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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.

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  • $\begingroup$ What happens when you actually put $R_a$ or $R_b$ equal to zero? $\endgroup$ – AV23 May 15 '15 at 16:34
  • $\begingroup$ well you as can see in the formula you'd be dividing by zero $\endgroup$ – Yuri Borges May 15 '15 at 16:36
  • $\begingroup$ 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. $\endgroup$ – AV23 May 15 '15 at 16:37
  • $\begingroup$ thank you already! what about opposite charges? $\endgroup$ – Yuri Borges May 15 '15 at 16:46
  • $\begingroup$ There's only one charge in the situation you've mentioned. $\endgroup$ – AV23 May 15 '15 at 16:51
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In reality, "point charges" don't exist. When you get close enough, you find that charge is spread out in a little cloud - just think of the way electrons behave around the nucleus of an atom, or about the Heisenberg uncertainty principle.

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.

That's how an apparent paradox like this can be resolved. Do you need to see the rest of the math, or is it clear like this?

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  • $\begingroup$ Thank you for your answer! Where should I picture the two points in your example? $\endgroup$ – Yuri Borges May 15 '15 at 17:34
  • $\begingroup$ The two clouds can now overlap. Even if they overlap perfectly, the potential will not become infinite. $\endgroup$ – Floris May 15 '15 at 17:35
  • $\begingroup$ I'm actually interested in the formula of the potential between two points in your example! $\endgroup$ – Yuri Borges May 15 '15 at 17:38
  • $\begingroup$ OK - I don't have time now. Will try to do it later. $\endgroup$ – Floris May 15 '15 at 17:38
  • $\begingroup$ take your time! $\endgroup$ – Yuri Borges May 15 '15 at 17:40

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