# For the electric field inside a uniformly charged sphere, why are external charges outside the Gaussian surface not taken into account for the field?

Consider a uniformly charged sphere with a charge Q. To find the electric field outside the sphere, a Gaussian surface of radius say 'r' (such that r ≥ R) can be imagined, and the result found is that the electric field is the same as if all the charge of the sphere was concentrated at the center of the sphere. In other words, the field is proportional to the charge inside the Gaussian surface, which is the charge of the sphere, and that's intuitive

However, now suppose that we are interested on the electric field inside the sphere. Using Gauss' Law, the argument that is usually made is that, if a Gaussian surface of radius 'r' is drawn (such that r ≤ R), then the electric field at any point on the Gaussian surface is proportional to the charge inside that Gaussian surface (which is the fraction of the charge of the sphere (Q) enclosed by the Gaussian surface). However, something doesn't make sense to me. What I don't understand is the following: why does only the charges inside this Gaussian surface contribute to the electric field at the surface and the charges outside of the surface which are still part of the real physical sphere are ignored?

To explain myself I did this diagram:

The charges along the y-axis of the sphere are depicted with the electric fields that they release. The charges inside the Gaussian surface cause the radially outwards electric field, that's ok. But the charges in between the distances r and R cause also electric fields. The positive charge outside the Gaussian sphere located above would cause an inwards electric field on the point closest to it, and an outwards electric field on the point farthest from it. The positive charge outside the Gaussian sphere located below would also cause the same, an inwards electric field closest to it and an outwards electric field at the point farthest from it. One would say that the inwards electric field produced by the proton above is cancelled by the outwards electric field produced by the proton below, but this would not be true of course because the outwards electric field would be diminished by the square of the distance.

I don't know if I explained myself correctly, if not, please let me know. The summarized question would be then: why are the external charges to a Gaussian surface inside a solid sphere not considered? Thanks in advance.

• Do you know what Gauss's law actually says?
– d_b
Feb 7 at 0:03
• The field is $0$ because the net field due to the outside charges adds to $0$. Feb 7 at 0:07
• @d_b What do you mean? I know that Gauss' Law states that the flux of an electric field E through a closed surface is equal to the net charge inside the surface divided by e0. So charges outside the surface cause 0 flux, but that doesn't necessarily mean that they don't contribute to the electric field, because I could take an electric dipole and just enclose one charge, obviously the electric field obtained due to that charge won't be the net electric field at the Gaussian surface. Feb 7 at 0:09
• @ZeroTheHero but is there by any chance a proof of that? I obviously can assume that the net electric field due to the outside charges will add up to 0, but I don't want to just take it for granted. Feb 7 at 0:15
• Of course. It is $0$ by careful consideration of the symmetry of the problem. See this related answer physics.stackexchange.com/a/457759/36194 for the correct geometry. Feb 7 at 1:28

Gauss's law says that the electric flux through a closed surface is determined by the charge enclosed by the surface. In other words, there is no flux due to charges outside the surface, even though charges outside do contribute to the electric field on the surface. It's not that the flux due to outside charges is cancelled by the effect of other charges. Rather, each charge outside the surface contributes some flux into the surface and some flux out of the surface, and when we add these contributions we get a net flux of zero.

In general, this doesn't tell us that we can find the electric field on the surface by only considering the charges inside the surface, since in general the electric field will change from one point to another. However, with enough symmetry there will be simple relationship between the flux and the electric field components, and this lets us find the electric field in terms of the charge inside the surface.

For the case of a uniform sphere of charge, we have spherical symmetry, so we know the electric field is the constant on a spherical gaussian surface concentric with the sphere. This means we can say \begin{align} \Phi_E = \int E_r\,dA = E_r A \end{align} so the electric field is completely determined by the flux and hence by the charge enclosed.

• Okay I think I finally got it. If E(out) and E(in) are the fields due to the charges outside and inside the Gaussian surface, respectively, then, the total flux through the surface is: ∫𝐸(out)⋅𝑑𝐴 + ∫𝐸(in)⋅𝑑𝐴 = Q(in)/e0. The flux due to the charges outside is 0, but supposing that the E(out)≠0, then we can say that: ∫(𝐸(out)+E(in))⋅𝑑𝐴 due to the superposition of electric field, but the flux is still Q(in)/e0, and since the situation is symmetrical and the E(net) is constant (and radial), then E(net)∫𝑑𝐴 = Q(in)/e0. Is the previous reasoning right? Feb 7 at 1:07
• @prado5083 Yes, that looks right to me.
– d_b
Feb 7 at 1:44

Newton's Shell Theorem states that the gravitational force exerted by a spherical shell of uniform density upon a mass interior to that shell is 0.

Coulomb's law is analogous to Newton's law of gravitational attraction.

$$F = k\frac{q_1q_2}{r^2}$$

vs

$$F = G\frac{m_1m_2}{r^2}$$

So, the force exerted by a spherical shell of uniform charge density upon a charge interior to that shell will also be zero.

Thus, the electric field inside a spherical shell of uniform charge density at a given point interior to that shell will have the same magnitude as if the shell a uniformly $$0$$ charge density.