Why can you not use Gauss law to calculate the electric field a distance $D$ away from a charged cube? Why can one not use Gauss law to calculate the electric field a distance $D$ away from a charged cube? I have read that the electric field at the face of the cube is not perpendicular or parallel to the surface. Why is it not perpendicular to the surface as in the case when you have a charged sphere?
And if you are not supposed to use Gauss law, how would one then find an expression for the electric field a distance $D$ away from a charged cube?
 A: I assume that by 'Gauss' Law,' you mean the integral form
$$\int_{\partial \Omega} \mathbf{E} \cdot d\mathbf{S} = \frac{Q}{\varepsilon_0} = \int_\Omega \frac{\rho(\mathbf{x})}{\epsilon_0} dV$$
(which is equivalent to the differential form $\nabla\cdot \mathbf{E} = \frac{1}{\epsilon_0}\rho(\mathbf{x})$ ).
In the classic argument for electric field outside of a charged sphere using the integral form, most of the physics comes from symmetry: the charge distribution is symmetric under the action of rotations, and so it follows that the generated field must exhibit the same symmetry. Any deviation of $\mathbf{E}$ from the radial direction would break this rotational symmetry, so it is not allowed - this allows one to simplify $\mathbf{E}\cdot d\mathbf{S} = E_r dS$
A cube, on the other hand, does not exhibit this structure. Its electric field will be direction dependent, meaning that one must instead solve the PDE given above. The other issue is what the cube is made of: if it is indeed a uniformly charged cube, it is presumably an insulator, and the electric field at the surface is not necessarily normal to the surface. (If it were conducting, that would be a different situation - charges could rearrange themselves within the cube).
To see this, imagine abstracting the cube to one large positive sphere in the centre, and 8 smaller 'edge spheres' at the vertices. The electric field will only be normal where symmetry allows - at vertices and edges.
P.S. Just the distance $D$ is not enough information to tell you the electric field outside the cube (unless it is far enough away that it looks like a sphere). You also need the orientation: are you facing a flat part of a vertex?
