electromagnetism - When to use method of images in Electrostatics?

I am a bit confused about when to use the method of images in E&M? For example, in Griffith's Electrodynamics Example 3.2, the problem reads:

A point charge $q$ is situated a distance $a$ from the center of a grounded conducting sphere of radius $R$. Find the potential outside the sphere."

But in Example 2.7, which says,

Find the potential of a uniformly charged spherical shell of radius $R$

or Problem 3.1, which says,

Find the average potential over a spherical surface of radius $R$ due to a point $q$ located inside

the method of image is not used to solve the problem. In Example 2.7, this equation for the potential is used instead: $$ V\left(\mathbf r\right)=\frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r}da' $$

Is it because the object is "conducting" that we use method of images? And if so, why? Could someone help me clear up my misunderstanding?

Answer

For problem solving, the method of images comes in handy in the following cases: conducting sphere, conducting cylinder, conducting ellipsoid, and conducting plane. Another example is two regions of dielectrics, with different $\epsilon$ (permittivity). And that is it. This general rule should save you some time.

So, the first problem you mentioned can be solved with images, the second and third cannot (there is no conductor).

A caveat about using the method of images, which I have seen frequently confuse students is as follows. In the method of images, there is always two distinct regions of space. For example, inside and outside the conductor, left and right of the conducting plane, or in dielectric region 1 and dielectric region 2. The imperative point is that, when you want to find the potential in one region, the image charges are not allowed to be in that region; they should be situated in the other region. For example, if there is a conducting sphere, not grounded, and there is a point charge inside of it (not at the center), and you are asked to find the potential for every point inside the sphere, then you should situate the image charge outside the sphere.

I suggest not confusing yourself too much with the method of images, and remembering that only the above-mentioned geometries have neat closed-form image solutions.