Electric potential of various points on a slightly conductive sheet connected to two electrodes There is a common experiment in which two electrodes are connected to a slightly conductive sheet of paper, and a multimeter is used to measure the potential at various points and even find equipotential lines.
 
I am trying to find the potential theoretically/mathematically to help my general understanding:
Suppose there is an infinite, slightly conductive sheet of paper that takes up the $(x,y)$ plane. 
Point $A$, e.g. $(-1,0)$, is the negative electrode, the reference point $0\text{ V}$. 
Point $B$, e.g. $(1,0)$, is the positive electrode and its potential, set by the power source, is $1\text{ V}$.
How do I go about finding the potential at any arbitrary point on the paper?
 
Thoughts so far:
I have a feeling the actual resistance of the paper does not have any effect as long as it is uniform, but no real proof -- i.e. if I connect 0V to 1V with two equal resistors, the midpoint is 0.5V regardless of what the resistances actually are.
Perhaps I could do $\int \mathbf{E}\cdot d\mathbf{s}$ from the 0-volt point to the point in question to evaluate the potential-- it seems easy but I don't know how to find $\mathbf{E}$ at any point either; so far my knowledge of $\mathbf{E}$ is from point charges or charge distributions.
Maybe this has something to do with laplace/dirac delta...
 
EDIT: maybe this is more feasible with, instead of point electrodes, some small spherical electrode with finite radius $r$
 A: If we assume that the paper is infinite, then the problem can be solved exactly.  However, this problem is often used in upper-level E&M classes, and the full solution is rather involved, so I'm only going to sketch how it's done.
In particular, this problem would be equivalent to a problem in 3D where the potential is held at two different potentials along two parallel "infinite" pipes.  For the sake of argument, I'm going to assume that these potentials are $+V_0$ and $-V_0$;  if the two potentials are not additive inverses of each other, you can always subtract a constant from both to make them so, solve the problem, and then add the constant back in.  (In your case, this would mean solving for the case where the electrodes are at $\pm 0.5$ V, and then adding 0.5 V to your solution.)
To solve this problem, we have to examine a similar problem:  imagine two line charges $\pm \lambda$ a distance $2a$ apart.  Let's define our coordinates so that the line charges are running along the $z$-axis, piercing the $xy$-plane at $(\pm a, 0)$.  The potential due to a line charge is a pretty standard problem in intro E&M, so I'll just cite the result as
$$
V(r) = -\frac{\lambda}{2 \pi \epsilon} \ln r + A
$$
where $r$ is the distance to the line charge, and $A$ is an arbitrary constant.  With two line charges $\pm \lambda$, the total potential is therefore
$$
V(r) = -\frac{\lambda}{2 \pi \epsilon} \ln r_+ + A_+ + \frac{\lambda}{2 \pi \epsilon} \ln r_- + A_- = \frac{\lambda}{2 \pi \epsilon} \ln \left( \frac{r_-}{r_+} \right) + A_+ + A_-,
$$ 
where $r_+$ is the distance to the positive line charge, $r_-$ is the distance to the negative line charges, and $A_+$ and $A_-$ are constants.  If we want the potential to go to zero at large distances, this implies that $A_+ + A_- = 0$.
Now, what's kind of remarkable about this is that the equipotential surfaces are also long "pipes", but that they're not concentric with the line charges.  By picking the values of $a$ and $\lambda$ correctly, you can always find a situation where two of these equipotentials have potential $\pm V_0$, have a given radius $r$, and are a distance $d$ apart.  (In your case, $d = 1$.)  But since solutions to Laplace's equation are unique, this means that the potential outside of these equipotentials surrounding the line charges is exactly the same as when you have two long pipes, each one held at potential $\pm V_0$.  
What does this mean if you're just trying to gain intuition?  Well the potential outside these pipes will "look" like the potential of two long line charges.  These "line charges" will not quite be concentric with the "pipes"—though if the separation of the electrodes is much larger than their size, they're pretty close to being aligned.   
