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Since the electrostatic field and the Newtonian gravitational field share a similar form: proportional to

$$ \frac{1}{r^2} $$

Is there any qualitative difference between motions under the influence of electric field and gravitational field?

Qmechanic
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soundslikefiziks
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  • I can't see how your assumption is justified. The gravitational field is a tensor field while the electromagnetic field is a vector field. The two are different in important respects. – John Rennie Sep 16 '15 at 16:16
  • Sorry, i re-edited my question excluding that assumption. i was really thinking about the kinetic energy – soundslikefiziks Sep 16 '15 at 16:17
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    The difference is that the gravitational field is a tensor field while the electromagnetic field is a vector field. At long range they both obey an inverse square law, but that's a pretty superficial similarity. Gravity is described by general relativity while the EM field is described by Maxwell's equations. – John Rennie Sep 16 '15 at 16:21
  • I see what you mean, But if eventually they both push\accelerate an object(mass\electron) why is the extra "charge" definition needed ? why can't it be said that they simply different in the ways you mentioned. – soundslikefiziks Sep 16 '15 at 16:28
  • I think, can't be sure, someone just assumed the existence of "magnetic" field of gravity, for a very similar reason of why there is magnetic field. Anyway, I don't like it is a bad question, but you have asked it in a way confusing and easy to misunderstand.(including myself at first sign) – Shing Sep 16 '15 at 16:28
  • I'm sorry , i did not intend to confuse, i just had no other way to ask it , if i would ask it in a different way it would mean i already had an answer, which i don't. i am still not seeing why accelerating decelerating masses are so greatly different , why can't it all be said that they are all just forms of gravity with the exception that electric force only effects electrons. – soundslikefiziks Sep 16 '15 at 16:31
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    I'm voting to close this question as off-topic because it shows dearth of any research-effort; questions like this can be easily solved by quick googling or so. –  Sep 16 '15 at 16:32
  • I have helped to edit this question as the way I understand, feel free to tell me if I get it wrong. On the other hand, it is highly suggested here to show some research effort in asking question (doesn't have to be perfect) – Shing Sep 16 '15 at 16:38
  • @user36790, My misconception is not an indication of lack of research , a question occurs when someone just "doesn't get it" and i don't. this is the only reason i decided to ask. – soundslikefiziks Sep 16 '15 at 16:41
  • @Shing , thanks for the edit, much better than my version. – soundslikefiziks Sep 16 '15 at 16:44
  • @soundslikefiziks No problem, it takes practice to ask, and discuss physics. I am still learning same as you. (but English is not my native language, I think there are some english mistakes there) – Shing Sep 16 '15 at 16:46
  • Related: http://physics.stackexchange.com/q/54942/2451 , http://physics.stackexchange.com/q/179401/2451 , http://physics.stackexchange.com/q/944/2451 and links therein. – Qmechanic Sep 16 '15 at 20:41

1 Answers1

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When we focus on classical mechanics and only take charged particles (mass/ actual electric charge) there is only one difference between the trajectories of the particle in an electrical/ gravitational field: in the electric fields particles can have positive/ negatice charge thus move towards/ away of the source (or to put it that way: in the electric field there are charges which never results in bounded solutions).

But even in classical mechanics there are differents between the different charged particles. In gravitational fields, there is only the "$1/r^2" law, but electric fields have a much more forms, depending on the source (dipols ) etc. Also electro dynamics aren't galilei invariant, which a counts for another difference.

As others already said, classical mechanics aren't a good frame to compare both fields and beyond classical mechanics they are completle different.

manthano
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  • So why can't charge be simply a (-) or (+) why must it have the coulomb value? if it's only a matter of the direction of motion (I couldn't Up-vote your answer since my reputation is lower than 15) – soundslikefiziks Sep 16 '15 at 17:50
  • @soundslikefiziks I am upvoting manthano's answer on your behalf if you don't mind. – SchrodingersCat Sep 16 '15 at 18:41