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.... Thanks for any help you can offer!