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Both electrons and protons have mass, so it is easy to conclude that they exert a gravitational pull on each other. But also, they are charged, so an electrical force is applied. Would we expect a "double dipping" reaction? Or does the K coulomb constant takes that in consideration?

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    $\begingroup$ You can easily calculate the gravitational attraction. All the masses and constants you need are just a Google away. Your question would be better if you did that. $\endgroup$ – John Rennie Sep 26 '17 at 17:20
  • $\begingroup$ @John Rennie With all due respect, no, it does not. I really want to understand whether Coulomb took that in consideration and how? Beside that, it does not seems very savvy to drive people away from this website to another. We come here to find answers, not as a last resort, but as easy to find and collaborate, or this site only is suited to Physicists. If so, I will be glad to never return. $\endgroup$ – Ernesto Melo Sep 26 '17 at 17:25
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    $\begingroup$ I'm not trying to drive you away. If you did the calculation you'd find the gravitational force is so ridiculously small as to be entirely negligible. So the Coulomb force law doesn't take the gravitational force into account because it doesn't need to. $\endgroup$ – John Rennie Sep 26 '17 at 17:28
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    $\begingroup$ This site isn't just for physicists, but we expect the same sort of effort from all posts that we would from a physicist. And the very first thing a physicist would do is calculate the force just out of curiosity to see what it was. I tried to encourage you to do that because if you had you'd have realised how unimportant the gravitational force is for an electron and proton. Learning to do this sort of thing is what makes physics so fascinating for us physicists. $\endgroup$ – John Rennie Sep 26 '17 at 17:31
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    $\begingroup$ @ErnestoMelo It's survival of the fittest here, but you show a promise :) $\endgroup$ – safesphere Sep 27 '17 at 6:07
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As far as I know, the smallest mass that was shown to have a gravitational attraction on another mass was 706 mg - and that was with a ridiculously careful experiment.

Coulomb attraction of a proton and electron is so many orders of magnitude greater (especially since the mass of the electron is so small) that there's no way it affects the measured force. Only with electrically neutral objects can you begin to measure the force of gravity.

The ratio of the two forces is of course simply given by

$$R = \frac{4\pi\epsilon_0 Gm_pm_e}{Q_p Q_e}$$

Wolfram alpha tells us the ratio is $-4\cdot 10^{-40}$.

That's a lot of orders of magnitude. Well outside the scope of any experiment I can think of.

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This question stems from a misunderstanding, that the coulomb force was determined by measuring the force attracting a positive and a negative charge in a vacuum or something, and attributing all of it to the electric force.

In reality, that's not how Coulomb's constant $k = \frac{1}{4\pi \epsilon_0}$ was determined. More importantly, coulomb's constant is so much stronger than the gravitational constant that it wouldn't have mattered if it was done that way.

It would be like trying to adjust the 'top speed' rating of a car for the gravitational force between the car and all the surrounding buildings when its top speed was tested.

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    $\begingroup$ I just wanted to see the numbers on this, The gravitational force between an electron and a proton is around $1\times 10^{-47} {Kg} m/s^2$ , whereas the electric force is about $2\times 10^{-8} {Kg} m/s^2$. Different by 39 orders of magnitude. $\endgroup$ – JMLCarter Sep 26 '17 at 17:47

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