# Confusion regarding gauss law

Gauss law states that flux due to net electric field on a Gaussian surface is charge enclosed divided by permittivity.

$$\Phi = \oint _ { A } \vec { E } \cdot \mathrm { d } \vec { A } = \frac { Q } { \varepsilon _ { 0 } }$$

Let's take a spherical Gaussian surface with $$+q$$ charge at centre. We can simply calculate the electric field at the Gaussian surface. Now if we introduce a charge $$-q$$ outside the surface at a point outside the surface and $$r$$ away from surface ($$r$$ is radius of Gaussian surface). The electric field at the surface would be same by Gauss law, but how can it be. Won't the point equidistant from the charges and on the surface experience $$0$$ electric field?

• "The electric field at the surface would be same by gauss law" no: its integral will be the same, but the actual value of $E$ at every point will not. Take for example the sum $3=1+1+1=0.5+2+0.5$. The summands changed but the sum did not. Apr 14, 2016 at 20:12
• Also, if the point is equidistant from charge +q and -q would only experience a net zero field from the two charges if they were in the same place, not possible if one is inside the Gaussian surface and the other isn't. Apr 14, 2016 at 20:14
• So the basic assumption often told carelessly that electric field is only due the charge enclosed by the Gaussian surface is false? Or it just means that the field may or may not be uniform at all points but their integral is the same? Apr 14, 2016 at 20:59
• If you have spherical symmetry, then it is true that the electric field at any distance $r$ from the center depends only on the charge enclosed in a sphere of radius $r$. But otherwise, the field on a surface will vary from point to point and will depend on the distribution of charges outside the surface. Apr 14, 2016 at 21:23
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– user10851
Apr 16, 2016 at 18:31

The fact is the when you add the charge $-q$ the electric field changes. In particular it has no more spherical symmetry and, consequently, you cannot compute the flux through the gaussian surface in a simple manner, as you did when you had only one charge. Thus, in this case, Gauss theorem is, of course, still valid, but it's of little use: simply because you cannot assume certain symmetry of the electric field (as you did having only one charge) without knowing its value.