Using Coulomb's law with atomic nucleus

Question

In my text book they stated:

As an example, the wave function of an electron that is in position $2p$ in a hydrogen atom given by: $$ \varPhi(r,\theta,\varphi)=\frac{1}{4\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0} e^\frac{-r}{2a_0}\cos\theta $$ If we want to calculate the average electrostatic force of the electron (Assume that the atomic nucleus is at rest) then we need to use the Coulomb's law: $$ \vec{F}=-\frac{kq_e^2}{r^2}\hat{r} $$

I don't understand why the Coulomb's law is between two electrons and not between the electron and proton inside the atomic nucleus. did they meant the proton but because $|q_e|=|q_p|$ then they just wrote $q_e$?

Answer 1

$$q_e$$ just means elementary charge, not charge of an electron.

Answer 2

It seems your textbook skipped a step in the explanation.

In a hydrogen atom the Coulomb force between proton ($p$) and electron ($e$) is $$\vec{F}=\frac{kq_pq_e}{r^2}\hat{r}$$

Now the proton charge is the opposite of the electron charge: $$q_p=-q_e$$

Therefore you can rewrite the Coulomb force between proton and electron as $$\vec{F}=-\frac{kq_e^2}{r^2}\hat{r}$$