# Kinetic energy of two charged balls at infinite distance between them

If I have two balls with masses and charges $m_1, q_1^{+}$, $m_2, q_2^{+}$, initially held at distance $d$, and then released, how can I know the kinetic energies of each of the balls at infinite distance between them? I'm quite stuck on that, because they both have the same potential energy at the beginning, and it decreases not in the same pattern, as if one of the balls was stationary. So it not only falls like $1/R$, because at the same time, the other ball that is causing this potential energy, is also being repelled. So how can I really find out the energies? I tried to apply the conservation of energy law, because I know that at infinite distance from each other they'll have zero potential energy, thus all the initial was transformed into kinetic form, however I'm stuck with the initial potential energy (they both have it, so should I put $2U_p$?), and even so, I can't find their kinetic energies separately, without having another equation.

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Maybe conservation of momentum will help? Also, it doesn't really matter "how" they're getting to infinity, because energy doesn't depend on that history. – Lagerbaer May 20 '13 at 20:25
@Lagerbaer - I thought about it. However, a quantitative question in my book, leads to a conclusion that even the conservation of energy equation that I wrote, is wrong ($2U_p=E_{k,1}+E_{k,2}$). That's why I want to know what am I missing here. How should I treat the potential energy of the whole system? – www May 20 '13 at 20:31

Potential energy is a property of the system, not any one object. Thus there should only be one copy of the typical $1/r$ potential energy between two charges (plus an analogous gravitational term if that can't be neglected).

The easiest way to see this is to start from "infinite" separation. Instead of pushing the two charges together, hold one fixed and move the other toward it. The moving charge must fight the standard Coulomb force (with a little help from gravity) to get closer to the stationary one, so the potential energy obtained here is just the integral of this force over the distance traversed ($d$ to $\infty$).

But what about the stationary object? Well, sure, we need to exert a force on it to keep it from being repelled by the approaching charge. But it is not moving, so the change in $\vec{F} \cdot \vec{x}$ energy vanishes.

The fact that at some point in the future we will let both objects move doesn't change the potential energy, so you should get the same potential energy as if the problem were stated:

A point mass $m_1$ with charge $q_1$ is fixed at the origin. Another point mass $m_2$ with charge $q_2$ is brought in from infinity. What is the potential energy of the system?

It may also help to remember that "$2\infty = \infty$." Moving objects from $x = -\infty$ and $x = \infty$ to the origin covers the same distance as moving one object from $x = \infty$ to the origin.

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It might just be me, but I don't think that your last sentence about moving objects to and from infinity and so on makes sense. Infinity isn't a number, so you can't multiply by two. Just a thought. – Greg May 20 '13 at 22:00
@Greg Indeed you can't, hence the quotes. This is just shorthand for $\lim_{x\to\infty} f(2x) = \lim_{x\to\infty} f(x)$. That is, there is no factor of two we are missing by considering one stationary + one "going to infinity" object, rather than two "going to infinity in opposite directions" objects. The potential energy doesn't care. – Chris White May 20 '13 at 22:05

I would suggest you to use this equation: $$W=\int_C{Fdx},$$ where $F$ is the force on an object and $W$ the work done by this force.

In this case there are two types of forces acting on the two objects, gravitation and Coulomb force: $$F_{result}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}-G\frac{m_1m_2}{r^2}=\left(\frac{q_1q_2}{4\pi\epsilon_0}-Gm_1m_2\right)\frac{1}{r^2},$$ with $r$ the distance between the two objects. So:$$W_{total}=\left(\frac{q_1q_2}{4\pi\epsilon_0}-Gm_1m_2\right)\int^\infty_{d}{\frac{1}{r^2}dr}=\left(\frac{q_1q_2}{4\pi\epsilon_0}-Gm_1m_2\right)\frac{1}{d}$$

Edit: This isn't entirely correct, since I am assuming that this is a symmetric situation, so $m_1=m_2$ and therefore $W_1=W_2=\frac{W_{total}}{2}$. This will affect the ratio of distance to the origin of the two objects and therefore the amount of work done on each object.

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First of all, thanks for the reply. This integration simply leads to a potential energy formula (by the way, we neglect gravity). My question was a bit different. I asked, what is the initial potential energy of the system, and how can I formulate the conservation laws with respect to this problem. – www May 20 '13 at 21:13
You can choose your initial potential energy (choose at which distance your potential energy is zero). In this sort of situation it is common to choose the position at which the potential energy is zero at infinity. – fibonatic May 20 '13 at 21:20