Why is the electric field inside a conductor zero at every point?

there are a couple of arguments on how the electric field inside a conductor is zero.

The electric field inside a conductor in which there is NO current flowing is 0.

If there is current flowing in a conductor, then it may be a useful approximation to the truth to neglect the electric field inside of a conductor. That is, it may be useful to treat that field as negligible, because it is "small" relative to other things we may be focused on. However, if there is current flowing in the conductor (and the conductor is not a super-conductor), the electric field is not exactly equal to 0. Rather the "microscopic" version of Ohm's law states

$$vec{J} = sigmavec{E}$$

where $vec{J}$ is the current density, $sigma$ is the conductivity, and $vec{E}$ is the electric field.

Now, for this configuration, the vector sum of all electric fields of all charges in the centre of said sphere would be exactly zero, quite straight forward so far.

This almost certainly is referring to the electric field in a conductive sphere after that sphere is in static equilibrium, i.e. there is no current. So, we can proceed with that assumption.

Since there is no current, there is no current density. Since there is no current density, there is no electric field.

We can go further, and show that there is no net electric charge inside the sphere; that it is electrically neutral.

Since the electric field uniformly 0 inside the conductive sphere with no current, the divergence of the electric field is also 0.

$$nabla cdot vec{E} = 0$$

However by Gauss's Law

$$nabla cdot vec{E} = frac{rho}{epsilon_0}$$

where $rho$ is the (net) charge density, and $epsilon_0$ is a constant.

So, the (net) charge density $rho$ must also be 0. When the conductor has reached a steady state with no current, there is no charge within it's interior. Any net charge must be located on it's surface only.

and another common explanation is the one involving gauss's law. but i still dont find it satisfactory as in my freshman-level electromagnetism course they didn't really give rigorous proof of it.

The explanation I gave relies upon Gauss's Law. There is no deductive proof of Gauss's Law. It is a basic law that is not derived from some other laws. It's "proof" consists in the fact that it has been successfully used in the highly accurate calculation of electromagnetic phenomena for many years. That is, it has been empirically validated.

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Addendum.

Ján Lalinský asked:

This argument only shows that electric field vanishes in the conductor making up the sphere. What about the electric field in vacuum inside the sphere?

Although the original question did not ask about vacuums inside a sphere, we can extend the argument above to the situation where there is a conductive body which contains a cavity within it, such that any net charge within the cavity is mobile. There need not be any charge in the cavity, it may be a complete vacuum.

In the argument above using the microscopic version of Ohm's law, no reference was made to the shape of the conductive body. Therefore in any uniform conductive body in electrostatic equilibrium, there can be no electric field.

Furthermore, this will be true even if the "conductive body" is not a classical conductor. It could be a super-conductor, a plasma, or even an ionic liquid, as long as charges are free to move. We can use the Lorentz force to show this. If a body is in electro-static equilibrium, then there is not only no current present, but also there is no net acceleration of charges. The Lorentz force is given by

$$vec{F} = q(vec{E} + (vec{v} times vec{B}))$$

where $q$ is a unit charge, $vec{v}$ is the velocity of that charge, and $vec{E}$ and $vec{B}$ are the electric and magnetic fields respectively.

When there is no current, the contribution of $vec{v} times vec{B}$ can be eliminated. If the charge is in electrostatic equilibrium, there is neither charge flow nor charge acceleration, so the net force on it must be 0. Hence the $vec{E}$ field must be 0. So, non-classical conductors in electrostatic equilibrium have no electric field in their interior either.

Now let's consider a conductive body with a cavity within it. If the cavity contains a non-classical conductor, we already know that in it's interior, there is no electric field. What if there is a vacuum in the cavity?

Since we are discussing a vacuum, with no charges within it, we can appeal once again to Gauss's law.

$$nabla cdot vec{E} = frac{rho}{epsilon_0}$$

Since there is no charges present, the charge density $rho$ is $0$, so the divergence of the $vec{E}$ field, $nabla cdot vec{E}$ must also be $0$.

However, if there is a volume (the cavity) in which the divergence of the $vec{E}$ field is 0, and the $vec{E}$ field itself is 0 on the surface of this volume, then the $vec{E}$ field itself must be 0 throughout the volume.

Therefore, in electrostatic equilibrium, there is no electric field within an empty (vacuous) cavity within a conductor.

[Now, one further point. Suppose the "cavity" is filled with a conductor which is different from the enclosing conductor. Although neither the "cavity" conductor, nor the enclosing conductor will have an electric field within their "bodies", it is possible for there to be an electric field at their boundaries. For example if the conductors are two different metals, or two types of semiconductor with opposite polarity doping. Due to the ambiguity of language, the inner boundary of the enclosing conductor might be considered part of the "interior" of that conductor. If that is what is meant, there could be an electric field in the "interior" of that conductor. However, if we consider "interior" to exclude the inside boundary, then we can say that there is no electric field in the interior of the enclosing conductor.]

How is the electric field inside a conductor zero?

In the electrostatic case, the electric field within a conductor is necessarily zero. The reasoning is as follows:

(1) within a conductor, electric charge is free to move (accelerate) under the influence of a non-zero electric field

(2) in the electrostatic case, electric charge is (by definition) at rest

(3) if there is a non-zero electric field within a conductor, electric charge within will accelerate under its influence which is inconsistent with the electrostatic condition

Thus, if the electrostatic condition holds, the electric field within a conductor is necessarily zero.

While it is not generally true that the electric field within a conductor is zero, the electric field within an idealized, perfect conductor is zero always.

What Math Keeps Me Busy said is true, but there is a simple intuitive way to see it.

In a conductor like a metal, electrons can easily move. There are positive nuclei that can't move.

If there was an electric field inside a conductor, electric forces would push the electrons away from their nuclei. The nuclei would create attractive forces that would pull the electrons back. Electrons would flow until enough charge had separated to cancel the original electric field. That is electrons would flow until the total force became zero.

When there is a current, electrons are flowing. It takes a battery to create that field and keep the electrons flowing. The field would speed electrons up. Electrons bump into things, which tends to slow them down. This is oversimplified, but it is the origin of resistance.

Electric field vanishes inside conductor only when the system is static. If electric current is present at some point in the conductor, then electric field at that point does not vanish.

In the electrostatic case, the field inside has to vanish because of Coulomb's law (or Gauss' law).

Proof: Due to Coulomb's law, electrostatic potential obeys the so-called Poisson equation $$ Delta V = -rho. $$

This equation implies that $V$ can have local maximum or minimum at some point of conductor only if $rho$ at that point is non-zero. If $rho$ is zero there, then $V$ has to either 1) decrease when moving in one direction and increase in other direction (a saddle point) or 2) stay the same when moving in all directions.

In an electrostatic system, $rho$ has to be zero everywhere inside the conductors. If there was some non-zero charge density at some point, it would not be stable and the charged particle would start repelling each other and the charge density would decrease in time.

So we have conductor with zero charge density everywhere inside. And according the the Poisson equation, the potential $V$ has no maximum or minimum anywhere inside.

Now we use a theorem from mathematics: if a scalar function of position is constant on a closed surface, and has no extremes inside, then it has to have the same value everywhere inside as it has on the surface. This is the case for the Coulomb potential function. Since it is the same everywhere on conductor's surface and has no extremes inside, it has to have the same value throughout the conductor.

Thus potential has zero gradient at all points inside the conductor. Thus electric field vanishes everywhere inside the conductor. The surface is a special place, because charge density there does not need to vanish, and the charges there also experience electric force that is pushing them out of the conductor in direction perpendicular to conductor's surface. However, unless this force is very strong, the charges stay bound to the surface by the conductor's surface microscopic forces (the potential well for the electrons is sometimes called the Fermi energy of the metal).

It’s simple. If there is an electric field, then the free electrons inside the conductor will migrate creating an opposite field thus cancelling the original one and hence maintaining the net zero field inside the conductor. This is the reason why there can’t be a net electric field inside a conductor and no net charge can exist inside a conductor