**Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. Find the magnitude of the electric field at a point P, a distance r from the center of the sphere.**

Using Gauss’s Law for $r \geq R$,

$$\begin{aligned} EA &= \frac{Q}{\epsilon_{0}} \\ E (4 \pi r^{2}) &= \frac{Q}{\epsilon_{0}} \\ E &= \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}} \end{aligned}$$

For $r < R$,

$$\begin{aligned} EA &= \frac{q}{\epsilon_{0}} \\ E \times 4 \pi r^{2} &= \frac{q}{\epsilon_{0}} \end{aligned}$$

q is just the net charge enclosed by a spherical Gaussian surface at radius r. Hence, we can find out q from volume charge density, $\rho$

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

$$\begin{aligned} q &= \rho \times \frac{4}{3} \pi r^{3} \\ &= Q \frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}} \\ &= Q \frac{r^{3}}{R^{3}}\end{aligned}$$

Hence, sub. q into the expression for E to get:

$$E = \frac{Q}{4 \pi \epsilon_{0}} \frac{r}{R^{3}}$$

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