Take the title in the figure below as an example to discuss the potential distribution of the conductor in the ground cavity.
First of all, it is necessary to know that the hollow conductor shell shields the influence of the internal and external charges on the corresponding area. That is to say, if the charge is located in the hollow conductor shell, there is no potential distribution outside the shell, and the same is true outside the shell.
Between the conductor shells is an equipotential body, and the potential of the spherical shell is 0.
Then we directly solve the third question
First take the area outside the spherical shell (r>b) as the solution area, then the charge in the shell is not considered, and the image charge of the charge outside the shell in the shell is obtained, and the shell (r=b) is used to solve it. According to the solution of the image charge of the electric field near the conductor shell, it can be known that the position and size of the image charge of Q in the shell have such a relationship:
Then it is only necessary to solve the potential distribution of the image charge and the original charge in the outer region of the spherical shell, and the potential distribution is
(r>b)
Then solve the potential distribution in the spherical shell area (r<a), shield Q, and consider the distribution of q's potential in the shell by considering the image charge of q, the method is the same as above.
Finally, the electric potential distribution in the spherical shell is solved, the conductor is an equipotential body, and the electric potential in the spherical shell is 0.
For various problems and specific solutions of the mirror image method, you can check the following link, which is quite comprehensive: https://max.book118.com/html/2018/0121/149918870.shtm