# Why can we write $\iint_S \vec{E}\cdot d\vec{a}=\frac{Q_{enc}}{\epsilon_0}=0$ when we are considering a Gaussian surface within a spherical shell?

Gauss' Law tells us that for a point charge q at the origin and a spherical surface of radius r centered at the origin, we have

$$\iint_S \vec{E}\cdot d\vec{a}=\frac{q}{\epsilon_0}\tag{1}$$

If we have $$n$$ charges $$q_i$$ within the spherical surface, then using the more general result that "the flux through any closed surface is $$\frac{q}{\epsilon_0}$$", we have $$\vec{E}=\sum_{i=1}^n\vec{E}_i$$ and Gauss' Law becomes

$$\iint_S \vec{E}\cdot d\vec{a}=\frac{Q_{enc}}{\epsilon_0}\tag{2}$$

where $$Q_{enc}=\sum_{i=1}^n q_i$$.

Now consider the task of trying to find the electric field inside of a uniformly charged spherical shell.

The derivations I see all use the reasoning that for a Gaussian spherical surface inside the charged spherical shell we have

$$\iint_S \vec{E}\cdot d\vec{a}=\frac{Q_{enc}}{\epsilon_0}=0\tag{3}$$

and because of symmetry we have a constant electric field on the Gaussian surface and so

$$E \iint_S da= E4\pi r^2=\frac{Q_{enc}}{\epsilon_0}=0\tag{4}$$

$$\implies E=0$$

What I don't understand is why we can write (3).

To obtain (1) and (2) we conjectured that there was a point charge and point charges, respectively, inside of a closed surface.

But in the task of finding the electric field inside of a spherical shell, we have charge outside the Gaussian surface and the electric field on the Gaussian surface, if it is nonzero, is coming from a charge located outside the region enclosed by the Gaussian surface.

(3) and (4) seem to be true for the charge 0 inside the spherical Gaussian surface, which generates a zero electric field on the surface. But then we have the charge outside.

I seem to be missing something related to the application of the divergence theorem in this context.

One thing I noticed is that for a charge outside of a closed surface, the flux through the surface is always zero.

Thus, for a system of charges and a closed surface, it seems that the equation

$$\iint_S\vec{E}\cdot d\vec{a}=\frac{q_{enc}}{\epsilon_0}$$

is true because only the charges within the closed surface can have flux through the surface.

But then, as far as I can tell, the argument for the electric field being zero inside a spherical shell would be as simple as arguing that the charge is outside any closed surface inside the shell and thus the electric field is zero on any such closed surface, which means it must be zero on the entire inside of the spherical shell.

the flux through any closed surface is $$\frac{q}{\epsilon_0}$$

This is correct and is a result of $$\iint_S E\cdot dS=\frac{q}{\epsilon_0}$$ But using the words net flux (instead of flux) through the closed surface is more accurate.

in the task of finding the electric field inside of a spherical shell, we have charge outside the Gaussian surface and the electric field on the Gaussian surface, if it is nonzero, is coming from a charge located outside the region enclosed by the Gaussian surface

The electric field lines, if coming from charge outside the Gaussian surface, still contribute nothing to the net flux through the Gaussian surface. They enter and exit the closed surface, so the net flux is still zero. Of course this is not the case for charge inside the surface, meaning the electric field cannot be zero.

Thus, for a system of charges and a closed surface, it seems that the equation $$\iint_S\vec{E}\cdot d\vec{a}=\frac{q_{enc}}{\epsilon_0}$$ is true because only the charges within the closed surface can have flux through the surface

Right, but because we can have a net flux through the surface enclosing the charge. There can be electric field lines from external charges entering the Gaussian surface. But since these lines enter then leave the closed surface, the net flux due to external sources is still zero.

(3) and (4) seem to be true for the charge 0 inside the spherical Gaussian surface, which generates a zero electric field on the surface. But then we have the charge outside.

As stated, external sources cannot contribute to the net flux if they are outside the Gaussian surface.

But then, as far as I can tell, the argument for the electric field being zero inside a spherical shell would be as simple as arguing that the charge is outside any closed surface inside the shell and thus the electric field is zero on any such closed surface, which means it must be zero on the entire inside of the spherical shell.

You should think about it as being zero because the amount of charge inside the Gaussian surface is also zero. Charge inside the surface will always lead to a nonzero field.

If we construct a Gaussian surface enclosing no charge, the total flux through that surface is zero.

$$\oint_S \vec{E}\cdot d\vec{a} = \frac{Q_{\text{enc}}}{\epsilon_0} = 0$$

However, this does not mean we can conclude the electric field is zero at every point on the surface. In fact the electric field may be nonzero over the entire surface. Instead, it's the total electric flux through that surface which is zero. Consider a Gaussian sphere enclosing no charge, with a point charge situated outside the surface. The electric field is nonzero at every point on the sphere. Meanwhile the flux is negative on some portions of the surface and positive for others. Gauss' law guarantees the total flux is zero, but the flux and field at individual points is generally nonzero.

We can make claims about the electric field itself (and not just the total flux) when the system possesses high degrees of symmetry. Say we construct a Gaussian sphere inside a spherical shell of uniform charge density. Symmetry guarantees the electric field has uniform magnitude and is oriented parallel (or anit-parallel) to the area elements $$d\vec{a}$$ over the Gaussian surface. Then we may "pull the electric field out of the integral"

$$\oint_S \vec{E}\cdot d\vec{a} = \oint_S E da = E\oint_S da = E\cdot A = \frac{Q_\text{enc}}{\epsilon_0} = 0$$

As the area of the Gaussian surface $$A$$ is nonzero, we can say the electric field must be zero over the entire surface. Contrast this to the point charge example where we could only say the total electric flux is zero.