# If no net charge is in a Gaussian surface, is the electric field zero?

Gauss’ Law states that that the electric flux of a Gaussian surface with no charge enclosed is zero $$\oint {\bf E} \cdot d{\bf A} = \frac{q_e}{\epsilon_{0}}.$$ If you do some algebra you can rearrange it like so: $$E = q_e /A \epsilon_{0}.$$ What this means is that the electric field in a Gaussian surface with no charge is zero. But how is this so? If I have an electric charge $$q_{e}$$ outside of a Gaussian sphere, won’t the electric field in the sphere be equal to whatever the electric field is in the space that the sphere occupies?

• That is correct, the flux across the sphere is equal to the flux it finds itself in. And the value of that flux is zero. The integral evaluates to zero. I'll let you think about that. Commented Mar 24, 2022 at 2:01
• Your "rearangment" does not work like this. You cannot replace the integral like that, in general.
– nasu
Commented Mar 24, 2022 at 3:02
• Try writing out the steps contained in your statement "If you do some algebra" explicitly. What assumptions do you need to make for those steps to be valid? Are those assumptions always true? Commented Mar 24, 2022 at 3:56

The issue with doing this is that $$\oint\mathbf{E}\cdot d\mathbf{A}= AE$$ only holds when $$E=|\mathbf{E}|$$ is constant on the surface, and is always orthogonal to the surface element $$d\mathbf{A}$$, so that $$\mathbf{E}\cdot d\mathbf{A}$$ is constant. For example, if your charge is external to the sphere on which you are integrating, the electric field is almost never perpendicular to $$d\mathbf{A}$$, changes in magnitude all the time, and more importantly it is sometimes opposite in direction to $$d\mathbf{A}$$. The net result is that the continuous sum of all the $$\mathbf{E}\cdot d\mathbf{A}$$ will have positive and negative terms and we end up with an integral of zero.