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The question Can a Neutron Star become charged? and the associated comments and answers started me wondering; Are there measurements of, or experimental limits to the residual charge of the Sun?

Because the Sun has a complex atmosphere and a net solar wind, I there may be some subtleties related to what sphere is used to establish the net charge Q inside, so rather than me define a radius it may be better to see what evidence is available.

Since there is an ionized component of the solar wind and protons are 2,000 times heavier than electrons, I'm thinking that an initially neutral Sun would loose electrons more rapidly than protons, until a strong enough static field was produced that would add the extra "push" for the protons to leave at the same rate, but that's an extremely simplistic model.

I'm asking primarily for some kind of measurement or experimental data rather than purely rationalization or hand-waving. Has there ever been an attempted measurement of the static residual charge of the Sun?

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  • $\begingroup$ Before discarding any rationalizations, you have to realise that any attempt at measuring what you propose will go through rationalizations first. Particularly the rationalization about the size of the Debye-sphere (en.wikipedia.org/wiki/Debye_length) of the solar plasma will thwart any thought of doing what you propose in it's infancy. $\endgroup$ Oct 28 '17 at 14:34
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    $\begingroup$ @honeste_vivere thanks for your comment. If there is some accepted, peer-reviewed work that demonstrates that the presence of a supersonic wind makes a measurement of a residual charge impossible, or a measurable charge impossible, that might be a good answer here. Of course a better answer would be "yes there has been a measurement or experimental limit", or "no there hasn't", because I didn't ask if a measurement should or could be made, but instead, asked if one has been. $\endgroup$
    – uhoh
    Oct 30 '17 at 0:58
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    $\begingroup$ @AtmosphericPrisonEscape I don't believe that saying "there exists a Debye length so the question is moot" is correct, and I don't believe the solar wind can mask an arbitrarily large amount of charge in 10 meters. If you can show otherwise, post it as an answer where it can be voted on. Or consider posting an answer to the question as asked. $\endgroup$
    – uhoh
    Oct 30 '17 at 1:00
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    $\begingroup$ @honeste_vivere I hadn't seen that before. OK I'll take a look and give them a read, thanks! $\endgroup$
    – uhoh
    Oct 30 '17 at 18:23
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    $\begingroup$ I am guessing that the move to chat option has been offered to each of you by now. This exchange is not really helpful (or interesting) for anyone else to read. Why not move it to a chat room, delete here, and when it's resolved just add some kind of conclusion instead; ideally as an answer to my question. Comments here should not be used for an extended discussion between two other users, that's what chat is for. Thanks!! $\endgroup$
    – uhoh
    Dec 5 '18 at 4:15
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Some of the comments here seem to be suggesting that there should not be any residual charge of the Sun at all because of the fact that in a conducting medium no electric fields can exist. This argument ignores the crucial point here, namely that there are unequal numbers of positive and negative charges, because electrons, unlike ions, can easily escape from the gravitational field of the sun (in fact, virtually all of them would escape without an electric field holding them back). And any object with an unequal number of positive and negative charges will appear charged from the outside.

The net charge resulting from the escape can easily be calculated from the fact that any particle with a kinetic energy higher than the absolute value of the combined potential energy due to gravity and any net charge $Q$ inside a sphere of radius $R$ will escape from the Sun. So for an electron this applies for energies

$$K_e > \phi_e =\frac{GMm_e+Qe}{R}$$

and for ions for energies

$$K_I > \phi_I = \frac{GMm_I-Qe}{R}$$

where $G$ is the gravitational constant, $M$ the mass of the sun and $e$ the absolute value of the elementary charge (using cgs-units here)

In order to achieve a kind of steady state, we must have the same amounts of positive and negative charges escaping, i.e. we must have for the energy distribution functions

$$f_e(K/\phi_e)=f_I(K/\phi_I)$$

where $K$ is now taken as a general energy variable.

In thermal equilibrium, the distribution functions of the electrons and ion will be the same i.e. $f_e=f_I=f$ (it should be given by the Maxwell-Boltzmann distribution, but knowing the exact form is not even required here), which means we have the condition $\phi_e=\phi_I$ (in other words, for electrons and ions of the same kinetic energy to have the same escape rates, they must have the same potential energy) i.e.

$$GMm_e+Qe = GMm_I-Qe$$

and thus

$$Q=\frac{GM}{2e}(m_I-m_e)$$

Inserting the constants for this (with $m_I$ the proton mass) and converting to SI units gives $Q=77$ Coulombs for a star the mass of the sun (this value is identical to the one derived in the paper by Neslusan (which has already been mentioned a couple of times on SE), but I think my derivation here is more straightforward and easier to understand).

It is remarkable that the charge does only depend on the mass of the star and not for instance on the plasma energy.

For the electric field near the surface of the sun at radius $R$ we get therefore from Coulomb's law

$$E=\frac{Q}{R^2} = 1.4\times 10^{-6} \frac{V}{m}$$

(after again converting from cgs to SI units).

This electric field is very small. It means that over the size of an atomic orbit the corresponding electric potential energy varies only by about $10^{-16} eV$. This would change the wavelength of spectral lines only by an amount which is 12 orders of magnitude smaller than the observed width of spectral lines, so spectroscopically this is impossible to detect.

However, as you mentioned the solar wind: the fact that this is observed to be quasi-neutral to a high degree shows trivially that the sun must be positively charged by the amount derived above. If the sun would be perfectly neutral there would be a massive excess of electrons in the solar wind (of course, this would then in turn charge up the sun, so such an assumption would be logically inconsistent in the first place).

It should also be of observable relevance for modelling the solar atmosphere, because the electric field, although very small, effectively halves the gravitational acceleration for ionized atoms, hence resulting in twice the density scale height compared to the neutral atmosphere (a fact that is also well known from observations of the earth's ionosphere).

As far as direct direct experimental verification is concerned, one should not overlook the fact that the electrostatic force on an ion is not only $-1/2\times$ the gravitational force near the sun (as follows from the above theoretical consideration) but in principle also at any other distance. At the earth, both should be about a factor $2\times 10^{-5}$ smaller, so the electric field would lead to an acceleration of an ion of about $3\times 10^{-4}$ the earth's gravitational acceleration of $9.81 m/s$. Within a couple of minutes, an ion initially at rest would therefore be accelerated to a speed of the order of $1m/s$ due to the Sun's charge. The problem will obviously be to eliminate any other electric fields whilst avoiding shielding the field by the experimental setup. I don't know whether this is technically feasible in practice, but in principle it should be possible. Gravimeters are considerably more sensitive than this these days, so at least the effect of both the Earth's and the Sun's gravity could easily be subtracted from the observed acceleration.

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  • $\begingroup$ What is the voltage and capacitance of the Sun? And where do all the electrons go, from all the stars in the universe? $\endgroup$ Jul 12 at 2:31
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    $\begingroup$ @KeithMcClary The voltage is given by $1/2$ the gravitational potential energy for a proton near the surface of the sun which is about 2 keV, so the sun must at a voltage of +1kV. With a charge of 77 Coulomb, this results in a capacitance of 0.077 farad. The surplus escaped electrons of all the stars will be in interstellar space. It is only a relatively very small amount though, so probably not detectable in practice. $\endgroup$
    – Thomas
    Jul 13 at 21:35
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    $\begingroup$ This just in: Physicists describe sun's electric field $\endgroup$
    – uhoh
    Jul 14 at 14:56
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    $\begingroup$ @uhoh Right on cue then. The paper is not accessible yet though on the ApJ website(the last issue listed there is from July 10). It doesn't seem to be on arXiv either. $\endgroup$
    – Thomas
    Jul 14 at 20:09

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