High School Resistance Between Two Small Conducting Spheres

[ergospherical]

Here's a little snack; what is the resistance between two small conducting spheres, each of radius $r$, separated by a distance $d \gg r$ within a material of resistivity $\rho$ (of infinite expanse)?

 

[bob012345]

In an infinite network of unit resistors $R$ in 3D the resistance between two points far apart is $R≈0.5055Ω$ becoming independent of distance. I suspect this will also be a constant for large separation for two small spheres at $d\gg r$ independent of $d$. Dimensional analysis gives $R = \Large\frac{\rho}{r}$.

 

[ergospherical]

Nice intuition! However, dimensional analysis tells you that $R = c \cdot \dfrac{\rho}{r}$, where $c$ is some dimensionless constant to be determined. A hint is to use the principle of superposition.

 

[bob012345]

Nice intuition! However, dimensional analysis tells you that $R = c \cdot \dfrac{\rho}{r}$, where $c$ is some dimensionless constant to be determined. A hint is to use the principle of superposition.

Well, that was research guided intuition, I don't think I could have come up with it totally out of the blue.

Given your hint, if we look at one sphere and integrate the total resistance to infinity it is;
$$R = \int_{r}^{∞} \large \frac{\rho}{4\pi r'^2} \,dr' = \large \frac{\rho}{4\pi r}$$ (Edit: where $r$ is the radius of the sphere).

But the other sphere also sees the same resistance out to infinity so if we think of a hypothetical current source injecting current at one sphere which goes to infinity and then returns from infinity into the other sphere, we have double the resistance, they are in series.

$$R = R_{in} + R_{out} = \large \frac{\rho}{4\pi r} + \large \frac{\rho}{4\pi r} = \large \frac{\rho}{2\pi r}$$

So the constant is $\large\frac{1}{2\pi}$.

 

[ergospherical]

That looks like the correct factor, yeah! You can think of the situation as the result of superposing the radial current fields $\mathbf{j}_{\pm} = \dfrac{1}{\rho} \mathbf{E}_{\pm} = \pm \dfrac{V}{2 \rho r} \mathbf{e}_r$ of two spherical current sources of potential $\pm \dfrac{V}{2}$ and separated by a distance $d$. In the limit of $d \gg \tilde{r}$, the current emanating from the positive one (and entering the negative one) can be taken to have a contribution from only that particular source, $I = \displaystyle{\int} \mathbf{j} \cdot d\mathbf{S} = \dfrac{2 \pi \tilde{r} V}{\rho}$ (with the integral taken over a surface just outside the sphere), and the resistance is $V/I = \dfrac{\rho}{2\pi \tilde{r}}$.

(N.B. edit: changed the radius of the spheres to $\tilde{r}$ to avoid confusion with the radial coordinate $r$)