Induced surface charge distribution

[Raihan amin]

Two identical metalic spherical conductor of radii $R$ are at a distance $d$ apart.One of the conductor has charge $Q$ while the another one is neutral.What will be the induced charge on the other conductor ?

If we put an image charge $q$ inside the neutral one. Then the potential at the surface is $$V = {\frac {Q} {r_1} }+\frac {q}{r_2}$$
So $$V=\frac {Q}{\sqrt {R^2 +d^2 -2Rdcos {\theta}}} + \frac {q}{\sqrt {R^2 + d'^2 - 2Rd' cos{\theta}}} $$
Now I know that the suface of the spherical conductor will be equipotential,independent of $\theta$
Here i consider the other charge as a point particle.but I can't find V.Still i don't know whether the image charge method will do or not.
Please help me someone to find it.Thanks.

 

[haruspex]

The general solutionn for two conducting spheres is quite complicated. See e.g. https://arxiv.org/pdf/0906.1617.pdf.
I recall it is a lot simpler if one is grounded, but I gather that is not the case here. Don't know whether it helps that one has net zero charge.