Find electric field $\vec{E}$ on the axis of a uniformly polarized cylinder

Question

I have a uniformly polarized infinite cylinder of radius $a$ whose axis lies along the $z$ axis with polarization vector $\vec{P} = P \vec{u_x}$, where $\vec{u_x}$ is the unit vector towards the $x$ direction, as shown in the figure, and I want to know the electric field $\vec{E}$ along the $z$ axis.

I first attempted using Gauss Law and considering the cylinder as the superposition of two cylinders separated a distance $\vec{\delta}$ with volumetric charge densities $+\rho$ and $-\rho$ respectively, and I obtained that the electric field at any point along the $z$ axis is:

$$\vec{E} = -\frac{\vec{P}}{2\epsilon_0}$$

Well, along the $z$ axis and for all positions inside the cylinder. However, I want to solve this problem as well by direct integration of the potential $\phi$ and then take its gradient to obtain the electric field, so I first obtained the superficial bound charge density as follows:

$$\sigma_b = \vec{P}\vec{u_n} = \vec{P}\vec{u{\rho}} = P\cos{\phi}$$

Where $\phi$ here means the angle between $\rho$ and the $x$ axis. Then I take the integral over the potential:

$$\vec{r} = z \vec{u_{z}}$$ $$\vec{r'} = a \vec{u_{\rho}} + z' \vec{u_{z}}$$ $$\vec{R} = \vec{r}- \vec{r'}$$ $$R^2 = (a)^2 + (z-z')^2$$

$$\phi = \frac{1}{4\pi\epsilon_0}\int\int \frac{\sigma_b}{R}da = \frac{1}{4\pi\epsilon_0}\int^{+\inf}_{-\inf}\int^{2\pi}_{0} \frac{P\cos{\phi'}}{\sqrt{(a)^2 + (z-z')^2}}d\phi'dz'$$

But then, as the integral of $\cos{\phi'}$ is zero, then the whole integral gets zero and the potential is zero, so there is no electric field. I know that this is not correct because I have calculated the electric field before with Gauss Law, but I don't understand what I am doing wrong so that I get these results. Please, could somebody tell me where is the mistake?

Thanks in advance!

Answer 1

I'm pretty sure you have over-simplified the integral.

The separation between a general point in the cylinder, $(\rho, \phi, 0)$, where we can set $z=0$ by symmetry, and a point on the surface of the cylinder, $(a, \phi', z')$, in cylindrical coordinates is: $$ R^2 = a^2+\rho^2-2a\rho\cos(\phi-\phi') + z'^2$$

Thus $$V(\rho, \phi) = \frac{1}{4\pi \epsilon_0} \int^{\infty}_{-\infty} \int^{2\pi}_{0} \frac{ P \cos \phi'}{(a^2+\rho^2-2a\rho\cos(\phi-\phi') + z'^2)^{1/2}}\ a d\phi'\ dz'\ .$$

Looks tricky, but presumably must give the solution $$V(\rho, \phi, z) = \frac{P}{2\epsilon_0} \rho\cos \phi$$

The potential at $\rho=0$ then is zero, because the integral collapses to what you have in your question (bar the fact that the surface element is $a d\phi'\ dz'$ and not just $d\phi'\ dz'$).

I believe it would be more usual to tackle this using the general solution to Laplace's equation in cylindrical coordinates.