Electric field of uniformly polarized cylinder

[Felesinho]

Hi all,

I have a doubt when calculating the electric field of a uniformly polarized cylinder P along its longest axis. The cylinder has length L and radius a.

Using Gauss's law:

$$\int D\cdot ds = \rho_{f} =0 \, \, (eq .1)$$

The electric field inside of cylinder would be: $$E =- \dfrac{1}{\epsilon_0} P$$

The electric field outside of the cylinder would be zero, but in Griffiths book, they say that's not correct.

I understand that D must be nonzero outside the cylinder (contradictory to eq 1), which is because P is discontinuous on the surface of the mantle.

Why can't Gauss's law be used in this problem?

 

[Charles Link]

$\rho_p=-\nabla \cdot P$ gives the result that $\sigma_p=P \cdot \hat{n}=\pm P$ on the endfaces. The electric field is found by Coulomb's law from the surface polarization charge on the two endfaces.
Note: $\rho_p$ is zero for the uniform P, but when P is discontinuous, we get a surface charge density forming.

 

[Charles Link]

Note for the above $\nabla \cdot D=0$ results in $\nabla \cdot E=-\nabla \cdot P/\epsilon_o$. The integral solution for this is $E(x)=\int \frac{ \rho_p(x') (x-x') \, d^3x'}{4 \pi \epsilon_o |x-x'|^3}$, where $\rho_p=-\nabla \cdot P$. You might think you can simply drop the $\nabla \cdot$ on both sides, (and write $E=-P/\epsilon_o$), but it does not work that way.

[Edit:(The reason this logic does not work is you can add to $E$ and $P$ solutions where $\nabla \cdot E=0$ or $\nabla \cdot P=0$, and it is also a solution that will be mathematically consistent. The solution that is the right one for $E$ with a given $P$ is found from Coulomb's law. There are other solutions for $E$ that will satisfy $\nabla \cdot E=-\nabla \cdot P/\epsilon_o$, including $E=-P/\epsilon_o$. In differential equation language, $E=-P/\epsilon_o$ is a particular solution, but it needs to be modified by adding a solution to the homogeneous equation $\nabla \cdot E=0$ to be the correct one. Edit': This one may be slightly more complex in that one could argue that $E=-P/\epsilon_o$ consists of a couple of homogeneous solutions, but in any case it is not the correct solution.).

It may also be of interest that the same thing results for the analogous magnetized cylinder problem, where $\nabla \cdot B=0$, and $B=\mu_o H +M$.The $H$ needs to be found just like the $E$ above, with $M$ replacing $P$, and $\mu_o$ replacing $\epsilon_o$. Even though $\nabla \cdot H=-\nabla \cdot M/\mu_o$, the solution $H=-M/\mu_o$ is not the correct solution. The correct solution is $H(x)=\int \frac{ \rho_m(x') (x-x') \, d^3x'}{4 \pi \mu_o |x-x'|^3}$, where $\rho_m=-\nabla \cdot M$. The magnetic field $B$ is then solved for everywhere from this $H$ with $B=\mu_o H+M$.
Note also that $\nabla \cdot B=0$ does not imply that $B=0$.]

Note that the integral here is Coulomb's law, which in principle is also Gauss's law. This problem, however, doesn't have the symmetry to simplify the flux integral that you get from Gauss's law.

Note also that $x$ and $x'$ are three dimensional vectors=I used a simplified notation.

 

[Felesinho]

Thanks for the answer. I must add the charge generated by the polarization.

I have a doubt with the value $\nabla \cdotp P$. Due to the polarization is uniform, I expect that the polarization creates surface charges but not a volume charge.

To use the gauss law i need to consider that $Q_{tot} = \int \sigma_b ds$, is this right?

 

[Charles Link]

Thanks for the answer. I must add the charge generated by the polarization.

I have a doubt with the value $\nabla \cdotp P$. Due to the polarization is uniform, I expect that the polarization creates surface charges but not a volume charge.

To use the gauss law i need to consider that $Q_{tot} = \int \sigma_b ds$, is this right?

Yes, that is correct. See also post 2 above, and the "Edit" of post 3.

 

[Charles Link]

Note above that with $\rho_p=-\nabla \cdot P$, you can use Gauss's law at a surface boundary to show that the surface polarization charge density $\sigma_p=P \cdot \hat{n}$.

Note also (considering one single layer of surface charge) that the identical amplitude (pointing away from the surface charge) electric field will appear on both sides of the surface charge, while the polarization goes from $P$ in the material to zero in the vacuum, so we know immediately that the solution $E=-P/\epsilon_o$ simply can not be correct.