# Flux of electric field through a closed surface with no charge inside? [duplicate]

I'm reading the Feynman lectures on electromagnetism and in Vol II, Chapter 1.4 Vol. II, Chapter 1-4 he talks about the flux of the electric field and says that flux of $E$ through and closed surface is equal to the net charge inside divided by $\epsilon_0$.

If there are no charges inside the surface, even though there are charges nearby outside the surface, the average normal component of $E$ is zero, so there is no net flux through the surface

I cannot see why the net flux is zero here. Say we have a closed unit sphere at the origin with no charge inside it and at the point $(2, 0, 0)$ we have some charge $q$.

1. Well doesn't this charge then define the electric field $E$ for the system and it will flow into the unit sphere on the right hand side, and out of the unit sphere on the left hand side?

2. Furthermore, as the strength of the electric field decreases with distance from $q$ won't we have more flux going into the right hand side which is closer to the charge $q$, and less flux leaving through the left hand side as it is further away - and hence we should have a non-zero flux?

Can someone please explain what I am misinterpreting here?

## marked as duplicate by Rob Jeffries, John Rennie, Kyle Kanos, Martin, ACuriousMind♦Sep 17 '15 at 12:44

You have more flux per unit area going into the right side, but the area on the right side is smaller. These two balance out so that the total flux is the same going in as going out.

The part of the sphere which has electric flux going in, traced in red, is less than half the area of the sphere.

Incidentally, flux per unit area is just the electric field.

• Thanks...but although the flux "spreads out" on the left hand side means that less is going out there....is there not EVEN less again going out due to the fall-off in electrical force with distance from the source? – Riggs Sep 16 '15 at 15:59
• @Riggs I think you're double-counting the effect. The "spreading out" of the flux is exactly the same physical effect as the $1/r^2$ decrease of the electric field. – David Z Sep 16 '15 at 16:15
1. Yes.
2. No.

The strength of the field near the charge is higher (because it is closer to the charge) but this electric field is entering through a smaller area $S_1$ whereas the electric field leaving the sphere is relatively weaker (as it is further away from the charge) but leaves through a larger area $S_2$ as visualised below. Hence, the flux going in is exactly equal to the flux going out, and the net flux is 0.

To understand this better, you might see the proof of Gauss's Law using solid angles. You'll see that the area through which the field leaves or enters is proportional to $r^2$ whereas the field itself is proportional to $\frac{1}{r^2}$ and hence, the flux leaving or entering doesn't depend on the position of the charge. (I mean, it does not depend on where it is placed inside or where it is placed outside)

Note: This picture isn't that of the sphere and charge described exactly, it's sort of flipped but that won't be a problem, I think.