Gauss's Law (electrostatics)

Use Gauss' law to determine the electric field around a solid conducting sphere of radius R carrying a charge Q.

So I've constructed a spherical Gaussian surface of radius r > R and centre at the centre of the sphere. The electric flux passing through the surface
$\displaystyle \Phi _{E} = \oint \mathbf{E}. \mathbf{dA} = 4 \pi r^{2} E$
but $\Phi _{E} = \frac{Q}{\epsilon {0}}$ by Gauss's Law
$\Rightarrow E= \dfrac{Q}{4 \pi r^{2} \epsilon _{0}}$

So what I've found (I think) is the electric field strength at an arbitrary distance r from the centre of the sphere (provided it is outside the sphere). But in the model solutions I have, the Gaussian surface has been constructed specifically at radius R, which is the radius of the sphere. Does that not mean that the field strength has only been found for the surface of a sphere of radius R (i.e. the general case as in the question hasn't been answered) or am I just being stupid?

Thanks

So what I've found (I think) is the electric field strength at an arbitrary distance r from the centre of the sphere (provided it is outside the sphere). But in the model solutions I have, the Gaussian surface has been constructed specifically at radius R, which is the radius of the sphere. Does that not mean that the field strength has only been found for the surface of a sphere of radius R (i.e. the general case as in the question hasn't been answered) or am I just being stupid?

Thanks