B Resistance Between Two Small Conducting Spheres

  • B
  • Thread starter Thread starter ergospherical
  • Start date Start date
  • Tags Tags
    Resistance
AI Thread Summary
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.
ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,097
Reaction score
1,384
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)?
 
Last edited:
  • Like
Likes Hamiltonian, Keith_McClary and Baluncore
Physics news on Phys.org
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}##.
 
Last edited:
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}##.
 
Last edited:
  • Like
Likes 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##)
 
Last edited:
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
I passed a motorcycle on the highway going the opposite direction. I know I was doing 125/km/h. I estimated that the frequency of his motor dropped by an entire octave, so that's a doubling of the wavelength. My intuition is telling me that's extremely unlikely. I can't actually calculate how fast he was going with just that information, can I? It seems to me, I have to know the absolute frequency of one of those tones, either shifted up or down or unshifted, yes? I tried to mimic the...
Back
Top