Putting a charged sphere in a uniform electric field - can we superpose the individual fields to find the resulting one? Let's say we want to find the potential outside a charged metal sphere with total charge $q$ when it is placed in a uniform magnetic field $\vec{E}_0$. Infinitely far from all other charge, the charge $q$ on the sphere will be uniformly distributed over its surface and its field $\vec{E}_q$ will be radially symmetric, as shown in the diagram below.

When the sphere is placed in the uniform field, positive charge will be moved to the top of the sphere and some negative charge will be induced at the bottom, with the net charge of course still $q$. The field will look something like this:

Does the final field $\vec{E}=\vec{E}_0+\vec{E}_q$? Griffiths, in the solution to one of his problems, assumes this without justification. However, I'm not convinced this can be done by superposition.If we use the principle of superposition we should be superposing the uniform field and the field of the redistributed charge, the latter of which is not $\vec{E}_q$. Is this understanding correct? If so, how would we go about calculating $\vec{E}$?
 A: The Maxwell equations are linear in the charges and the currents, meaning that if you superpose two sets of charges and currents, the fields will just add up. So, assuming that the charges causing your uniform field will not move due to the addition of extra charges, you can add up the uniform field with the field of your additional charges.
Important detail however is whether your sphere is conducting or not. In case of a non-conducting sphere, the uniform field will not lead to a rearrangement of your charges on the sphere, so you can superimpose the radial field of your sphere with the uniform field. If your sphere is conducting on the other hand, your charges will rearrange as correctly showed in your drawing.
For calculating the field caused by the charges on your conducting sphere, you need to start from Laplace's field equation for the electric potential in a spherical geometry, using separation of variables. As you have axial symmetry (no $\phi$ dependence), you only need to worry about the $r$ and $\theta$ dependence. The solutions to the differential equations for the $r$ and $\theta$ dependence need to be fitted to the boundary conditions, which are basically the condition the the potential caused by the sphere goes to zero as $r$ goes to infinity and the fact the electrical field should perpendicular to the sphere surface.
