Flux with a non-uniform electric field

[FS98]

If the electric field is uniform, the electric flux passing through a surface of vector area S is

,

where E is the electric field (having units of V/m), E is its magnitude, S is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to S.

This is from Wikipedia. Why does it say that the equation above applies if the electric field is uniform. Many of the examples I’ve seen apply the same equation with non-uniform electric fields. For example, a sphere with a charged particle q in the center has an electric field E = kq/r^2 multiply this by the surface area of a sphere 4pi(r^2) to get 4kq(pi) or q/Eo. I assume the electric fields dependence on r^2 means that it’s not uniform.

 

[cnh1995]

Why does it say that the equation above applies if the electric field is uniform

Because in case of a non-uniform electric field, the flux is given by Φ=∫E^→⋅ds^→ and this is a surface integral.
When the field is uniform, E can be treated as a constant and the above equation becomes
Φ=E^→⋅∫ds^→=E^→⋅S^→=EScosθ.

I assume the electric fields dependence on r^2 means that it’s not uniform.

The field is non-uniform in the radial direction. But at any radius r, the field on the surface is uniform (normal to the surface, hence parallel to the area vector).

 

[FS98]

Because in case of a non-uniform electric field, the flux is given by Φ=∫E^→⋅ds^→ and this is a surface integral.
When the field is uniform, E can be treated as a constant and the above equation becomes
Φ=E^→⋅∫ds^→=E^→⋅S^→=EScosθ.

The field is non-uniform in the radial direction. But at any radius r, the field on the surface is uniform (normal to the surface, hence parallel to the area vector).

So the electric field doesn’t need to be constant for this equation to work? There just needs to be a constant value of theta to plug into the equation phi = EScos(theta)?