# Why can't I measure electric potential from source to point according to this formula

The formula for electric potential of points A and B in the presence of an electric field due to a point charge where $R_a$ and $R_b$ are the distances from source to point A and B respectively is: $${V_{AB}}=kq\left({1\over{R_B}}-{1\over{R_A}}\right)$$
But this formula only works when you are measuring distances $R_a$ and $R_b\,\neq0$
I don't understand the physical meaning of the fact that $R_a$ and $R_b$ cannot be zero.

• What happens when you actually put $R_a$ or $R_b$ equal to zero? – AV23 May 15 '15 at 16:34
• well you as can see in the formula you'd be dividing by zero – Yuri Borges May 15 '15 at 16:36
• That's why. The potential gets arbitrarily large when you get closer to the charge. When one of your points is on the charge, the potential difference with another point would simply be infinite. – AV23 May 15 '15 at 16:37
• thank you already! what about opposite charges? – Yuri Borges May 15 '15 at 16:46
• There's only one charge in the situation you've mentioned. – AV23 May 15 '15 at 16:51

Once you accept that there is no such thing as a "point source", but instead everything is a "charge cloud" of finite size, then the singularity goes away. Imagine for a moment two "charge clouds", each of small radius $r$ and with total charge $Q$. Now the charge density at a given point scales with $\frac{Q}{r^3}$ , and if you now look at the potential between any two points, you see that since the charge in a volume element goes down with the dimension of that element cubed, you don't end up with a singularity.