# Why do outside charges do not contribute to net flux of a Gaussian Surface?

I don't quite understand why external charges can be ignored when calculating the net flux of a Gaussian surface. I understand that $\nabla \cdot \vec{E}$ of any point charge equals $0$ and I can reason using equations, but I can't find an intuitive physical understanding. Most arguments I have heard mention that all electric field lines that enter a Gaussian surface must then leave it, and so an external charge has no effect on net flux. But doesn't the flux also depend on the magnitude of the field?

For instance, Let's say I had a particle next to a Gaussian sphere and I look at the electric field line which pierces the sphere at its closest point. Wouldn't the field vector's magnitude be greater when it enters the sphere compared to when it exits because it is farther away when it leaves? And by the equation for flux,

$$\int \vec{E} \cdot \mathrm{d}\vec{A} = \int E \cos (\theta) \ \mathrm{d}A$$

which depends on $E$, wouldn't this have an affect on the net flux?

I'm not sure where my misunderstanding of flux is, but I know that I am clearly missing something huge. Perhaps is it that I have to consider all electric field lines and not just a single one? Or am I incorrectly assuming something about the relationship between the magnitude of the field and the flux through the surface?

• Duplicate? physics.stackexchange.com/q/63710 – Farcher Jan 31 '17 at 8:53
• It's a purely mathematical result. The reason is because the infinitesimal solid angle pointing to a closed surface, looking from outside the surface, crossed the surface in a way that for any positive surface element, there exists exactly one negative that compensates its effect. – AHB Jan 31 '17 at 12:59
• you pretty much answer the question in the question, unless your intuition about what a divergence is is lacking. – JEB Sep 27 '18 at 23:04

If a charge is kept near a sphere, the charge will not affect the flux of the sphere because flux is dependent on magnitude of electric field and area it pass through. So when the field enters the near end of the sphere the magnitude of electric field is high and the surface pass through is low but when the filed comes out the magnitude of electric field is low but the area it pass through is high. Hence, it compensate and don't affect the flux of the sphere.

There is a more intuitive view. Each field line of the flux created by an internal charge crosses the surface only once.

However, any external charge's line will either not pass over the surface or cross it twice.

• If the line doesn't meet the surface, iit doesn't contribute.
• If the line crosses the surface, as there are no sinks inside the surface, the line will have to exit as well. The input and the output cancel out.

Hence, only internal charge contributes to the flux.

As @AHB has already said, its just pure mathematical result. Unlike fields flux is not actually per se a physical phenomenon. The $$\cos\left(\theta\right)$$ is the cosine of angle between the field at that point and the area element $$\mathrm{d}A .$$ So if its a uniform field, its also true for non-uniform ones considering gaussian surfaces with uniform fields makes the calculations easier, for example a sphere in uniform field from left to right, the $$\theta$$ will be less than $${90}^{\circ}$$ on the right and the $$\theta$$ on the left side will be greater than $${90}^{\circ}$$ and between $${180}^{\circ} .$$ So the flux equation for net flux will become $$E \int{\cos\left(\theta\right) \, \mathrm{d}A} - E \int{\cos\left(θ\right) \, \mathrm{d}A} ~=~ 0$$ since $$\cos\left(\theta\right)$$ is negative in $${90}^{\circ} \le \theta \le {180}^{\circ} .$$

Suppose you enclose a positive charge with a Gaussian surface, then you put another positive charge near it but outside the surface. The field lines might look something like this: Remember, you choose your Gaussian surface because you want to find the charge inside it. You can pick a surface arbitrarily close to one of the positive charges, and until it becomes big enough to actually enclose the second charge, the field lines passing through the surface will cancel out. Since the magnitude of the electric field due to the charge inside the surface is dependent only on the enclosed charge ($E = \frac{kQ}{r^2}$), there will be no increase in the magnitude of the field exiting the surface by putting more charges outside, and so the flux will remain the same.