Question

Magnitude and direction of the electric field.

Answer 1

By spherical symmetry, it should be fairly clear that the total charge inside any Gaussian sphere will be positive, so the electric field should point radially outward.

You can find the magnitude of the electric field at r=R by integrating the charge density from r = 0 to R

Answer 2

I am not sure I understand about the direction. You said that the charge inside any Gaussian sphere will be positive. I dont get that part.

Answer 3

Firstly, the field should be radial due to symmetry. Do you agree? There's no angular component to this distribution. The reason it should be outward is because there's only positive charge involved.

Gauss' law says that the total electric flux through any imaginary bubble you make will equal the charge inside, or
\[ \int \vec{E}\cdot d\vec{S} = \int \frac{\rho}{\epsilon_0} dV \]
Because the field is spherically symmetric and radial, that first integral becomes
\[E\cdot 4\pi R^2 \]
Because there's no angular component to the distribution, the integral on the right reducs to
\[\frac{4\pi}{\epsilon_0} \int_0^R r^2 \rho(r) dr\]
So finally,
\[ \vec{E}(r) = \frac{\hat{r}}{\epsilon_0 R^2} \int_0^R r^2 \rho(r) dr \]