Are two charges attracting each other constantly accelerating?

Using Coulomb's Law: $\frac{k_eq_1q_2}{r^2}$, say we have $2$ charges of $1\ \rm C$ each, separated by a distance of $1\ \rm m$. The force would be $8,987,551,787.3681767\ \rm N$, considering $k_e = 8,987,551,787.3681767\ \rm N m^2 C^{-2}$. So if we were to now set the distance between the $2$ coulombs at $1\ \rm mm$, the force would then be $8,987,551,787,368,176.7\ \rm N$. So my question is, if we were to release the $2$ charges from a distance of $1\ \rm m$, would the acceleration be constant, and more importantly, would that mean the force is also constantly increasing? What would happen when they collide? Does this have to do with what Coulomb's Constant actually represents?

• You are using WAY TOO MUCH precision in your numbers. "Calculator vomit" is useless in physics. In addition, what are the conditions that these charges are exposed to? Is this done in a vacuum? Another comment: narrow your questioning down to one or two questions per post. Sep 13, 2018 at 1:19
• In order to say $F=8,987,551,787.3681767\ \rm N$, you need to have a similar number of significant figures in the charges. Why don't you just cut it down to 3 significant figures? It doesn't look like that kind of precision is relevant anyways.
– user191954
Sep 13, 2018 at 5:12
• @Chair In my case, specificity is important, so please bear with me. As the gentleman who left the answer had said that the electrostatic force does increase as the 2 charges get closer, how could I then find the formula for work being done? It doesn't seem possible with the simple W = Fd, considering the force is changing. Would you happen to know the equation by any chance? Sep 13, 2018 at 5:32
• Integrate the force over the relevant distance, like $\int^B_A F\ dr$. That's the real formula for work which can be used almost anywhere (unlike the simplification $W=Fs$), and it makes sense dimensionally too. We know $F$ is a function of $r$. But I still don't really see how that many sig figs are necessary... the only calculation is changing the position of the decimal point, and you can't use so many sigfigs for $F$ if you have only 1 for each of the charges.
– user191954
Sep 13, 2018 at 5:34
• @SamBTz Sig figs are an indication of how much you know. If the charges are actually $1.4\ \rm C$, it's still acceptable to say that they're $1\ \rm C$ if you can't measure very precisely and you can achieve only 1 significant figure. It's like the fun example: "A tour guide at a museum says a dinosaur skeleton is 100,000,005 years old, because an expert told him that it was 100 million years old when he started working there 5 years ago."
– user191954
Sep 13, 2018 at 6:09