If r=0 then will the electric charge of two points=infinity?

[Hami Hashmi]

I found that the electric field at r=0 equals infinity. What if two negative charges were put infinitely close together so the electric field was infinite, then would the charge of those two points be -infinity as well?

 

[Nugatory]

The infinity that you found is telling you that the $F=Cq_1q_2/r^2$ formula doesn't apply and shouldn't be used when $r=0$. It works anywhere outside of a charged sphere (a very large number of interesting and problems, which is why we use it), but $r$ is never zero if you are outside the charged sphere.

 

[Hami Hashmi]

I thought the formula for an electric field E=k*Q/d^2 (where k=constant, Q=source charge, and d=distance between the centers of the two objects)?
http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Intensity

 

[Nugatory]

I thought the formula for an electric field E=k*Q/d^2 (where k=constant, Q=source charge, and d=distance between the centers of the two objects)?
http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Intensity

That is the equation for the field. I gave the equation for the force, which is the product of the field and the charge the field is acting on. So replace the $C$ in my formula with $k$ - it's Coulomb's constant either way - and identify $Q_1$ as the source charge and the two equations are saying the same thing.

 

[PeroK]

I found that the electric field at r=0 equals infinity. What if two negative charges were put infinitely close together so the electric field was infinite, then would the charge of those two points be -infinity as well?

$\frac10 \ne +\infty$

$\frac10$ is undefined.

What you can say is that:

As $r \rightarrow 0, \ \frac1r \rightarrow +\infty$

What this means that as $r$ gets smaller, $1/r$ increases without bound. But, at no time do you arithmetically arrive at $+\infty$.