Let's say I have a capacitor defined by two perfectly conducting plates of area x by z separated by some distance y. If my understanding is correct, the Poisson equation describing the charge density of the region between the plates is given by: $$\nabla^2 V=-\frac{\rho}{\epsilon}$$ Where $V$ is the electric potential, $\rho$ is the charge density, and $\epsilon$ is the permittivity of the space between each plate. If we solve for the potential field $V$ we can then take the gradient of this field to get the electric field vectors in x, y, and z: $$\nabla V=-\vec{E}$$ If there is nothing between the plates then the charge density $\rho$ simplifies to a set of boundary conditions where the charge on one plate is larger than the other in volts and $\epsilon$ is a constant which is just the vacuum permittivity $\epsilon_0 =8.85 \cdot 10^{-12} F \text{ }m^{-1}$. To find capacitance, we can integrate over the permittivity times the gradient of the potential $V$ between the plates to get the total charge $Q$: $$Q=\lVert \iiint \epsilon \nabla V \text{ }dxdydz \rVert$$ And dividng $Q$ by the voltage differene between the plates gives us capacitance.

If I have misunderstood anything thus far please point it out to me! Okay, say that there is some dielectric material which takes up some 3D space between each plate. If that is the case, then the permittivity $\epsilon$ in the first equation now varies in x,y,z as a piecewise function $\epsilon (x,y,z)$. Here's where my first question arises. When solving the Poisson equation, since $\epsilon (x,y,z)$ varies in space, should it be moved to the left side of the equation thusly?: $$\nabla_{(x,y,z)}\text{ } \epsilon (x,y,z) \text{ } \nabla_{(x,y,z)} V (x,y,z)=-\rho$$ Where this first $\nabla$ is the divergence operator and the second is the gradient operator. Assuming the charge density is still defined as the same boundary conditions as previously, we can solve this equation to get the potential field and then vis a vis the capacitance. Here is where my second question arises: how would you set the dielectric boundary conditions on this equation so that the electric field vectors $\vec{E}$ tangential to the dielectric surface at any point are unchanged and that the electric flux vectors $\vec{D}$ normal to the dielectric surface remain unchanged at any point? Furthermore, do these boundary conditions only hold when calculating the electric field vectors, or when solving the Poisson equation, or both?

I am using Mathematica to run a finite element analysis simulation of a capacitor and I want my capacitance results to converge with the parallel plate model. Thus far I have been unable to get them to converge except in the case where the permittivity is constant throughout the 3D mesh.

  • $\begingroup$ The parallel plate capacitor model is an approximation, so your (hopefully correct) FE code will not converge towards it. $\endgroup$ Apr 13, 2023 at 11:41

1 Answer 1


You have many small mistakes here and there.

The integral of the electric field does NOT give you the total charge. I mean, let's consider the infinite parallel capacitor plates situation with vacuum as the dielectric. Then the E field is perfectly constant, integral along the perpendicular line between the capacitor plates gives you voltage, and any other integral gives you just area, and hence no charge, only voltage times area, a meaningless integral. You have better chance with E field squared, but still not what you think you are integrating.

If you wanted to solve this with a dielectric, then you should be using $$\vec \nabla \cdot \vec D = \vec \nabla \cdot \left ( \vec{\vec \epsilon} \cdot \vec E \right ) = \rho_f$$ $$\vec E = \vec{\nabla \phi} $$ This is actually a complicated problem that couples a lot of nonsense together. It is NOT equivalent to a Poisson equation because the constitutive relation can be actually a lot more complicated than even the assumption of a tensorial dependence (that can vary everywhere in spacetime!) of $\vec D = \vec{\vec \epsilon} \cdot \vec E$, by which I mean that you can even have hysteresis, say. Of course, in the most simple case where the relative electric constant is just a constant, maybe this thing simplifies back to a Poisson equation, but the boundary conditions are horrible even in this simplest case.

You might want to find as much of simplification and of comparable to real world data as you can. Physics can grow hairy really quickly!

  • $\begingroup$ I really appreciate the feedback! It's been years since I took electromagnetism physics in undergrad so I'm kinda rusty on these things. $\endgroup$ Apr 13, 2023 at 6:21
  • $\begingroup$ Same here. It is, however, something we learn by horror of having seen the catastrophe. It carves a fear directly onto the soul, leaving all of us scarred. At least, those of us who have tried to witness graduate level physics. We can only usher the innocent undergrads away from the abyss $\endgroup$ Apr 13, 2023 at 6:27

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