you don't need two charges, you can use only one charge and the answer will be the same, you only should note to add a second image charge in the centre of the sphere to make the sphere neutral.
in both cases, when you take the limit, you get same electric dipole in the centre of sphere.
but because the symmetry of two charges, calculations are somehow easier.
with any relevant boundary condition, the sphere should remain neutral. to see this suppose it gains charge $q$. if you change $E \to -E$ then the charge of sphere should change to $-q$ (you can see this by dimensional analysis, since the $q$ should be proportional to $E$). but the new setup are the same as the first one, because letting $E\to -E$ is the same as rotating the setup by 180 degrees. it certainly shouldn't change the sphere's charge. so $q=-q=0$.
This doesn't give you the distribution of the charge on the surface, but it's an elegant way of getting the total induced charge on the sphere.
A couple of things I notice:
- The larger $r$, the smaller the induced charge
- The sign is correct (the charge must be the opposite of $q$)
So yes, I "find this OK". Plus - it matches the expression in your rigorous derivation.
Best Answer
In order to find the potential function, one usually employs the methods used here. Then, once you have this function for the potential:
$V(r)=\frac{2q}{4\pi\epsilon_0 a^2}(r-\frac{R^3}{r^2})cos(\theta)$
Find the electric field at the surface of the grounded sphere of radius R
$E(R)=\frac{\partial{V}}{\partial r}\bigg|_{r=R}=\frac{2q*cos(\theta)}{4\pi\epsilon_0 a^2}*3=\frac{6q*cos(\theta)}{4\pi\epsilon_0 a^2}$
$\frac{\sigma}{\epsilon_0}=E(R) $
$\Rightarrow \sigma=\frac{6q*cos(\theta)}{4\pi a^2}$
That should answer your first question.
As for your second question, no, the potential inside each sphere will not change. Why?
Let's see what happens when a charged sphere is brought near another charged conducting sphere, or really any charge configuration changes the electric field in the environment of the charged conductiong sphere in question. The elctric field everywhere will change (except, of course inside the sphere) and there are lots of nifty tricks to figure out the new electric field, the method of image charges being the most prominent. Now, consider a regular conducting sphere with a charge q and a radius R sitting somewhere with no external fields. The potential at its surface and inside it $=\frac{q}{4\pi\epsilon_0R}$, a constant. That means that you can add this constant to the (new) potential function for the region outside without changing the electric field outside, because the gradient of a constant is $0$. Now, inside the sphere, $E=0$, which means the potential inside is a constant function (gradient of a constant is zero). You can set that constant to be whatever you want and $=\frac{q}{4\pi\epsilon_0R}$ looks pretty good, unless you want a discontinuity in the potential function.