My question is more about how Gauss' Law can be used to find the strength of the electric field inside an insulator when the Gaussian surface used only contains a portion of the total charge.

From my understanding, the reason Gauss' Law works is because the field from a charge located outside the Gaussian surface fluxes into (negative flux) and out of (positive flux) the surface chosen, and as a result it contributes nothing to the total flux.

In cases like finding the field inside a ball (Note: not a shell) as the distance (r) from the center varies, all of the solutions I see involve choosing a small spherical surface of radius r concentric to the ball's surface to apply Gauss' Law on, and using $EA=4\pi kQ$ to find E.

Since the charge outside the Gaussian surface is essentially "invisible" to Gauss' Law, aren't these solutions leaving out the portion of the total field created by this charge? As you move out from the center of the ball, doesn't the charge "still ahead of you" exert a field in the same way that the charge "behind you" does? Or am I missing something?

An example of one of these solutions can be found here:



1 Answer 1


While it is true that the field of one charge located outside your ball would not be nil inside the ball, the field of a spherical shell of such charges would cancel out by symmetry. So it’s not so much that the charge outside the Gaussian surface is invisible so much as a (uniform) shell of such charges outside the Gaussian surface is invisible.

The symmetry argument here is essential for this to work. If you select one charge outside your ball you break the spherical symmetry, but if you choose one shell of charges you maintain the symmetry.

Roughly speaking, one can make the following geometrical argument. Imagine a point inside your ball, and a shell of charge completely outside your ball, as illustrated in the figure (which I can't quite resize correctly). enter image description here

Then you can see there are fewer charges contained in the surface area contained by the arc closer to the point but they are closer than the greater number of charges on the corresponding farther area. Since the area of the sphere grows like $r^2$ but the strength of the electric field decreases like $1/r^2$, the effect of the smaller number of closer charges exactly balances (i.e cancels) the larger number of farther charges. (This argument is not very mathematically rigorous but gives a sense of situation.)

  • $\begingroup$ Thanks! The only part about that that I'm finding hard to grasp, is that it would have to mean that the field inside each uniform shell outside the gaussian surface cancels at every point inside that shell, and when I visualize it I only see it cancelling at the center of each shell. $\endgroup$
    – AlexP
    Commented Oct 29, 2018 at 22:34
  • $\begingroup$ @AlexP I added a bit which will hopefully be helpful. $\endgroup$ Commented Oct 29, 2018 at 23:50

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