Step 1: Write down what Gauss's law provides.
Gauss's law links the outward electric flux of a closed surface to the charge it traps:
\[ \Phi_E = \oint \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}. \]
Step 2: Exploit the symmetry to pick the surface.
A uniformly charged shell of radius \(R\) and charge \(q\) looks the same from every direction. So the field at an outside point must be radial and its size can depend only on the distance \(r\) from the centre. The natural Gaussian surface is a sphere of radius \(r\) (\(r>R\)) centred on the shell.
Step 3: Simplify the surface integral.
On this sphere \(\vec{E}\) is constant in magnitude and everywhere perpendicular to the surface, so the dot product becomes a plain product and \(E\) comes out of the integral:
\[ \oint \vec{E}\cdot d\vec{S} = E\oint dS = E\cdot 4\pi r^{2}. \]
Step 4: Insert the trapped charge.
Because \(r>R\), the sphere encloses the complete charge of the shell, \(q_{enc}=q\). Substituting into Gauss's law:
\[ E\cdot 4\pi r^{2} = \frac{q}{\varepsilon_0}. \]
Step 5: Isolate the field.
\[ E = \frac{q}{4\pi\varepsilon_0 r^{2}} = \frac{k\,q}{r^{2}},\qquad k=\frac{1}{4\pi\varepsilon_0}. \]
This is an inverse-square field, identical to that of a point charge \(q\) sitting at the centre of the shell.
\[\boxed{\ E = \dfrac{q}{4\pi\varepsilon_0 r^{2}}\quad (r>R)\ }\]