# Electrostatics: Charge Distribution and Energy - Confusion

Consider the following situation:

Between two metallic plates, are two dielectrics of dielectric constants $$K$$1 and $$K$$2. The surface charge density on the upper metallic plate is $$\sigma$$, and that on the lower one is -$$\sigma$$. The dielectrics have width $$d$$1 and $$d$$2 as shown. 1. Now, how is the charge distributed?

2. How can we find the surface charge densities at the interface of the two dielectrics?

3. How does the field between the plates change, after insertion of dielectrics? What is the new value?

4. Qualitatively, from what I understand, the energy stored between the plates should increase. How do we find the energy stored in each dielectric slab?

From what I know, I tried using Gauss' Law at the interfaces, where the field abruptly changes. This is clearly due to accumulation of some charge at the boundary of dielectrics, which we should be able to find. However, I'm really confused about charges that come into picture due to polarisation, and the ones that are already there. What's the net field now, and which charges contribute to it?

It'd be great if someone could explain what's going on in detail, so I could probably try figuring out answers to the problems I've posted.

Thanks a lot!

• When you ask how the charge is distributed, are you referring to the free charge or the bound charge? – probably_someone Oct 13 '18 at 11:02
• Both! I'm really new to handling dielectrics, so it'd be great if you could tell me about both of them – arya_stark Oct 13 '18 at 13:23

When an electric field acts on a dielectric, it induces polarization within that dielectric (the creation of electric dipoles that oppose the external field). The total dipole moment per unit volume induced by the field is called the polarization, and is a vector that usually carries the symbol $$\mathbf{P}$$.

The overall effect of the dipoles in the medium can be shown to be equivalent to the effect of placing charge in a certain distribution in the surface and bulk of the dielectric (this fictitious charge is called bound charge, as opposed to free charge). The surface density $$\sigma_b$$ and bulk density $$\rho_b$$ of bound charge is completely determined by the polarization vector:

$$\sigma_b=\mathbf{P}\cdot \hat{\mathbf{n}}$$

$$\rho_b=-\nabla\cdot\mathbf{P}$$

where $$\hat{\mathbf{n}}$$ is the normal vector to the surface. Once you have replaced the dielectric medium with these point charges, you can just use the ordinary Maxwell's equations (including both bound and free charges) to calculate various quantities.

So the only thing left is to get the polarization vector. In general, there does not have to be any particular relation between the induced polarization and the external electric field, and this relationship can be highly nontrivial for certain materials. However, many materials are linear dielectrics, which means that there is a linear relationship between applied field and induced polarization. Specifically, for linear dielectrics:

$$\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$$

where $$\chi_e=K-1$$ is the electric susceptibility of the material. So this completely determines the polarization in terms of the applied electric field, and therefore determines the bound charges in terms of the same.

When you insert the dielectric, the amount of free charge doesn't change, but the net effect of the dielectrics is to 1) modify the amount of charge on the plates that is "seen" by the area between the plates, and 2) create a layer of bound charge in the middle of the two plates that will create a discontinuity in the electric field "seen" in each of the two regions. This is a very cursory treatment of the topic (for example, I have not even introduced the electric displacement), and textbooks usually devote an entire chapter to this material, so it is highly recommended to seek out further details, for example in Griffiths' Introduction to Electrodynamics.

• So, what's the bound charge density? And I don't understand how polarisation vector will help us find out the required stuff - could you please elaborate and if possible provide a solution to the problem? It'll be really helpful – arya_stark Oct 13 '18 at 15:38

$$\let\eps=\varepsilon \let\rho=\varrho \let\sig=\sigma \def\vD{\vec D} \def\vE{\vec E} \def\ns#1#2{#1_{\rm#2}} \def\cE{{\cal E}} \def\cEa{\ns\cE a} \def\cEb{\ns\cE b}$$ Given that @probably_someone choose to answer with no reference to displacement $$\vec D$$, I will take the alternate way. I agree that using polarization vector a deeper understanding is got as to what happens in the dielectric. But if electric displacement still survives, almost 150 years after Maxwell, there are good reasons: using it in problems with dielectrics is generally much simpler.

Now, how is the charge distributed?

If the plates are much larger than their distance, over most of the surface (borders excepted) $$\sig$$ is constant.

Let me summarize the relevant equations. In electrostatics they are three: two Maxwell's equations $$\nabla \times \vE = 0$$ $$\nabla \cdot \vD = \rho$$ and one constitutive relation, which for isotropic dieletrics is $$\vD = \eps \vE$$ with $$\eps$$ a scalar quantity. I assume your dielectrics are homogeneous, so that only two values are needed. You will excuse me if I prefer $$\eps_1$$, $$\eps_2$$ to your $$K_1$$, $$K_2$$. AFAIK, $$\eps$$ is a universal symbol for dielectric constant.

Given the problem's symmetry, it is obvious that both $$\vE$$ and $$\vD$$ are wherever perpendicular to the plates, and are uniform within each dielectric separately. They could be different between the upper and the lower dielectric, but use of Gauss' theorem for a small cylinder put across the interface shows that $$\vD$$ stays the same, as there is no free charge there. (Free charges are located on plates only.)

Continuing with $$\vD$$, we can see its advantages: we can reason when dielectrics are present exactly like we do in vacuum. Between the plates $$\vD$$ is directed from positive to negative plate and its modulus equals $$\sig$$. Moreover, $$\vD=0$$ outside (always neglecting border effects).

How can we find the surface charge densities at the interface of the two dielectrics?

This cannot be answered using $$\vD$$ alone, as it ignores bound charges. But once you know $$\vD$$, $$\vE$$ follows immediately: $$E_1=D/\eps_1$$, $$E_2=D/\eps_2$$. The same cylinder we used before tells us that $$\sig' = \eps_0 (E_2 - E_1) = \eps_0\,D \left(\!{1 \over \eps_2} - {1 \over \eps_1}\!\right) = \eps_0\,\sig \left(\!{1 \over \eps_2} - {1 \over \eps_1}\!\right).$$

Qualitatively, from what I understand, the energy stored between the plates should increase. How do we find the energy stored in each dielectric slab?

This is an interesting question. I can't see why energy should increase, in your opinion. Actually the opposite happens. Let me recall the general expression for energy density within a dielectric: $$U = \frac12\,\vE \cdot \vD.$$ Then total energy before dielectric insertion is $$\cEb = \frac1{2 \eps_0}\,\sig^2 S\,(d_1 + d_2)$$ where $$S$$ is plate's area. After insertion we have $$\cEa = \frac1{2 \eps_1}\,\sig^2 S\,d_1 + \frac1{2 \eps_2}\,\sig^2 S\,d_2 = \frac12\,\sig^2 S \left(\!{d_1 \over \eps_1} + {d_2 \over \eps_2}\!\right)\!.$$ Since both $$\eps_1$$ and $$\eps_2$$ are greater than $$\eps_0$$, it's clear that $$\cEa < \cEb.$$ As a matter of fact, it is known that dielectrics are pulled inside electric fields. In other words, by inserting a dielectric in a capacitor useful work can be obtained.