# Do repelling charged objects reach a finite velocity?

Let's imagine I have 2 spheres made out of metal, each with a mass of exactly 1 kg. These spheres are sitting in a place in the universe devoid of any significant masses - meaning there is negligible gravitational force on my beautiful spheres.

I give each sphere a charge of 1/100000th of a Coulomb and move their centers of mass exactly 1 meter away from each other. Then I release them.

By Coulomb's law, they begin to accelerate away from each other, and the force on each one is given by:

$F=kQq/r^2$

I iterated through this calculation in Excel 10,000 times at 1/100th of a second intervals, and came up with an acceleration at t=100s of 2.55*10^-5 m/s^2, and the spheres ended at 187.77 meters away from each other.

My question is, will my beautiful spheres ever reach an essentially constant velocity?

I've tried to approach this theoretically - integrating acceleration with respect to time from t=0 to t=infinity should tell me whether or not there is a finite ending velocity. However, one value for acceleration depends on previous values, so I am not sure where to go.

So, at the end of time, will my spheres be moving at a finite velocity or will they be moving away from each other infinitely fast?

• Hint: use conservation of energy Oct 17, 2018 at 11:05
• The force of interaction (hence the acceleration) tends to 0, as the massive spheres reach higher and higher separation. In the limit $r\to 0$, the acceleration of any of them is 0, thus the velocity will no longer increase, but reach a definite value. See the answer below. Oct 17, 2018 at 11:19

Yes, they will have a finite velocity, and this is because of the nature of the inverse square law. The force decays rapidly enough so that there is an asymptotic approach to a finite velocity. You can solve the equations of motion to see that, but it is much simpler to see it using conservation of energy.

$$E_{initial}=E_{final}$$ $$\frac {K*Q*q}{r}=\frac 1 2*m*v^2$$ $$v=\sqrt \frac {2*K*Q*q}{m*r}$$

I would just like to add that this is not simply because the acceleration goes to $$0$$ as $$r$$ goes to $$\infty$$, which is clear from Newton's Law applied to this case, that is: $$F=ma=\frac {KQq}{r^2}$$

$$\lim_{r\to \infty}a=\lim_{r\to \infty}\frac {KQq}{mr^2}=0$$

Even with the acceleration going to $$0$$ you can still have an infinite velocity at $$r\to\infty$$. This is the case for a force that instead of being $$F \propto \frac1{r^2}$$, is actually $$F \propto \frac1 r$$. In this case it would still be true that $$\lim_{r\to \infty}a=0$$, but it would no longer be true that the final velocity would be finite. That is easily checked by the fact that $$\Delta U \propto ln(r)$$, then in the example given:

$$v \propto \Delta K=\Delta U \propto ln(r)$$ $$\lim_{r\to \infty}v \propto \lim_{r\to \infty}ln(r)\to\infty$$

So in this case, even though $$\lim_{r\to \infty}a=0$$, you still have $$\lim_{r\to \infty}v \to \infty$$

• I think you have a typo in the formula for the initial potential energy (not consequential for this answer): it should have $r$ - not $r^2$ - in the denominator.
– V.F.
Oct 17, 2018 at 12:31