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Consider charge particle $q$ in electric field $\mathbf{E}$. The force on the charge is given by $$\mathbf{F}=q\mathbf{E}$$

Now we know that charge $q$ will also produce an electric field. Due to this field, the field already present in the space should be modify. And thus we should use the modified version of the field. But we don't? (atleast I didn't see).

So the question is if the above reasoning is correct what should be the correct expression? If it's wrong why?

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  • $\begingroup$ I think the question would sound much cooler if it was like of the form " How does the charge of a particle effect the prexisting field of space?" or something $\endgroup$ Commented Nov 5, 2020 at 21:33
  • $\begingroup$ In attempts to answer your question, I went around stack hopping and read this awesome answer by Andrew steane See here $\endgroup$ Commented Nov 5, 2020 at 21:43
  • $\begingroup$ I asked what I think is the same or very similar question here: physics.stackexchange.com/questions/589313/… $\endgroup$ Commented Nov 6, 2020 at 0:56

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Now we know that charge q will also produce an electric field. Due to this field, the field already present in the space should be modify.

Yes, the total electric field has contribution due to the charged particle, in the sense that at all points where total field is defined, its value is different than it would be if the particle wasn't there.

However, modifying the standard equation $$ \mathbf F_{electric~on~particle~a} = q_a\mathbf E_{ext}(\mathbf r_a) $$ which is the basis for experimental definition of external electric field at any given point, into $$ \mathbf F_{electric~on~particle~a} = q_a\mathbf E_{total}(\mathbf r_a) $$

is not warranted.

Why? In case the particle is a point, its field diverges when approaching that point and has no meaningful value at that point - so total field is not defined at that point either. So the second equation would be meaningless.

In case the particle is a ball or sphere or other extended body with finite charge density, total field is defined everywhere, but its has different directions at different parts of the particle. So the equation doesn't really look like the second one above, but more like

$$ \mathbf F_{electric~on~particle~a} = \int \rho_a(\mathbf r_a + \mathbf x') \mathbf E_{total}(\mathbf r_a+\mathbf x')\,d^3\mathbf x' $$ where the integration goes over region containing the whole particle. We can integrate all those little parts and get net electromagnetic self-force as a function of internal structure of the particle, its position and all position derivatives, but only approximately.

This was done by Abraham and Lorentz at the beginning of 20th century. The resulting self-force dependence on position derivatives is complicated, but it has two interesting properties:

  1. there is a component of self-force of the form $-\mu_{EM} \mathbf a$ where $\mu_{EM}$ is some positive constant factor, which for single sign charge distribution depends on total charge and its distribution (size of the particle) and $\mathbf a$ is acceleration of the particle; this effectively increases inertial mass of the particle;

  2. there is component of self-force of the form $k\dot{\mathbf a}$ where $k$ is a positive prefactor that depends only on total charge, it doesn't depend on the size of the particle.

So the equation this leads is something like

$$ \mathbf F_{electric~on~particle~a} = q_a\mathbf E_{ext}(\mathbf r_a) - \mu_{EM} \mathbf a + k \dot{\mathbf a}~+ $$ $$+~\text{other terms depending on motion of the particle}. $$

Effects similar to those of the two additional terms are observed in reality in macroscopic coils and emitting antennas: in coils, increased effective mass of electrons due to mutual interaction between all the accelerating electrons is responsible for the effect of self-inductance; and in an emitting antenna, in addition to the first effect, the force $k\dot{\mathbf a}$ for oscillating current behaves as a friction force $-k\omega^2 \mathbf v$, so this is the force of radiation resistance, sucking energy away from the antenna.

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  • $\begingroup$ Great, thanks. Can you give some reference for the proof? $\endgroup$ Commented Nov 7, 2020 at 12:53
  • $\begingroup$ H.A. Lorentz, The theory of electrons, Notes, section 18. $\endgroup$ Commented Nov 7, 2020 at 15:53
  • $\begingroup$ Is that mean the particle is pushing itself or exerting a force on itself in some sense? $\endgroup$ Commented Nov 7, 2020 at 16:15
  • $\begingroup$ Yes. More accurately, one part of particle acts on another one in such a way that the net resulting force is non-zero. $\endgroup$ Commented Nov 7, 2020 at 21:02
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The electric field that appears in this expression (and in Lorentz force equation more generally) is the total electric field, meaning the field as contributed from all sources. The reason the field due to the point particle in question (which would make it interact with itself) is usually ignored because the field of a point particle diverges at the location of the point particle and it becomes impossible to get anything approaching a reasonable result.

The exact calculation including the interaction of a particle with its own field can be performed but is extremely intensive from a technical standpoint (the calculation appears near the back of Classical Electrodynamics by Jackson). This issue exact issue with point particles can in some respects be pinpointed as a signal that point particles are probably only useful as an approximation than that a better description might involve fields that change in time instead of particles that move around (fields do not run into the same issues with divergent quantities).

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In the absence of particle acceleration the only way I know to alter the force on the particle is by proximity to a (neutral) conductor which would induce asymmetric surface charge on the conductor of the opposite polarity to the particle's.

The quantitative calculations are beyond intro physics. I'll let others delve into Legendre polynomials etc.

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