electricity - How do you model the resistance across a symmetric sheet of plastic, streched (think ceran wrap) between a circular anode and cathode?

Here is the problem. You need to model the resistance across a circular sheet of polypropylene (plastic). The resistivity of polypropylene is 1E15 Ohms per meter. The expression for resistance on a wire is:

But, this expression does not help with a sheet of plastic. The geometry of the sheet is shown below. It runs from a 6 mm diameter (along the edge of the anode) to 2 mm inner diameter (along the edge of the cathode).

Here is what I tried. I modeled the sheet like a series of tiny wires. I said each wire was 1E-8 meters wide. Doing this gave me a cross section area for my little wires. Then I multiplied by length and got the resistance along my small wire. Then I multiplied by the number of wires, which was the average diameter divided by 1E-8 meters.

I thought I was pretty clever.

But, this did not work, because the system should converge to a final solution. As the small wires got thinner and thinner, the solution should reach some final number… but it does not.

How do you model the resistance across a symmetric sheet of plastic, streched (think ceran wrap) between a circular anode and cathode?

Answer

Basically, you want to find the proportionality between the total current and the voltage difference between cathode and anode. Let's assume that the current flow is radial under steady-state conditions, which basically allows me to ignore the $z$-direction throughout. In a steady-state solution, we will have $\nabla \cdot \vec{J} = 0$; moreover, if we have a medium of constant conductivity, then the fact that $\vec{J} = \sigma \vec{E}$ implies that $\nabla \cdot \vec{E} = 0$ as well. This means that the potential $V$ will satisfy Laplace's equation, $\nabla^2 V = 0$. Working in cylindrical coordinates, Laplace's equation $$ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} = 0 $$ and since we are assuming no $z$ or $\phi$ dependence, this becomes $$ \frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) = 0 \quad \Rightarrow \quad r \frac{\partial V}{\partial r} = C \quad \Rightarrow \quad V(r) = C \ln r + D, $$ where $C$ and $D$ are constants of integration. These latter constants will be determined by the boundary conditions; if we require $V(r_o) = 0$ and $V(r_i) = V_0$, we get $$ C \ln r_o + D = 0 \quad \text{and} \quad C \ln r_i + D = V_0, $$ which can be solved to yield $$ C = \frac{1}{\ln(r_i/r_o)} \quad \text{and} \quad D = - \frac{\ln r_o}{\ln(r_i/r_o)}. $$ Plugging this back in, we get the solution for $V$: $$ V(r) = V_0 \frac{\ln (r/r_o)}{\ln (r_i/r_o)} $$

Now that we've done this, the rest is more straightforward. The electric field is $$ \vec{E}(r) = - \nabla V = \frac{V_0}{\ln (r_o/r_i)} \frac{1}{r} \hat{r} $$ and so the magnitude of the current density is $$ J(r) = \frac{\sigma V_0}{\ln (r_o/r_i)} \frac{1}{r} $$ (flowing radially.) This means that the total current flowing from anode to cathode will be $$ I = 2 \pi d \frac{\sigma V_0}{\ln (r_o/r_i)} $$ where $d$ is the thickness of the filament; this means, finally, that $$ R = \frac{V_0}{I} = \frac{\ln (r_o/r_i)}{2 \pi d \sigma} = \frac{\rho \ln (r_o/r_i)}{2 \pi d}. $$