For the image below, lets say the top charge is 5C and the bottom 1C. And consider this observation frozen at $t=0$, consider an imaginary horizontal line between the 2 charges denoting the centre of the distance between them They both exert the same equal and opposite forces on each other, equal and opposite forces imply no acceleration of change in velocity. Unfrozen at $t=1$, from $t=1$ to $t=2$ do both charges move the same distance from this horizontal line assuming the mass of both are the same?

enter image description here

  • $\begingroup$ The rules for drawing field likes require that the charge is proportional to the number of field lines surrounding that charge. These are two equal positive charges. $\endgroup$ Oct 25, 2018 at 22:47
  • $\begingroup$ I know but i couldn't find an image, but consider that the top charge is 5C and bottom is 1C $\endgroup$
    – zenarthra
    Oct 25, 2018 at 22:59
  • $\begingroup$ If they are the same mass then they would get equal velocities by F=ma and v=at. $\endgroup$ Oct 26, 2018 at 0:41

1 Answer 1


Coulomb's Law:

$$F=\frac{q_1q_2}{4\pi\epsilon_0 r^2}$$

dictates that the force felt by each of the two charges is equal. Given that their masses are equal and $F=ma$, this also means their accelerations are equal, and thus their distance from the horizontal line will be the same, as a function of time.


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