0
$\begingroup$

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).

enter image description here

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Commented Jan 20, 2021 at 5:29
  • 1
    $\begingroup$ @Young Kindaichi Well, i get your point, but since I spent 15 min of my time answering to you, you could be more polite. $\endgroup$ Commented Jan 20, 2021 at 7:58
  • $\begingroup$ @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. $\endgroup$ Commented Jan 20, 2021 at 8:08
  • $\begingroup$ 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. $\endgroup$ Commented Jan 20, 2021 at 11:29
0
$\begingroup$

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

$\endgroup$
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.