# Surface charge on dielectric interface

What I have in mind is a system that consists of a capacitor filled with two dielectric materials, one of them with dielectric constant $$\kappa_1$$ and the other with $$\kappa_2$$. The capacitor is connected to a battery of voltage, say $$V$$. As the dielectric constant of the plates are different, I would expect a surface density at the interface of the two dielectric materials. I want to calculate this surface charge density.

I have the following figure in my mind but with two different slabs of dielectrics (area and length are the same).

Now let the $$Q_0$$ be the charge on the plates when the dielectric material was absent. Now when the dielectric slabs are fitted, then the charge on the upper plate would be $$Q_1=\kappa_1 Q_0$$ and on the lower plate $$Q_2=\kappa_2 Q_0$$ but this doesn't seem right to me! as I have never seen a capacitor with different charges on plates. That means charges must redistribute themselves to make it equal. But that would mean $$Q_1=Q_2 \Rightarrow \kappa_1=\kappa_2$$ but that is incorrect.

Now if I take some equivalent constant for the two plates, say $$\kappa$$ and say that the charges on the upper and lower plate is $$Q=\kappa Q_0$$. Then the charge on the upper dielectric plate would be $$Q'_1=(1-\kappa )Q_0$$ and on the lower part of the second dielectric would be $$Q'_2=Q'_1$$. But that's the same as one plate case and I don't know what to do.

Can someone point out the flaw in my thinking and give me a right approach to the problem.

I'm a huge fan of the field D displacement.
Since $$\nabla \cdot D = \rho_{free}$$ , using gauss law with a cilinder passing trough the planes, you get that $$D$$ inside the capacitor is uniform (Similar argument to the classic capacitor).
Now you have that for linear dielectrics $$D=\epsilon E$$

to get $$D$$ from the $$\Delta V$$ you compute the integral

$$\Delta V= -\int_{a}^{c} E \,dx =-\int_{a}^{b} E \,dx -\int_{b}^{c} E \,dx=$$
$$=-\int_{a}^{b} \frac{D}{\epsilon_1} \,dx-\int_{b}^{c} \frac{D}{\epsilon_2} \,dx$$

where a, b, c are points on the tree surfaces (b being the middle one) so (since D is uniform)

$$D=\frac{\Delta V}{\frac{(b-a)}{\epsilon_1}+\frac{(c-b)}{\epsilon_2}}$$

I did not put too much attention on signs but you get my point.

Now from $$D$$ trough $$D=\epsilon E$$ you get the two fields in the to regions, and their difference is $$\frac{\rho}{\epsilon_0}$$ (Gauss law), so you can evaluate $$\rho$$ between the two dielectrics.

• And I hate when someone doesn't tell me my fault nor tell me How to proceed the way I did and give another way to solve it. In the present case, I m not interested in doing it with the displacement field. Commented Jan 20, 2021 at 5:29
• @Young Kindaichi Well, i get your point, but since I spent 15 min of my time answering to you, you could be more polite. Commented Jan 20, 2021 at 7:58
• @Young Kindaichi anyway, when you say that there shoud hold $Q_1=k_1Q_0$, think about how this result is obtained and see that it just hold for a capacitor filled with ONE and only one dielectric. So that is a wrong approach to the problem. Commented Jan 20, 2021 at 8:08
• I appreciate your efforts and sorry If I were harsh. How do I proceed without introducing the displacement field? How the picture would look like for two dielectrics? Like the one shown. Commented Jan 20, 2021 at 11:29

By solving with the help of this formula when the different dielectric of same length is used it is giving answer with(+ve) sing but when we are solving it with the basics then it is ging the (-ve) sing so I think that there should be an (-ve) sing in this formula