# Why is the charge of electron not taken as negative when calculating the electrostatic force between the nucleus and electron in Bohr's model of atom?

In one of the postulates of Bohr's model of hydrogen atom it is said that "While the electron revolves, the electrostatic force between the electron and nucleus provides centripetal force. The derivation then follows:

$$F_e = F_c$$

$$\implies \frac{\textsf{charge of electron x charge of nucleus}}{4 \pi \epsilon_0 \times r^2} = \frac{mv^2}{r}$$

$$\implies \frac{e \times +Ze}{4 \pi \epsilon_0 \times r^2} = \frac{mv^2}{r}$$

In this derivation, we take the nucleus's charge as positive, but why don't we take the elementary charge as negative if we are talking about the charge of electron here?

The simple explanation is that we only need to consider magnitudes in the equations, as directions are trivial in this situation (there is only one force, and it is clearly in the required direction). Now think of $$F_e$$ and $$F_c$$ not as forces, but as magnitudes of forces. This is consistent with the lack of vector notation.
• $F_c$ is the centre-seeking component of the net force. In this case, it is the net force. Therefore, it belongs on the other side of the equation: $F_1 + F_2 + \cdots + F_n = F_c$. Trying to move it to the left side of the equation is just confusing, and I think 5.1pyrso’s comment above just expresses that confusion. Nov 15, 2020 at 9:54