# Using Gauss's law to prove that a field is zero

I have seen countless examples where Gauss' law is applied to state that the electric field in a region is zero. For example, taking any Gaussian surface inside a spherically conducting shell where there are no charges inside, we find the the surface integral is zero by Gauss' law, but then it is always claimed that this implies the electric field is zero. Why is this the case?

Couldn't it be that there is a uniform electric field through the Gaussian surface such that the total surface integral comes out as zero but individual sections of the surface will have a non-zero field value?

• You mention a conducting shell. In this case, the useful Gaussian surface surrounds the cavity but lies within the body of the conductor, where we know the E field is 0. This shows that there is no net charge on the inner surface of the conductor. Then you use (assuming electrostatics) that the integral of E around any closed loop is 0 to show that there's no way to have an E field inside the cavity (since a field line would have to start and end on the inner surface, take the integral path following a field line through the cavity and closing the path within the body of the conductor). – Alex Nov 9 at 3:47
• The above argument holds for conductors of any shape with cavities: E=0 inside the cavity (as long as you have placed no extra charges inside the cavity). The accepted answer is kind of off track if you're asking about conductors. – Alex Nov 9 at 3:52
• @Alex Please post answers as answers, not comments. – Aaron Stevens Nov 9 at 15:04

## 3 Answers

You are exactly correct. In general $$\iint\mathbf E\cdot\text d\mathbf a=0\nrightarrow\mathbf E=0$$

However, in special instances of symmetry (like your spherical cavity), you can argue that the field must have the same magnitude and point radially outwards at all points on your Gaussian surface, then you do get $$\iint\mathbf E\cdot\text d\mathbf a=\iint E\,\text d a=E\iint\text da=EA=0\to \mathbf E=0$$

This is something to understand about the integral form of Gauss's law: You can only use it to find the field when specific symmetries are present in the system. Without the symmetries, then there is nothing to be done (as the value of an integral does not uniquely determine an integrand in general).

The argument uses a spherical Gaussian surface whose center coincides with that of the sphere. Spherical symmetry then means that any nonzero field would be radial and only a function of the radial coordinate. Thus the flux integral reduces to $$4\pi E_r$$. The only way it can be zero is for the field to be zero.

In general,if the flux of electric field through a closed surface is zero doesn't always mean that the electric field through that closed surface is zero.For spherical symmetric bodies is true that if electric flux through a closed surface is zero then electric field is also zero. If, $$\iint\mathbf E\cdot\text d\mathbf a=0$$ Then there are two possibilities $$E=0$$(in general for spherical symmetry) and $$E \neq 0$$(for any other closed surface).