Potential Problem solution for a Charged Sphere and a Point charge

Question

In this problem I have an uniformly charged sphere of radius a, charge Q, at a distance x from the point charge q. I want to find the solution to Poisson's equation with the method of images.

What I don't understand is a part of the solution given by my professor.

"the external potential of the sphere is solved by placing a charge $q'=-q\frac{a}{x}$ at a distance $x'=\frac{a^2}{x}$ and a charge $q'=-q\frac{a}{x}$ (to conserve the charge of the sphere) at the center of the sphere. This way the potential outside is the same as that produced by a charge $Q+q\frac{a}{x}$ at the center of the sphere."

Now this second to last part is the one that baffles me. Why do we need to "conserve" the charge of the sphere, since we usually do not do this when neutral, uncharged conductors are involved?

For example in the case of this same problem with a neutral sphere, the potential is solved by putting a charge equal to $q'$ at $x'$. In that case the charge inside the sphere would be $-q'$ but in the case that charge conservation would be a requirement, that cannot happen since the total charge has to be equal to $0$.

Also I don't understand how to choose(find?) the bound conditions in non-grounded cases, and it is unclear how my professor acted in this regard.

Thank you for your attention and an even bigger thank you for any clues!

Answer 1

See Jackson, sec. 2.3, p.60. To summarise the argument, you could think of it like this: a conducting sphere of radius $a$ is held at $0$ potential, and a point charge $q$ is brought near it at a distance $x$. This has the effect of inducing a charge of $q'=-q\frac{a}{x}$ in the sphere, with a certain distribution across its surface. The sphere is still an equipotential surface of $0~V$.

Now, the sphere is disconnected from ground, and injected with an additional amount of charge $Q-q'$ (so as to bring its total charge up to $Q$, as the problem statement demands). Where does this extra charge go? The existing $q$ and $q'$ have the effect of 'keeping each other in check', so to speak, so the extra $Q-q'$ charge is unaffected by the $q$, basically thinking it's entering a 'fresh' conducting sphere of radius $a$, and thus spreads out evenly all over the surface. Due to symmetry, this is equivalent to considering a virtual charge $Q-q'=Q+q\frac{a}{x}$ to have been placed at the centre of the sphere (at least, when considering the area outside the sphere).

This will also have the effect of changing the potential of the sphere, from $0$ to $\frac{Q+q\frac{a}{x}}{4\pi\epsilon_0 a}$. Note that the earlier problem we do in this chapter when solving for a grounded sphere is for a given potential, while this one is for a given charge.