Method of images and induced charge I've read the section in Introduction to Electrodynamics by Griffiths over and over and I still don't understand the concept of the method of images. This problem builds on the concept of using an image charge inside a spherical conducting shell (neutral).
My confusion arises from a number of things.

*

*Why do we even need to pretend to have a charge inside the conducting shell?


*How can I geometrically perceive the concept of potential? Currently, I imagine putting a tack on a board. Then, I loop one end of a rubber band to the tack and begin to pull the rubber band to the left. As I move farther and farther to the left, the tension in the rubber band grows to infinity. If I were to let go, the rubber band would fly towards the right with an energy related to the amount of tension in the band just before release. I imagine this when moving a charge closer to another charge. As I move one charge closer and closer to another one, the amount of energy that the charge will have once released becomes greater and greater. This is my understanding of potential. Is this correct?


*What does adding another charge do in the problem below?

For a grounded sphere ($V=0$) a single image charge is sufficient. But with the addition of a second image charge, the same basic model will handle the case of a sphere at any potential $V_0$ (relative to infinity). What charge should you use, and where should you put it? Find the force of attraction between a point charge $q$ and a neutral conducting sphere.

Update: I solved for the potential on the outside of the sphere and on the surface. If I'm wrong, please feel free to help me understand better.
Potential outside the sphere:
$V=\displaystyle\frac{kq}{R}$ when $r>R$.
Potential on the surface:
$V=\displaystyle\frac{kq}{R} + kq(\frac{1}{r} - \frac{1}{R})$ when $r<R$.
 A: You do know where $\nabla$$^2$V = -$\rho$/$\epsilon$$_0$ comes from. it comes directly from the divergence of the electric field and from the relation E = -$\nabla$V. There is a proof in Griffiths which shows why the solution of V is unique, if at all the boundaries of the system the potential is known, and the charge within the system remains the same. Given the uniqueness of this system, if one can intuitively guess a situation in which these situations hold, then ANY problem having the same initial conditions will have the same solution for potential. That is the basic idea behind image charge. To reduce the problem.
A: Presumably you understand how to solve the problem of finding the potential outside a conducting sphere of radius $R$ (held at $V = 0$) when a point charge $q$ is placed at some distance $a$ outside the sphere ($a > R$), using the method of images (placing an image charge $q'$). To be sure, this is Example 3.2 in the textbook by Griffiths. 
The problem you have posed asks you what you must do to solve this problem when the conducting sphere is held at a potential $V = V_0$. The answer is that you place a point charge $q''$ at the center: doing so keeps the sphere an equipotential, but just raises the potential at the surface to $V_0$ if 
$$q'' = \frac{V_0 R}{k}$$
($k = 1/(4\pi\epsilon_0)$ in the notation you used in the comments.)
For the sphere to stay neutral, the two image charges $q'$ and $q''$ must add up to zero: 
$$q' + q'' = 0$$
Then the force on external charge $q$ (which is the physical charge, in the region where you're effectively solving Laplace equation if you like) is due to $q'$ and $q''$ and is given by
$$F = kq\left[\frac{q''}{a^2} + \frac{q'}{\left(a - (R^2/a)\right)^2}\right]$$

Note that what you have done is solve Laplace equation in the region outside the spherical cavity (where the physical charge $q$ is placed). You cannot ask (in the above problem) what the potential or field inside the cavity are.

What you asked in the comments was therefore a different problem: consider a grounded conducing spherical cavity of radius $R$, with a charge $q$ placed somewhere inside it. What is the electric field and potential inside? Note that this is a different problem and should not be confused with the image problem above because for this problem you will effectively solve the Laplace equation inside the sphere. For more on this problem, see Problem 2.2 in http://users-phys.au.dk/thostrup/pdf/chapter2.pdf.
I will edit this answer to resolve any outstanding doubts/concerns you might have.
