# Electric Potential of an Electron Orbiting a Nucleus

Based on my understanding, electric potential is $$\frac{kg}{r}$$. Why is the electric potential felt by an electron orbiting a nucleus is quantitatively described by the equation in image shown below?

Source: Quantum Physics by Robert Eisberg

• Are you confused by the use of the letter $V$? It's not voltage. Mar 10 '21 at 2:56
• What is $g$? Did you mean $q$? Mar 10 '21 at 6:22

This is the potential energy. First of all, $$k = 1/(4 \pi \varepsilon_0)$$. The potential energy is given by $$V = -e\phi$$ where $$\phi$$ is the electrostatic potential, which is the formula that you write (I think). So we have: $$V(r) = -e\phi(r) = -e(\frac{ke}{r}) = \frac{-ke^2}{r} = \frac{-e^2}{4\pi\varepsilon_0}$$ $$Z$$ is the atomic number. If our electron is orbiting a nucleus with $$Z \neq 1$$, then we just multiply the above by $$Z$$.