How can a charged conductor be influenced by its *own* electric field?

Question

Consider two isolated, spherical, conducting shells: one carrying charge $+Q$ and the other $-Q$. The electrostatic potential just outside the surface of a conductor (which equals the conductor’s potential) is given by:
$$ V(a) = -\int_{\infty}^{a} \vec{E} \cdot d\vec{l}, $$ where $a$ is the radius of the spherical conductor.

Now, imagine placing a conducting path between the two spheres. The positive sphere has an outward-directed electric field, and the negative sphere has an inward-directed field. When connected, the field from the $+Q$ sphere attracts electrons, and the field from the $-Q$sphere seems to "repel" them.

Here’s where my confusion arises:

The electric field near the negatively charged conductor is produced by its own excess electrons. How, then, can this same field act on those charges to push them around? In the case of the positive sphere, it's easy to imagine its field attracting negative charges from elsewhere. But in the negative sphere's case, it feels paradoxical to say that the field generated by the charges is also acting on them. This seems like self-interaction, which classical electrodynamics usually avoids.

In case also of the potential, it's the same thing like the field , where that same field is the one that created the potential already so it's the same point

Question:
How is it correct to describe the field of a negatively charged conductor as “repelling” its own surface charges? Doesn’t this imply that a charge is affected by the field it itself generates, which seems physically inconsistent?

Answer 1

Taking a non mathematical approach, imagine the electrons as heavily crowded people in a room. Now, each individual is applying force on another, and together they end up applying force on the walls. The total force on the wall could be imagined for the net field, the pushing as individual electronic force.

When you connect the two spheres you mentioned, you create a opening in the room. People before the opening end up getting pushed out, and with time, others behind as well. That's how they flow out.

• I feel thinking that the field itself applies force may not be the correct way to go about it. It may be better to think of the fields as aids for better visualization instead.

• The electron's field is not acting on itself. The repulsion between that electron and others is what is causing the conduction

I am a student too, so please forgive (and correct) mistakes if any!

Answer 2

Conductor is just a material containing lots of electrons and positive ions, and its field is just the sum (superposition) of the fields created by those. There is nothing wrong with one electron/ion in a conductor acting on another electron or ion, just like there is nothing wrong with two conductors acting on each other.

Now, electron interacting with itself may be somewhat of a problem, which in macroscopic electrodynamics is done by regularizing singularities.

Say, a potential of a charge distribution $\rho(x,y,z)$ is given by: $$ \phi(x,y,z)\propto\int dx'\int dy'\int dz' \frac{\rho(x',y',z')}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}} $$ and the field is $$ \mathbf{E}(x,y,z)\propto\int dx'\int dy'\int dz' \frac{\left[\mathbf{e}_x(x-x') + \mathbf{e}_y(y-y') + \mathbf{e}_z(z-z')\right]\rho(x',y',z')}{\left[(x-x')^2 + (y-y')^2 + (z-z')^2\right]^{3/2}} $$ These are just the Coulomb law integrated over the charge distribution, and the integrands of these expressions contain singularities... but the integrals converge in the mean sense.

The problems do arise when dealing with low-dimensional systems, like carbon nanotubes.

Note also that we can contain the self-interaction energy of this distribution: $$ E\propto\iiint dx dy dz\iiint dx' dy' dz' \frac{\rho(x,y,z)\rho(x',y',z')}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}}, $$ which is again finite, if the distribution has a finite net charge.

Update
After a discussion in the comments and important point was clarified: in continuous electrodynamics we are not talking about many small but finite electrons, but about infinitesimal charges of a continuous charge distribution. The reason why the above integrals converge is because the charge contained in a small region around the divergence point decreases faster than the electric field increases.

If we were dealing with discrete electrons, we would typically exclude explicitly the self-interaction: $$ E = \sum_{i,j, i\neq j}\frac{q^2}{|\mathbf{r}_i-\mathbf{r}_j|} $$ Although, if treating this with a mean-field theory approach (coarse-graining), we would revert back to ignoring the self-interaction.