how can gauss's law and electric flux help us calculate electric field

I recently started my study on electric flux and gauss's law in college, and I am currently really confused. So, I may ask some really stupid questions, please forgive me on that.

I don't understand how we could use electric flux to calculate the Electric Field.

Say we have a spherical surface with a whatever charge OUTSIDE that surface. Electric flux of that closed surface is Zero. However, what confuses me comes next. If the electric flux is zero, according to gauss's law and the formula ∮E⋅dS(Since it's a spherical surface, Flux = E*A), can I then conclude the electric field is ZERO?

What is that "electric field" calculated in this case? Why is it zero even though there is a charge out there?

Also, if I divide Flux by Area, I got a value of electric field? Is that the electric field for an area? Shouldn't electric field be described at a certain point?

• $\int \int E \cdot dS \neq \int \int EA$ in this case. The vectors are not parallel and the dot product wont always be positive / negative either, as half of the intersections of field lines and surface elements will have field lines pointing opposite to the surface element vectors. – user177179 Jan 27 '18 at 23:27
• Thanks. So, in this case, are there any ways or advanced maths that I can use that flux to calculate the electric field for the closed surface? I remember seeing a problem using flux to calculate the electric field if there is a charge inside. Can we do that if the charge is outside (which is in this case)? – Chun Hei Poon Jan 27 '18 at 23:30
• Any external charges do not contribute to the flux at all, so I'd be surprised if such a method existed for any charges outside. – user177179 Jan 27 '18 at 23:33

So in your equation, you can't reduce the integral to the product of the magnitudes $EA$.