Why Does Feynman Use a Single Electron Charge in Dipole Calculations?

[Seidhee]

Hello,

I am reading the volume 2 of the Feynman's Lectures on Physics, and something is bothering me when he calculates the dipole moment of a single atom induced by an extern field :

https://books.google.co.uk/books?id=uaQfAQAAQBAJ&pg=SA11-PA2&lpg=SA11-PA2&dq=feynman+dipole+single+atom&source=bl&ots=6nmZCqHDMk&sig=cd6wOGCKh9E9227ZX-a8KKhTqlM&hl=fr&sa=X&ei=NuI8VbHdI8v9UOSvgOAD&ved=0CDEQ6AEwAQ#v=onepage&q=feynman%20dipole%20single%20atom&f=false

Indeed, he states that : " p = q_ex "

But why ? I would use in general : " p = Zq_ex " where Z is the number of electrons in the atom.

x is the displacement of the center of charges of the electrons, and thus x is also the displacement of each electron.

Could you explain his reasoning ? It is not the first time he uses a single electron charge instead of Z in his calculations, and I do not understand.

Thanks.
PS : First, I thought that was because the square of the natural pulsation ω₀ depended on Z, which means that ω²₀(Z) = Zω²₀ (Z=1), which would simplify the Z replacing ω²₀(Z) by Zω²₀ (Z=1) ; but Feynman seems to use ω₀ = ω₀(Z) and not ω₀ (Z=1) everywhere, so it does not matter.

 

[Seidhee]

And q_e is obviously equal to the charge of an electron and note equal to Z*(charge), because he uses then e² = q²_e/4piε₀.

Moreover, when he the does his calculations for the Helium atom, he keeps using q_e = charge of one electron and does not use a new value for Z..