Q.Obtain the expression for the electric field due to a uniformly charged spherical shell.

Step 1. Symmetry. By the shell's spherical symmetry, the field can only be radial and can only depend on the distance $r$ from the centre.

Step 2. Outside the shell ($r>R$). A spherical Gaussian surface of radius $r$ encloses the full charge $Q$: $E(4\pi r^2)=Q/\varepsilon_0$, so $E=\dfrac{kQ}{r^2}$, exactly as if all the charge were concentrated at the centre.

Step 3. Inside the shell ($r<R$). A spherical Gaussian surface of radius $r$ encloses zero charge, since all the real charge lies on the shell itself, outside this smaller sphere: $E(4\pi r^2)=0$, so $E=0$ everywhere strictly inside. …

Apply Gauss's law with two separate spherical Gaussian surfaces -- one outside, one inside the shell -- …

• Assuming the field is zero everywhere (inside AND outside) -- it is only zero strictly inside the shell; …