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What is the word or way of referring to a neutral electrical potential. For example let's say you have a 10 cm sphere of electrically neutral mass that is 1 Kg mass and you separately have a different 10 cm sphere of 2 Kg mass. The electrostatic potential of either of these two spheres at any distance will be 0.

However the amounts of positive and negative charge will be twice as much in the more massive sphere in comparison to the less massive one. The neutral field density of the larger mass will be twice that of the smaller mass. What is the term for the neutral electric field density? Does anyone have useful references on this topic?

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  • $\begingroup$ You’re talking about the charge density. The spatially-averaged charge density for both spheres is zero. In each sphere, the positive charge density of the protons is balanced by the negative charge density of the electrons. $\endgroup$ – G. Smith Dec 11 '19 at 4:34
  • $\begingroup$ What's wrong with "the amount of positive or negative charge"? $\endgroup$ – knzhou Dec 11 '19 at 4:35
  • $\begingroup$ I'm talking about the field intensity at a distance. For a polarized charge it would be the positive or negative potential at any particular distance (q/r). For neutral electric fields I don't know how to refer to it so that people know what I am talking about. $\endgroup$ – DMac Dec 11 '19 at 4:45
  • $\begingroup$ What is the field that is measured at 2 or 3 meters distance from the sphere called that is twice as large for the neutral 2 Kg sphere than for the 1 Kg sphere? $\endgroup$ – DMac Dec 11 '19 at 4:48
  • $\begingroup$ It's called the "gravitational field". I know, that's a dumb answer, but the answer certainly isn't the electric field, which is zero. $\endgroup$ – knzhou Dec 11 '19 at 4:54
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let's say you have a 10 cm sphere of electrically neutral mass that is 1 Kg mass and you separately have a different 10 cm sphere of 2 Kg mass. The electrostatic potential of either of these two spheres at any distance will be 0.

Assuming you mean that these masses are uniformly charged and neutral, we'd say the field due to them is zero, and that there's no gradient in the potential around them (assuming they exist alone in the universe with nothing else to create electric fields around them).

If one of the masses is overall neutrally charged, but it has a positive charge on one half and a negative charge on the other, then you'd have an electric dipole and the field around it would not be zero.

As an aside, you shouldn't expect the charge on an object to be proportional to its mass. You can have a 100 kg object with 1 nC net charge, or a 1 mg object with 1 C net charge. Charge is, in the macroscopic world, an independent property from mass.

The neutral field density of the larger mass will be twice that of the smaller mass.

In the sense that two times zero is zero, sure.

What is the term for the neutral electric field density?

Since there's no electric field around these masses we don't have a special word for it; we just say "no electric field".

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  • $\begingroup$ Hi Photon, I'm aware of the elementary issues you mention above. I agree, the field is 0, but the concept of a polarized field doesn't capture what I'm trying to find out. I have stipulated that the spheres are neutral, thus the field is 0. One has twice the protons which are neutralized by twice the electrons. One of the spheres has twice the positive flux canceled by twice the negative flux. How do I refer to the difference of these two different magnitude fluxes. There is a flux propagating to infinity. What is the name of the neutral flux? There must be a jargon term quantifying this. $\endgroup$ – DMac Dec 11 '19 at 6:46
  • $\begingroup$ You keep saying fluxes, and yet there is zero flux. Were the spheres truly neutral, there is no field and thus no flux of anything. We have no name for this non-existent thing. $\endgroup$ – Triatticus Dec 11 '19 at 9:51
  • $\begingroup$ @DMac, if there's no field, there's no field. It doesn't matter if it's because of 1 nC neutralizing -1 nC, or 1 kC neutralizing -1 kC. Since there's no effect from the difference between these scenarios, there's no need for a special word to distinguish them. $\endgroup$ – The Photon Dec 11 '19 at 16:46

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