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Field Inside Uniformly Charged Spherical Shell

Electric field inside a uniformly charged spherical shell is exactly zero. A direct consequence of Gauss's law: no enclosed charge.
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Derivation

Gaussian surface inside the shell

For r<Rr < R, choose a concentric spherical Gaussian surface of radius rr.

By spherical symmetry, E⃗\vec{E} is radial and constant in magnitude on this surface:

∮SE⃗⋅dA⃗=E⋅4πr2\oint_S \vec{E} \cdot d\vec{A} = E \cdot 4\pi r^2

Enclosed charge

The Gaussian surface lies entirely inside the shell. No charge is enclosed:

qenc=0q_{enc} = 0

Result

By Gauss's law:

E⋅4πr2=0ε0=0E \cdot 4\pi r^2 = \frac{0}{\varepsilon_0} = 0 E=0(r<R)\boxed{E = 0 \quad (r < R)}

Why this is not obvious

It might seem that nearby parts of the shell would dominate. In fact, the solid angle argument shows that for any interior point, a cone in one direction intersects a small nearby patch of shell, while the same cone in the opposite direction intersects a large far patch. The greater area and the greater distance balance exactly — the 1/r21/r^2 falloff precisely cancels the r2r^2 growth of the intercepted area. The forces from both patches are equal and opposite.

Remember
This result has a direct consequence: a conductor in electrostatic equilibrium has zero field inside its bulk. Any excess charge resides on the outer surface only.