Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
(continued from the last comment) will make the insulator a conductor. I suggest you to take this as fact and move ahead in whatever book you are studying. You will get this second time around with a fresh mind. These types of things do happen a lot where we get stuck in simple things but get it easily when we are revising that concept.
I already told you. But let me rephrase a little bit this time. The electrons are within the grasp of it's core nucleus which exerts far greater force that an applied can. But that being said, the applied force do exert forces on the atoms i.e positive core (nucleus) and the electron cloud. Electron cloud being less massive in comparison to its nucleus do move towards the applied E field while the its massive core remains practically stationary. So we say the applied field has polarized the atom. If the applied field is very strong then it will completely ionize the atom and (see next comment)
The conductors and the insulators, BOTH, experience the same type of force due to an applied external field i.e. like an insulator, a conductor do face a force on it. The expressions might be different but they are the same type of force. Does that make any sense or not?
Actually, the applied E field keeps on stretching atoms till the applied force and the now induced force cancel. It just happens instantaneously, maybe of the order of 10−8secs. When you apply a very strong field to an insulator where the field is zero inside, the insulator becomes a conductor.
The applied electric field is just not strong enough to separate electrons from its nucleus to the point where the now induced field would cancel out the applied field completely. Under normal conditions insulators cancel some of the applied field inside. Apply a strong enough field you will get a zero field inside.
To your second paragraph: Yes, the coordinates are usually denoted by (r,θ,ϕ) but he has taken it as ψ in the original coordinate system too. Things would have been confusion-less had he taken that as θ in the figure 5.45.
I think you are mixing the dictionary definition of energy (work) and the physics definition of energy (work). They both are as different as it can be and that is why it is counter intuitive to you.