B Resistance Between Two Small Conducting Spheres

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The resistance between two small conducting spheres, each with radius r and separated by a distance d much greater than r, can be expressed as R = (ρ / (2πr)), where ρ is the resistivity of the material. Dimensional analysis suggests a constant factor c, which is determined to be 1/(2π) through integration of resistance to infinity. The discussion emphasizes the principle of superposition in analyzing the current fields generated by the spheres. The total resistance considers contributions from both spheres, leading to the conclusion that they are in series. This analysis provides a clear understanding of the resistance behavior in this configuration.
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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)?
 
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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}##.
 
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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. :smile:
 
ergospherical said:
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. :smile:
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}##.
 
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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##)
 
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