# Resistance in a Medium Between 2 Spheres

1. Oct 24, 2013

### mindarson

1. The problem statement, all variables and given/known data

Two small conducting spheres with radii a and b are imbedded in a medium of resistivity ρ and permittivity ε with their centers separated by a distance d >> a, b. What is the resistance R between
them?

2. Relevant equations

R = V/I

I = ∫JdS = (1/ρ)∫EdS

3. The attempt at a solution

I assume a charge +Q on the sphere of radius a and a charge of -Q on the sphere of radius b. (As you'll see, my answer for the resistance doesn't depend on the charge, which physically makes sense: why would the resistance of the medium depend on the charges of the spheres?)

The resistance is given by ΔV/I, where ΔV is the potential difference between the spheres and I is the current. I calculate the current using the equation

I = ∫JdS

where the integrand is a dot product. I can solve this using Gauss's law/Ohm's law:

I = ∫JdS = σ∫EdS = Q/(ρε0)

I'm not entirely sure of the move I just made. To be clear, I draw a gaussian surface around EITHER sphere, and compute the current via Gauss's law. Is this correct? It seems to me I can only have one current here, and that's the only way I can think to get it. But something does seem fishy, at the very least because one of the spheres is negatively charged and one is positively charged, so I'd get opposite currents by using Gauss's law on each sphere in turn.

Now I find the potential difference between the spheres. Using superposition, the potential on the surface of the sphere of radius a is

Va = (kQ/a) - [kQ/(d-a)]

where d-a is just the distance between the center of the sphere of radius b and the surface of the sphere of radius a. Similarly the potential on the surface of the sphere of radius b is

Vb = (-kQ/b) + [kQ/(d-b)]

Now I assume that a and b are both negligible compared with d (as implied in the problem statement). Thus d-a = d-b ≈ d. Then difference between the 2 potentials is

ΔV = Va - Vb = kQ(1/a + 1/b - 2/d)

Now I can calculate the resistance. After some algebra, I get

R = ΔV/I = (ρ/4∏)[(1/a) + (1/b) - (2/d)]

Is this looking pretty good?

I'm still very unsure about my current calculation....

2. Oct 24, 2013

### Staff: Mentor

You have to be careful here. You assume a constant EdS. For a spherical surface around one of the spheres with a distance <<d, this is a reasonable approximation, but you should make it clear that this is not exact. With one sphere, it would be, but with two spheres the field is not symmetric any more. Superposition will not help, as the charge distribution on the surface can depend on the other sphere.

Looks good, and you can neglect 2/d here.