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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?

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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.

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  • $\begingroup$ I see, thank you so much. But, is there as a reason as to why we take the electrostatic force as -ve while calculating the potential energy of the electron? What's the importance of directions in that context? Also, while studying a bit of Coulumb's law i saw that there's a modulus in the numerator q1 x q2, does that have anything to do with this? $\endgroup$
    – 5.1pyrso
    Nov 15, 2020 at 10:02
  • $\begingroup$ @5.1pyrso If you found the answer helpful, please give it an upvote by clicking the up arrow next to the answer. “is there as a reason as to why we take the electrostatic force as -ve while calculating the potential energy of the electron?” Short answer: Because it is negative? Long answer: Please create another post and include more details. “while studying a bit of Coulumb's law i saw that there's a modulus in the numerator q1 x q2, does that have anything to do with this?” Probably, but this question also lacks details. $\endgroup$ Nov 15, 2020 at 10:38
  • $\begingroup$ thank you! I upvoted but since I am a new contributer it doesnt show up :'(. Yes, I will try to make a different post about all that. Thank you again! $\endgroup$
    – 5.1pyrso
    Nov 15, 2020 at 12:34
  • $\begingroup$ @5.1pyrso Yes, I should have known better. You should still be able to mark the answer as accepted. $\endgroup$ Nov 15, 2020 at 12:38
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The force equation says -Fe + Fc = 0. So you do have a negative sign coming from the negative charge of electron.

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  • $\begingroup$ Doesn't that mean that the negative sign transfers over to mv^2/r? $\endgroup$
    – 5.1pyrso
    Nov 15, 2020 at 9:18
  • $\begingroup$ $F_c$ 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. $\endgroup$ Nov 15, 2020 at 9:54

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