I was given the following explanation of Faraday cage: There is no electric field inside a hollow, charged conducting shell because a Gaussian surface inside the shell must have zero flux through it since it encloses zero charge. In the picture below, ${\bf S}$ is the Gaussian surface.
But I am not convinced by this explanation. There is clearly a field outside the shell since the shell is charged. What if I consider a closed surface ${\bf T}$ outside the Faraday cage like in the image below:
Again, the charge enclosed by ${\bf T}$ is zero so shouldn't this mean that there is zero flux through ${\bf T}$ and hence no electric field inside ${\bf T}$?
Best Answer
All electric flux start at a positive charge (a.k.a source of flux) and end at a negative charge (a.k.a sink). Since the Gaussian surface you considered outside the charged shell houses neither positive nor negative charges, there can't be any net electric flux going into or coming out of the Gaussian surface of your choice.
That means all electric flux that goes in must come out.