# 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?

• Have a look at "curvilinear squares" - This may help and agh, wow – Russell McMahon Jul 13 '15 at 17:42
• No matter how thin you make your wires, they are not the same thickness near the cathode and near the anode. If you consider a series of circles instead (short fat wires instead of long skinny ones - put them in series) you will find the answer in just a couple of steps - and in fact you will see that there is a closed form solution. Do you want me to write it out, or is the hint enough for your? – Floris Jul 13 '15 at 17:48

## 2 Answers

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 $$R = \frac{V_0}{I} = \frac{\ln (r_o/r_i)}{2 \pi d \sigma} = \frac{\rho \ln (r_o/r_i)}{2 \pi d}.$$

• Very nice and complete answer. – Floris Jul 14 '15 at 15:46

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 $dl=dr$.

$$R=\int_{r_i}^{r_f}\frac{\rho}{2\pi rt}dr$$

$$R= \frac{\rho}{2\pi t}\ln(r_f/r_i)$$