I want to calculate the electric field of a charged solid sphere without using Gauss's law I can find the electric field from a charged solid sphere using Gauss's law but I am struggling to calculate this from Coulomb's law (I have seen examples of calculating e-field using Coulomb's law for a disk, a ring, a line etc. but not a solid sphere). 
If anyone could help me out I would be very grateful!
 A: 
Following on from Arturo's answer (thanks) I think I have figured out how to solve the integration over the sphere to find the electric field at a point using spherical coordinates and vectors.
A: First let me clarify something. I think what you mean by "Coulomb's law" is the solution to the electrostatic Poisson equation with the assumption (boundary condition) that it vanishes at spatial infinity:
$$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d^3\mathbf{r}'$$
You can get the electric field by taking the negative gradient of this.
$$\rightarrow\mathbf{E}(\mathbf{r})=-\nabla V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\mathbf{r}')\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}d^3\mathbf{r}'$$
When the only charge you have is a point charge sitting at the origin ($\rho(\mathbf{r'})=Q\delta^3(\mathbf{r}')$, where $\delta^3(\mathbf{r})$ is the 3-dimensional Dirac-delta function), then you get the regular Coulomb's law.
$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\frac{Q\mathbf{r}}{|\mathbf{r}|^3}$$
Now, for your problem, simply using Gauss's law will do. Gauss's law says
$$\iint\mathbf{E}\cdot d\mathbf{A}=\frac{Q_{enc}}{\epsilon_0}$$
Since you know the problem is spherically symmetric, not only can the electric field can only point radially outwards (prove this to yourself), but it is also spherically symmetric (also prove this to yourself). So, Gauss's law reduces to 
$$E(r)\cdot(4\pi r^2)=\frac{Q_{enc}}{\epsilon_0},~~~\textrm{where}~~~ \mathbf{E}(\mathbf{r})=E(r)\hat{r}$$
To solve for the electric field, you must calculate the charge enclosed within your "Gaussian surface", and for that you must know the charge distribution within the sphere.
$$Q_{enc}(r)=\int_0^r \rho(r')(4\pi r^2) dr$$
For example, if the solid sphere of radius $R$ and charge $Q$ has a uniform charge density, then the total amount of charge within the spherical Gaussian surface with radius $r<R$ scales with the cube of the radius - i.e. $Q_{enc}=Q(r^3/R^3)$. In that case, 
$$E(r)=\frac{Qr}{4\pi\epsilon_0 R^3},~~~r<R$$
For a spherical Gaussian surface with radius $r>R$, all the charge $Q$ is inside, so we simply get
$$E(r)=\frac{Q}{4\pi\epsilon_0 r^2}, ~~~r\geq R$$
A: Electric field by uniformly charged solid sphere using Coulomb's Law and not Gauss' Law

