I want to calculate $\vec{E}$ and $\vec{H}$ inside this capacitor:

enter image description here

So this would be a cylindrical capacitor but with two different dielectrics in it (they both occupy the same amount of space, excuse me if the drawing I made is not symmetrical). The dielectrics are not ideal, so there is current going through them. We could state then that each material has a $\epsilon$ and a $\sigma$. Let's say that these quantities are $\sigma_1$, $\epsilon_1$, $\sigma_2$ and $\epsilon_2$.

The potential difference in the plates is $V_o$ so the electric field will be the same in each material.

$$E = E_1 = E_2 = \frac{V_o}{d}$$

However, different current densities will flow in each material. They can be easily calculated using Ohm's law.

The problem arises when I try to find the magnetic field $\vec{H}$. There is no symmetry in this problem, so I can't use Ampére and take a circulation around the center of the capacitor because $H$ will not be constant for a given radius around the central axis. Also, the fact that different currents are flowing in each material also confuses me. I don't know how to calculate $H$ for a system like this.

Is there a way to find $\vec{H}$ inside this capacitor? Or maybe some software in which I could simulate this?

  • $\begingroup$ Just handwaving here, happy if it actually helps you. Did you try using $\nabla \times H$ = $J_f + \frac{\partial D}{\partial t}$? since you know $D$ and the free current density $J_f$ is the current density produced due to the voltage source alone(exclude magnetization). This immediately tells you that since the currents are longitudinal, the magnetic field $has$ to be circumferential. $\endgroup$ – Prasad Mani Oct 28 '16 at 15:46
  • $\begingroup$ @PrasadMani Hi there, I know that it has to be circumferential but I wanted to know its value (as a function of the conductivities, the permittivities and $V_o$) , not only its direction. $\endgroup$ – Tendero Oct 28 '16 at 15:48
  • $\begingroup$ Which is why you have to calculate the RHS of the above maxwell's equation; again, handwaving, sorry! $\endgroup$ – Prasad Mani Oct 28 '16 at 15:49
  • $\begingroup$ What is your question? Why do you need magnetic fields here? Where (inside or outside the capacitor) would you like to know the magnetic fields, if any? $\endgroup$ – freecharly Oct 28 '16 at 17:03
  • $\begingroup$ @freecharly It is an exercise our teacher gave us, and he told us to calculate $\vec{E}$ and $\vec{H}$ inside the capacitor. $\endgroup$ – Tendero Oct 28 '16 at 17:06

I don't think the electric field, E, is the same in both materials, rather the displacement, D, is--which is the whole point of "D"--it accounts for induced charge at the surface that is caused by divergence of the materials' polarization.

I would treat this problem as 2 uniform capacitors in parallel.

  • $\begingroup$ There is absolute no reference to magnetic field in your answer. Apart from that, you are wrong respecting E and D: E is the same and D differs. $\endgroup$ – Tendero Oct 31 '16 at 4:17

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