Why is E proportional to q/R^3 inside a uniformly charged sphere?

[tronter]

In a uniformly charged sphere, why inside the sphere is $$\bold{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{R^3} r \bold{\hat{r}}$$?

I don't see how this follows from Gauss's law.

 

[Snazzy]

The ratio of the charges between the inner sphere (Q') and the outer sphere (Q) is equal to the ratio of the volume between the inner and outer spheres.

$$\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=\frac{Q'}{Q}$$

Thus: $$Q' = Q\frac{r^3}{R^3}$$

Now use Gauss's Law to find the electric field inside the inner sphere.

 

[kkrizka]

Just to clarify what Snazzy said, the inner sphere is your Gaussian Surface.