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

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?

• Find your whole question confusing. You agree (below) that there is no $E_z$, yet you also claim it is $-P/2\epsilon_0$. Dec 2 '20 at 7:23
• In every point along z axis $\vec{E} = -P/2\epsilon_0 \vec{u_x}$, because that's the $\vec{P}$ directrion and because this is an isotropic medium. So the $E_z$ and $E_y$ components are zero. Dec 2 '20 at 16:52
• Even though $E_x$ exists inside the cylinder and its value is $E_x = -P/2\epsilon_0$ in every point of it, including the z axis. Dec 2 '20 at 16:58
• I see. I have made a couple of small edits. Dec 2 '20 at 18:05
• Thanks! It's much more clear now. Dec 2 '20 at 19:56

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.

• Thanks! Yes, the only component is $\vec{E_x}$, which is the result I've got from Gauss. What I don't understand is why the scalar potential is zero from direct integration while $\vec{E}$ isn't. Dec 1 '20 at 20:58