How does the force between two charges becomes zero? In case of Coulombs law, 
$$
F=k\frac{q_1q_2}{d^2}
$$
So, if $F=0$, then $q_1q_2$ must be 0. Is this true?
 A: What Jim is talking about in the comments is a limit: for $d$ becoming increasingly large, $F$ becomes increasingly small.
For example, if $q_1 = q_2 = 1\,\text{C}$ and the charges are a distance $d = 1\,\text{m}$ apart, we find that the force is
$$F = k \sim 10^{10}\,\text{N}.$$
If the distance is made 1000 times bigger, the force is
$$F = k/(1000\,\text{m})^2 \sim 10^4\,\text{N},$$
i.e. $10^6$ times smaller. If the distance is increased by another factor 1000, the force is decreased by a factor $10^6$ yet again and becomes $F \sim 10^{-2}\,\text{N}$.$^1$
So even though the distance between $q_1$ and $q_2$ can never actually reach infinity, it's clear that $d$ can become so big that the Coulomb force becomes negligible. At this point you can treat $F$ as being exactly zero without introducing any noticeable errors.

$^1$ Note that the Coulomb is a large unit and so $1\,\text{C}$ is already very extreme. More realistically you'll see values $|q| \sim 1\,\text{mC} = 10^{-3}\,\text{C}$ which would give you a force $F\sim 10^4\,\text{N}$ for charges only one meter apart already. For $d = 1000\,\text{m}$ then, you'd have $F\sim 10^{-2}\,\text{N}$. Since the Newton is a fairly small unit, this value for the force is already immensely small as compared to other forces that might be present.
For example, if the charges are tiny little identical metal cubes of mass $m = 1\,\text{kg}$ (a fairly realistic value for the given charges) lying on a plastic table, the static friction will be large enough to keep the cubes from sliding towards each other for a Coulomb attraction of up to $1$-$10\,\text{N}$. This is 2 to 3 orders of magnitude larger than the actual Coulomb force between the charges so that we can safely neglect the latter.
A: $d$ can be infinite only in math books.  Your statement about Coulomb's Law is true. if $F=0$ then $q_1q_2=0$.
As $d$ gets arbitrarily large $F$ will get arbitrarily small.  You can find a $d$ that makes $F$ as small as any number you choose ... except zero.
I'm interpreting your question as being about Coulomb's Law itself, not any practical application which would need to account for charge distributions, and the practical limits of measuring a very small force.
