(\vec{p}=\vec{P}V)$

$\vec{E_{net}}$=$\frac{\vec{P} R^2}{2 \epsilon_0 r^2}$