Dielectric in a uniform external field

[Kangaroos]

Homework Statement

A slab of dielectric of thickness a and relative permittivity ε is placed in a uniform external field E whose field lines make angle θ with normals to the surface of the slab. What is the density of polarization charge on the surface of the slab? Neglect end effects.

Homework Equations

P⋅dS[/B] = σ_p
D = P + ε₀E

The Attempt at a Solution

The correct answer is ± ε₀Ecosθ(1-1/ε).
I assume there is no free charge in the dielectric so D is zero. But I can't get the answer.
Does someone can help me?

 

[Charles Link]

Perhaps the simplest way to work this problem is to separate the applied electric field into two components: One normal to the plane of the slab and the other parallel to the plane of the slab. The parallel component of E will not introduce any surface polarization charges and will be the same inside the material as outside. Try solving the problem for the perpendicular component of the applied E. If you need additional hints, I'd be happy to try to assist. $\\$ Editing... In fact, I'll give you a hint: Since there are no free electrical charges in this problem, $D$ (perpendicular component) needs to be continuous in going from air/vacuum and into and across the material and again back into the air/vacuum. What does that tell you about $E_{in}$ (electric field inside the material)? (i.e. Can you compute it?). Knowing the electric field inside the material (i.e. the perpendicular component) and the perpendicular component outside the material ,$E_{out}$, you should be able to use Gauss' law to compute the surface charge density (which arises completely from polarization charges.) $\\$ One additional hint: Depending on the textbook, the dielectric constant $\epsilon$ can be defined differently. For the book you are using (in viewing the answer for $\sigma_p$ that you provided, this is apparent), $D=\epsilon \epsilon_o E$. (Some books (in SI units) may simply use $D=\epsilon E$ (with $\epsilon_o$ contained in the $\epsilon$), but with the form of the answer, your book is using the first form.) ( $\epsilon=1$ in vacuum and has some value $\epsilon >1$ in the material. ) $\\$ Additional note: Having no free charge in the dielectric implies $\nabla \cdot D=0$ but it doesn't mean that $D=0$. Having $\nabla \cdot D=0$ tells you that $D$ is continuous everywhere, including across the boundary.

 

[Charles Link]

I would like to add to the above because the concepts in this problem aren't always explained clearly in the E&M textbooks for what they represent. $\\$ The basic equation for the electric field is $\nabla \cdot E=\rho_{total}/\epsilon_o$ where $\rho_{total}$ is the total charge density. Meanwhile for linear dielectrics $P=\chi E$ where $\chi$ is the dielectric susceptibility. A polarization charge density results when there are gradients in the polarization vector $P$, i.e. any time that the polarization $P$ is non-uniform. The polarization charge density is given by $\rho_p=-\nabla \cdot P$. (If a material has uniform polarization $P$, the result is a discontinuity in $P$ at the surface boundary. The surface charge density $\sigma_p=\hat{n} \cdot P$ can readily be found by using Gauss law on the equation $-\nabla \cdot P=\rho_p$ around the surface boundary.) $\\$ Now going back to the equation $\nabla \cdot E=\rho_{total}/\epsilon_o$, we can make a mathematical construction $D=\epsilon_o E+P$ and take the divergence of both sides of this equation: $\nabla \cdot D=\epsilon_o \nabla \cdot E +\nabla \cdot P$. The result is $\nabla \cdot D=\rho_{total}-\rho_p$. Since $\rho_{total}= \rho_{free}+\rho_p$ the result is $\nabla \cdot D=\rho_{free}$ where $D$ is the mathematical construction defined by $D=\epsilon_o E+P$. (Note: the vector symbol has been left off of $D, \, E, \, and \, P$ in all of this for brevity). The $D$ formalism makes for a handy way of solving these problems, but the problem can be solved using just the $E$ and $P$ and the $\chi$ as follows: $\\$ $P=\chi E_i$ and let's assume uniform $P$ and $E_i$. $\sigma_p=\hat{n} \cdot E_i$ as previously mentioned. The result is a $\sigma_p=P$ on one surface and $\sigma_p=-P$ on the other. (Here $P$ is just the amplitude of vector $P$. ) The electric field in the material from these two layers of surface polarization charge is given by $E_p=-P/\epsilon_o$. Now $E_i=E_o+E_p$ where $E_o$ is the applied field. We can substitute for $E_i=P/\chi$ , so that $P/\chi=E_o-P/\epsilon_o$. Solving, we get $P=\epsilon_o E_o/(1+\epsilon_o/\chi )$. Since $P=\chi E$ and $D=\epsilon_o E+P$, we get $D=(\epsilon_o+\chi) E =\epsilon \epsilon_o E$. This makes $\epsilon=1+\chi/\epsilon_o$. Putting in just the amplitude of $E_o$ and $P$, since $P=\sigma_p$, a little algebra where we substitute for $\chi$ in terms of $\epsilon$ gives: $\\$ $\sigma_p=\epsilon_o E_o (1-1/\epsilon )$. $\\$ This solution took more effort than simply using the mathematical construction $D$ along with the equations that apply for it, but I wanted to show that the same result can be obtained in a more fundamental way, (without using the $D$), by working with just the $E$ and $P$ and $\chi$.

 

[Kangaroos]

I would like to add to the above because the concepts in this problem aren't always explained clearly in the E&M textbooks for what they represent. $\\$ The basic equation for the electric field is $\nabla \cdot E=\rho_{total}/\epsilon_o$ where $\rho_{total}$ is the total charge density. Meanwhile for linear dielectrics $P=\chi E$ where $\chi$ is the dielectric susceptibility. A polarization charge density results when there are gradients in the polarization vector $P$, i.e. any time that the polarization $P$ is non-uniform. The polarization charge density is given by $\rho_p=-\nabla \cdot P$. (If a material has uniform polarization $P$, the result is a discontinuity in $P$ at the surface boundary. The surface charge density $\sigma_p=\hat{n} \cdot P$ can readily be found by using Gauss law on the equation $-\nabla \cdot P=\rho_p$ around the surface boundary.) $\\$ Now going back to the equation $\nabla \cdot E=\rho_{total}/\epsilon_o$, we can make a mathematical construction $D=\epsilon_o E+P$ and take the divergence of both sides of this equation: $\nabla \cdot D=\epsilon_o \nabla \cdot E +\nabla \cdot P$. The result is $\nabla \cdot D=\rho_{total}-\rho_p$. Since $\rho_{total}= \rho_{free}+\rho_p$ the result is $\nabla \cdot D=\rho_{free}$ where $D$ is the mathematical construction defined by $D=\epsilon_o E+P$. (Note: the vector symbol has been left off of $D, \, E, \, and \, P$ in all of this for brevity). The $D$ formalism makes for a handy way of solving these problems, but the problem can be solved using just the $E$ and $P$ and the $\chi$ as follows: $\\$ $P=\chi E_i$ and let's assume uniform $P$ and $E_i$. $\sigma_p=\hat{n} \cdot E_i$ as previously mentioned. The result is a $\sigma_p=P$ on one surface and $\sigma_p=-P$ on the other. (Here $P$ is just the amplitude of vector $P$. ) The electric field in the material from these two layers of surface polarization charge is given by $E_p=-P/\epsilon_o$. Now $E_i=E_o+E_p$ where $E_o$ is the applied field. We can substitute for $E_i=P/\chi$ , so that $P/\chi=E_o-P/\epsilon_o$. Solving, we get $P=\epsilon_o E_o/(1+\epsilon_o/\chi )$. Since $P=\chi E$ and $D=\epsilon_o E+P$, we get $D=(\epsilon_o+\chi) E =\epsilon \epsilon_o E$. This makes $\epsilon=1+\chi/\epsilon_o$. Putting in just the amplitude of $E_o$ and $P$, since $P=\sigma_p$, a little algebra where we substitute for $\chi$ in terms of $\epsilon$ gives: $\\$ $\sigma_p=\epsilon_o E_o (1-1/\epsilon )$. $\\$ This solution took more effort than simply using the mathematical construction $D$ along with the equations that apply for it, but I wanted to show that the same result can be obtained in a more fundamental way, (without using the $D$), by working with just the $E$ and $P$ and $\chi$.

Charles! Thank you so much!

 

[Charles Link]

I think you probably have it figured out, but just one additional comment is that the parallel component of $D$ will not be continuous across the boundary(editing=see also the last paragraph of this post), because the parallel component of $E$ will be continuous across the boundary. This basically says that the parallel sources of $E$ lie outside of the sample which can be a square but somewhat thin platelet: e.g. 2"x2"x 1/4". Dealing with a completely infinite slab is somewhat unphysical, because how do you then apply the parallel component of the $E$ field across the material? The answer is that the sample must lie between two large capacitor plates that deliver the uniform parallel component of $E$ and that the edge effect can be ignored from any surface charge that forms on the edges. Meanwhile, in the other direction, normal to the sample, you would have another set of capacitor plates, and this $E$ field will create a surface polarization charge density (+ and -) on the broad faces, and the perpendicular $D$ is continuous across the material, but the perpendicular $E$ is discontinuous. $\\$ Additional item is that the interpretation that $D$ must be continuous everywhere because $\nabla \cdot D =0$ was correct, but what will happen is the parallel component of $E$ will take on the same value everywhere in the sample that is away from the edges. At the edges (of the 2"x2" platelet), the parallel component of $E$ will be discontinuous because of the edge surface polarization charge that forms, but the effect of ant $E_p$ from this edge surface charge will diminish and will be negligible in the interior of the sample where it will simply experience the applied field of the capacitor plates that generate the parallel component of $E$. This is a lot of extra detail, but hopefully it was helpful. Usually with problems such as this, the applied $E$ is presented as being normal to the sample face. $\\$ Editing... One further clarification is in order: If z is the direction normal to the face of the slab and (x and) y parallel to it, the equation $\nabla \cdot D=0$ says that $\frac{\partial{D_y}}{\partial{y}}=0$, etc., (additional editing: $\nabla \cdot D=0$ says the sum of the 3 partial derivatives equals 0, so that this last statement doesn't always hold true, but it is, in fact, the case for this problem that the 3 partial derivatives are independently zero), so that $D_y$ needs to be continuous along y, but $\frac{\partial{D_y}}{\partial{z}}$ has no requirement so that $D_y$ can jump discontinuously (and it does), in crossing into the material in the z direction, because $E_y$ is basically unchanged from just outside the material to just inside. Hopefully this extra detail is helpful and not too confusing.

 

[Charles Link]

One more follow-up post on the above might be in order here... $E_x$ and $E_y$ are basically uniform in this problem because any polarization charge density that appears on and near the edges of the sample can be assumed to have negligible effect anywhere near the center of the sample because their contributions to $E_p$ will fall off with an inverse square law. Thereby $D_x$ and $D_y$ are also basically uniform for this problem, so that $\frac{\partial{D_x}}{\partial{x}}=0$ and $\frac{\partial{D_y}}{\partial{y}}=0$. Since $\nabla \cdot D=0$, we must also have $\frac{\partial{D_z}}{\partial{z}}=0$. We had assumed earlier that $D_z$ is continuous across the boundary and uniform everywhere, but here is proof of it with a little more mathematical rigor. Notice that $E_z$ is discontinuous across the boundary(because of the layer of surface polarization charge density $\sigma_p$), so it took a little extra effort to show that $D_z$ is indeed uniform everywhere.