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What is the net electric charge (in magnitude and sign) of the Sun and its corona?

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According to this post and references there, the charge of the Sun is positive, the magnitude is estimated as 77 Coulombs, or about 1 electron per million tons of matter.

The reason for this is that

The net global charge on the Sun comes about because electrons, being rather less massive than protons, are more able to escape the sun as part of the solar wind. The net charge achieved is a result of the balance between the forces that eject the solar wind, which push electrons more efficiently then protons, and the attractive force on the electrons of the net positive charge that results. Equilibrium of these forces establishes the allowed net charge.

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    $\begingroup$ @honeste_vivere: I respectfully disagree with your estimate. An electric field of a point charge $q$ at a distance $r$ is $q/(4\pi\epsilon_0 r^2)$. When I substitute $q=77$, $\epsilon_0=8.85\times 10^{-12}$, $r=7\times 10^8$ (the Sun radius, SI units everywhere), I get $1.4 \mu V/m$. $\endgroup$ – akhmeteli May 3 '16 at 1:33
  • $\begingroup$ Yes, you are correct... I forgot to square the $8 \ R_{s}$ factor. My colleagues eventually corrected my mistake. $\endgroup$ – honeste_vivere May 3 '16 at 12:20
  • $\begingroup$ I have added some updates to my answer explaining why the Neslusan, [2001] paper in the link in your answer is incorrect at: http://physics.stackexchange.com/a/253491/59023. $\endgroup$ – honeste_vivere May 4 '16 at 13:18
  • $\begingroup$ @honeste_vivere: Your update does not look very convincing: so the model does not describe dynamic phenomena well, this does not mean it does not describe static phenomena. So what charge does the newer model predicts? So far I just see your conclusions about neutralization. Or is it the authors' conclusion? $\endgroup$ – akhmeteli May 5 '16 at 0:44
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    $\begingroup$ @lolmaus-AndreyMikhaylov : Yes, I do understand this, but that does not mean that there is a charge comparable to 1 C in a battery at any moment in time. The energy there is in the form of chemical energy. $\endgroup$ – akhmeteli Sep 7 '17 at 14:00
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Utilizing the accepted number 3.8427 x 10^26W. The charge of the Sun is 2.948124982 x 10^13 As. While the Amp is 1.00995716 x 10^12 A, the volt 3.804814 x 10^14 Kgm2/As3. VA = W and V/A = 376 Ohms. Which is the background impedance of vacuum space. The power is derived from the equation P= oeAT. From which the solar constant of 1366 W/m2 is derived. q = sq.rt.WF/m(4Pir)2/c = As A = qc/4Pir = Amp V = q/4PiF/mr = Kgm2/As3 W = q2c/F/m(4Pir)2 = Kgm2/s3

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    $\begingroup$ We do have MathJax here that makes your equations look better. You can search 'notation' in help center for details, if you aren't familiar with it. $\endgroup$ – Kyle Kanos Jan 3 at 21:46

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