Why isn't the electric field zero in the empty space? A spherical portion has been removed from solid sphere having a charge distributed uniformly in its volume the electric field inside the emptied space is? Isn't the electric field suppose to be zero in the empty space? But the answer is non-zero and uniform.
 A: If the sphere has a volume charge, it implies the sphere is a non conductor, in that case E field exists within the sphere(there being no scope for electrons to move under the effect of these field lines as it is a "non-conductor").
When you construct a cavity within this volume..
By gauss law, you get the closed Integeral of E wrt area to be 0 within this cavity. This does not mean that the e field is zero, it simply means the integral is zero.
You can think of it in this way,All the flux that enters this volume also leaves it.. implying thereby that there is no flux being contributed by enclosed charge and consequently leads to the result that the enclosed charge is zero.
Now that you know it is not zero, lets try to prove that it is uniform as well
By uniform you would mean that e field at apoint inside the cavity is independent of its position vector.
Proof:
The e field within the cavity is the superposition of the e field due to the original uncut sphere and a sphere of same volume and shpae as the cavity but having having a uniform negative charge density.
Let us denote the densities by p and -p respectively,
 E(net) = E1 + E2
where E1 = (p/3e)k where k is the vector in the radial direction from center of original sphere to point P
also , E2= (-p/3e)s here s is the radial vector from center of cavity to the point under investigation, say, P
ALSO k=a+s , vectorially. a is the vector from the center of the original sphere to the center of the cavity
therefore you have E)net) as ,
E(net)=pa/3e 
which for a given cavity is independent of the position of a point within the cavity



P.S.:Formatting edits welocomed.
