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?
 A: 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
$$
R = \frac{V_0}{I} = \frac{\ln (r_o/r_i)}{2 \pi d \sigma} = \frac{\rho \ln (r_o/r_i)}{2 \pi d}.
$$
A: This problem can be solved using a simple integral. First, we take the formula for resistance and rewrite it small lengths of wire. I also break up the cross sectional area in terms of thickness $t$ and circumference $2\pi r $
$$dR = \frac{\rho dl}{2\pi rt}$$
Think of this as the resistance of a single ring of plastic sheet. Now we integrate over the resistance and radius, recognizing that $dl=dr$.
$$R=\int_{r_i}^{r_f}\frac{\rho}{2\pi rt}dr$$
$$R= \frac{\rho}{2\pi t}\ln(r_f/r_i)$$
