Effect of a dielectric cylinder parallel to an external field

[dphysicsf]

Homework Statement

If a dielectric cylinder, say radius R, is placed with its axis parallel to a uniform electric field. What effect will it have on the field? Picture is included which makes it a bit clearer

Homework Equations

The Attempt at a Solution

I know that the D-field should be continuous over the z boundary because there is no free surface charge, so the E-field inside the cylinder should be the same as that on the outside but less by a factor of epsilon (permittivity of the cylinder). Therefore the polarization in the cylinder is along the z-direction, this means there is no bound charge on the curved surface of the cylinder (because the polarization is perpendicular to the surface vector) and therefore the field outside the cylinder (at r>a) is still E0.


Is this correct? Is there anything I've missed in this anaylsis?

 

[ShayanJ]

I don't remember it that much to help you but I can suggest you to study chapters 3 and 4 of Foundations of electromagnetic theory by Milford and Reitz.

 

[DNN]

@dphysicsf
I wonder if you managed to find the answer. Could you please share if so?

 

[Charles Link]

The solution is more complicated than what the OP presents. With no external electric field, a uniformly polarized cylinder does not generate a uniform electric field in its interior. For the case of uniform $\vec{P}$, there are two endfaces that have $\sigma_p=\vec{P} \cdot \hat{n}$. The electric field from these poles will drop off as an inverse square law, so that the electric field that points opposite $\vec{P}$, from this depolarizing electric field, is smaller towards the middle length of the cylinder than it is near the endfaces. $\\$ Thereby, the assumption that $\vec{E}$ and $\vec{P}$ are uniform is incorrect. $\\$ With the external field applied, the polarization $\vec{P}$ will be largest in the middle portion of the cylinder where it experiences less depolarizing electric field from the endfaces. The external electric field is enhanced somewhat on and near the axis of the cylinder, just outside of the cylinder, from the surface polarization charge that arises. Inside the cylinder, the electric field is reduced somewhat from what it would be without the material present. Towards the middle of the length of the cylinder, this reduction can be very minimal if it is a long cylinder.

 

[DNN]

The solution is more complicated than what the OP presents. With no external electric field, a uniformly polarized cylinder does not generate a uniform electric field in its interior. For the case of uniform $\vec{P}$, there are two endfaces that have $\sigma_p=\vec{P} \cdot \hat{n}$. The electric field from these poles will drop off as an inverse square law, so that the electric field that points opposite $\vec{P}$ is smaller towards the middle length of the cylinder than it is near the endfaces. $\\$ Thereby, the assumption that $\vec{E}$ and $\vec{P}$ are uniform is incorrect. $\\$ With the external field applied, the polarization $\vec{P}$ will be largest in the middle portion of the cylinder where it experiences less depolarizing electric field from the endfaces. The external electric field is enhanced somewhat on and near the axis of the cylinder, just outside of the cylinder, from the surface polarization charge that arises. Inside the cylinder, the electric field is reduced somewhat from what it would be without the material present. Towards the middle of the length of the cylinder, this reduction can be very minimal if it is a long cylinder.

Thank you very much Charles for your detailed explanation.
What I am interested in is the electric field at a point R which is d distance along z - direction and r distance away from its surface (attached is a sketch for better understanding of the scenario). Would you be able to help me in deriving an expression?

Could you recommend me a textbook to look for? I referred Griffiths and many other books, but they all consider the case where E field is perpendicular to the axis of the cylinder.

Appreciate your support very much.

 

[Charles Link]

@DNN An exact solution is quite difficult. The approximation can be made that $\vec{P}=\chi \epsilon_o \vec{E}_o$, (where $\vec{E}_o$ is the applied electric field), and that $\vec{P}$ is uniform everywhere inside the cylinder. $\\$ (This is different from the OP's assumption, where he used a continuous $\vec{D}$ across the endfaces, and that $\vec{D}_{external}=\epsilon_o \vec{E}_o$. (The second part of this is where my solution differs the most from the OP's. Because I'm working under somewhat of an approximation, I may not have $\vec{D}$ precisely continuous across the endface. I'm basically assuming that the depoarizing $\vec{E}_p$ from the endfaces can be ignored in computing $\vec{P}$)). $\\$ This approximation that I'm making here can be reasonably good, especially if the cylinder is reasonably long, e.g. a length $L>4d$, and/or if $\chi << 1$. (The dielectric constant $\epsilon=\epsilon_o \epsilon_r =\epsilon_o(1 +\chi)$). Under this assumption of uniform $\vec{P}$, $-\nabla \cdot \vec{P}=\rho_p=0$, (where $\rho_p$ is the polarization charge density), and the only polarization charge occurs on the endfaces, where surface polarization charge density $\sigma_p =\vec{P} \cdot \hat{n}$. That will make $\sigma_p=\pm P$ on the endfaces. Besides the applied $\vec{E}_o$, this is the only source of electric field in this problem. (Again, we made an approximation that $\vec{P}$ is perfectly uniform, but under this approximation, this is the result). $\\$ These endfaces that have area $A$ will have total charge $Q_p=\pm PA$. Up close to the endface, it looks like an infinite sheet of electric surface charge. Somewhat distant from the endface, you can simply write Coulomb's inverse square law: $\vec{E}_{endface}(\vec{r})=\frac{Q_p}{4 \pi \epsilon_o |\vec{r}-\vec{r}'|^3}(\vec{r}-\vec{r}')$ , where $\vec{r}'$ is the coordinate of the center of the one endface (remember there are two of them) . $\vec{E}(\vec{r})_{total}=\vec{E}_o+\vec{E}_{endfaces}$.

 

[DNN]

@Charles Link : Thank you for your prompt response.
So, under this argument, if the point R is located right above the middle of the cylinder (z coordinate is L/2),
E(r)_total = E_o ?

 

[Charles Link]

@Charles Link : Thank you for your prompt response.
So, under this argument, if the point R is located right above the middle of the cylinder (z coordinate is L/2),
E(r)_total = E_o ?

I presume you are putting the base at $z=0$ and the top of the cylinder at $z=L$. There is a $" +"$ surface charge density at $z=L$ of $\sigma_p=P$ and a $"-"$ surface charge density $\sigma_p=-P$ at $z=0$. It is somewhat objective whether to include their effect inside the cylinder, e.g. at $z=\frac{L}{2}$. For a long cylinder, and/or small $\chi$, the effect of $\vec{E}_p$ will be minimal at $z=\frac{L}{2}$. $\\$ If their effect is included in $\vec{E}_{total}$ then you get a $\vec{P}$ that is no longer uniform, if you assume $\vec{P}=\epsilon_o \chi \vec{E}_{total}$. $\\$ That non-uniform $\vec{P}$ will likely generate some internal polarization charge density $-\nabla \cdot \vec{P}=\rho_p$ that will be the source of additional electric field. In general, I don't think there is a closed form mathematical solution to solve this, and very refined numerical methods would have a "first guess" $\vec{E}$, and compute $\vec{P}(\vec{r})$ everywhere, and then compute any additional $\vec{E}_p$ that results, etc., in a rather difficult iterative process to try to get a complete $\vec{E}_{total}$ and $\vec{P}$ that is self-consistent=i.e. that it satisfies $\vec{P}=\epsilon_o \chi \vec{E}$ everywhere. The dielectric cylinder of finite length in a uniform electric field does not have an exact closed-form solution.

 

[DNN]

@Charles Link : I get your point now. Thank you very much for helping out.

 

[DNN]

@Charles Link : May I ask you of one more favour? Can you please give me a reference to this uniform P assumption? It would be really useful in my future work.
Thanks in advance.