Consider a system of point charges. To calculate the value of an electric field at a point, we consider the contribution of the electric field from all the charges at that point. Consider the following situation:

Say we have a system of two point charges, and I want to calculate the value of the electric field at the point where one of the point charges lies. I'll consider the contribution from the two charges using coulomb's law:

$E=$ $KQ\over r^2$

When I do this what I get is $E=\infty$ since $r=0$, since $ \lim_{x\to 0} f(x)=$ $1\over x^2$ $=\infty$.

Now if we were to calculate the force acting on that charge at that point which is given by:

$F=qE$ , since $E=\infty$, therefore $F=\infty$ and $a=\infty$ where $a$ is the acceleration.

Can someone please correct me?

  • $\begingroup$ Two issues here (1) Don't have a particle be affected by its own field (2) when you say "$r = 0$" you have to pick an origin, which is important if you have two charges because the relevant quantity will be their separation. $\endgroup$ – zeldredge Jun 2 '15 at 20:55
  • $\begingroup$ @zeldredge (1)can you please explain to me why a particle cannot be affected by its own field? (2)when I say $r=0$ I mean the distance between the charge and the point I want to calculate its field strength is zero. In another words, I want to calculate the field at the point at which the charge lies. $\endgroup$ – Omar Nagib Jun 2 '15 at 21:09
  • $\begingroup$ @OmarNagib: electric self-force is a nontrivial problem: math.utk.edu/~fernando/barrett/bwald1.pdf $\endgroup$ – Jerry Schirmer Jun 2 '15 at 22:10
  • $\begingroup$ @JerrySchirmer Does this mean the answer presented by Steven in which He argued that no self-force is possible owing to Newton's 3rd law is incorrect? $\endgroup$ – Omar Nagib Jun 2 '15 at 23:12
  • $\begingroup$ @OmarNagib: self-force is only present if the charge is accelerating. $\endgroup$ – Jerry Schirmer Jun 2 '15 at 23:23

The exposition you give is fine in classical physics. Note though that in classical physics a particle cannot be a point particle, because something has to carry the charge in classical physics formulations. So the fact that one finds infinity at r=0 just hits on this constraint. One could use the argument as a proof by reductio ad absurdum that particles should have a size .

Elementary particles are point particles but they also are quantum mechanical objects. The realm of quantum mechanics is the realm of the Heisenberg uncertainty principle. The location of the elementary particle is uncertain within the bounds given by the HUP. The microworld of point particles has different rules. The value of the field of an electron meeting a positron becomes irrelevant when they annihilate . All this becomes mathematically rigorous by the solution of the quantum mechanical equations.

In general whenever classical physics give infinities one finds that the quantum mechanical formulation eliminates them. And quantum mechanics is the underlying level of nature from which all classical fields and their equations emerge.


Can someone please correct me?

There is nothing to correct. You are not wrong.

Say your charge is negative.

If you put a positive charge exactly on the negative charge with distance $r=0$ between them, then yes, the force is quite large. It is more difficult to bring them appart than if there was a larger distance. So, there might happen some kind of merging now (which is for another topic), since the force is so great.

Surely, you cannot have exactly zero distance, but your theoretical thinking is correct. Consider a field drawing. Where field lines are closer, the field is stronger. When they meet (which will only in theory be exactly at the charge location) electric field would in theory be infinit.

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From the elaborated question in the comments:

Why this charge is not affected my its own field which in this case should be infinity since $r=0$?

Consider a charge like a planet. The planet flows around in space and just is there. It doesn't pull in itself; it's gravity doesn't make itself accelerate.

Now, if a large asteroide is near, then the planet starts to attract this astroide. It pulls in it. This will move the planet too. From Newton's 3rd law, the force it pulls with is also acting on itself, pulling itself forward with exactly the same force. But in the opposite direction.

Now, think what would happen if it pulled in itself with its gravitational force. It will pull itself in "some" direction. But at the same time, from Newton's 3rd law, the exact same magnitude of force will pull it forward in the opposite direction. There will be no net force.

Similarly, when I pull up in my own hair, I will not fly. The exact same force, with which I am pulling, is also pushing me downwards.

Similarly for any force from any "object". Similarly for charges. Nothing can exert a net force on itself.

  • $\begingroup$ I think I did not explain myself enough. Let's forget about the previous two charge systems and consider a physical system that consists of only one point charge. Why this charge is not affected my its own field which in this case should be infinity since $r=0$?(the distance between the point in consideration and the charge is zero). So can you please edit your answer to address that? $\endgroup$ – Omar Nagib Jun 2 '15 at 21:13
  • $\begingroup$ @OmarNagib If you place a charge on the table, if will attract other charges. If there are no other charges, nothing will happen. This is simply Newton's 3rd law. If you pull yourself up by the hair, you will not levitate; it doesn't matter how strong you are. $\endgroup$ – Steeven Jun 2 '15 at 21:18
  • $\begingroup$ I would be quite thankful if you elaborate more in detail on how it's related to Newton's 3rd law. $\endgroup$ – Omar Nagib Jun 2 '15 at 21:31
  • $\begingroup$ @OmarNagib Sure. I have made an addition to the answer. Please let me know, if this does not clear it out. $\endgroup$ – Steeven Jun 2 '15 at 22:16
  • $\begingroup$ Although your argument is convincing to me, are you sure the answer is that trivial? I'm saying this because I bumped into this: link And @JerrySchirmer gave me this link: link So what are your thoughts about that? $\endgroup$ – Omar Nagib Jun 2 '15 at 23:09

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