# Electric potential distribution in a volume of varying conductivity

1. Jun 6, 2012

### eman3

Hey,
Assume I have a very large tank of some liquid with uniform but medium conductivity.
I now insert two fine wires into the tank some distance (D) apart, and create an electrical potential difference between them (V).

1) If I were to take my voltmeter (input impedance very high, so we assume it draws no current itself), with its two fine-tipped probes fairly close together (<<0.1 * D), what would the voltage readings look like in different parts of the tank?

2) Now, let's say I drop a flat bar of metal (much more conductive than the liquid) somewhere in the tank between the original probes. How does it affect the voltmeter readings from 1) if the bar is very short? If the bar is long enough to occupy some significant fraction of D?

2. Jun 7, 2012

### Staff: Mentor

You can calculate the potential distribution by solving Poisson's equation in the tank and fixing the potential at the metal wires. Without the metal bar, you get a nice, analytic solution, and you can calculate the potential everywhere. The voltage depends on the positions of the probes.

With the metal bar, it is more complicated, and I think a numeric simulation of the system would be the easiest way, unless the bar follows some area of constant potential or has some other special positions. If it is very short, it might be negligible.

3. Jun 7, 2012

### eman3

Thanks. Having failed DiffEq twice when I took that class, is there e.g an online tool that will generate a nice graphic showing how the solution looks?

I don't need a precise solution, just an order of magnitude estimation...

4. Jun 7, 2012

### Staff: Mentor

If your system is two-dimensional (or does not have significant variations in the third dimension) and the wires are identical, the potential V(x) should be approximately $$V(x) = c\, \left( ln(|x-x_1|) - ln(|x-x_2|)\right)$$ where x1 and x2 are the positions of the wires and c is chosen to get the correct potential at the wire surfaces. You will need the diameter of them. The approximation assumes that the wire diameter is << D.

Excel can produce nice color gradients, and I don't know any online tools.

5. Jun 15, 2012

### eman3

mfb -- thanks, but I'm looking to figure out V(x,y). Your equation gives me an idea of what voltage measurement I might get as I vary the measuring point along the x-axis (assumed to be a straight line between the two wire probes). But what happens if I vary the measuring point along the y-axis as well?

6. Jun 15, 2012

### Staff: Mentor

x has two components in my equation. $|\vec{x}-\vec{x_1}|$ is a short version for $\sqrt{(x-x_1)^2+(y-y_1)^2}$ and I dropped the vector arrows. That was a bit sloppy.