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$$V(r)=k_eq/r$$

What is the electric potential when the test charge is placed very very close to the source charge, ie when $r=0$? Is it infinity?

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  • $\begingroup$ See answers for a similar question regarding potential in $1/r$. $\endgroup$ – user130529 Jan 30 '17 at 16:17
  • $\begingroup$ So, what if the charge is an electron? What is the potential very close to the surface of the electron? $\endgroup$ – Allen Jan 30 '17 at 16:31
  • $\begingroup$ In classical physics, assuming a point charge, the potential $\to \infty$ when $r \to0$. $\endgroup$ – user130529 Jan 30 '17 at 16:47
  • $\begingroup$ But that is physically impossible, right? Would you please explain the concept behind the argument and please do look at @AHB 's answer. $\endgroup$ – Allen Jan 30 '17 at 16:49
  • $\begingroup$ Yes, a point charge in classical physics is physically impossible, it is only a model which is very convenient to use when your are "far" from the source. You may want to use other models if you want to get closer, and you may finally have to enter QM. $\endgroup$ – user130529 Jan 30 '17 at 17:01
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In this case, why don't you just input actual values and see what happens?

You get the electric field by taking negative potential gradient and as you can see, it varies inversely with square of distance. So if the distance changes from 1m to 10^-3 m then the force will be 10^6 times greater. You see where I am going with this? It becomes stronger and stronger and hence you have to do more and more work. Thus the potential will approach infinity.

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