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Kindly refer to page 261 in the link below:

Click here enter image description here enter image description here Griffiths claims that for an electron orbiting an origin, the centripetal acceleration is sustained by the electrical forces $\frac{1}{4 \pi \epsilon_{0}}\frac{e^{2}}{R^{2}}=m_{e}\frac{v^{2}}{R}$

Note that he made no references as to what this electron $-e$ is orbiting with respect to. However, from the expression above with the $e^{2}$ and by Coloumb's law, it can be guessed that there is a charge at the origin. Otherwise, it is nonsensical to speak of electrical force of an electron. It is always the electric force between at least two charges. The charge, I think, should be $-e$ for if it were not, the expression $\frac{1}{4 \pi \epsilon_{0}}\frac{e^{2}}{R^{2}}=m_{e}\frac{v^{2}}{R}$ would be negative. The only caveat is if everything is in scalar form.

He goes on to hypotheses that if this orbit were in the presence of a magnetic field in the z-direction, this magnetic force would be added to the above expression $\frac{1}{4 \pi \epsilon_{0}}\frac{e^{2}}{R^{2}}=m_{e}\frac{v^{2}}{R}$.

Recall that the magnetic force is given by $-e\left ( \vec{v} \times \vec{B}\right )$ for a negative charge -e. The magnetic field is perpendicular to the orbit of -e and $\vec{v}$ the velocity vector of -e. Then we have -evB.

But adding this force to the expression $\frac{1}{4 \pi \epsilon_{0}}\frac{e^{2}}{R^{2}}$ would result in $\frac{1}{4 \pi \epsilon_{0}}\frac{e^{2}}{R^{2}} -evB$.

Griffiths, however, has$\frac{1}{4 \pi \epsilon_{0}}\frac{e^{2}}{R^{2}} +e\bar{v}B$ which I cannot understand.

Someone please help.

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2 Answers 2

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You're correct that the charges are equal in magnitude. However, this equality really just concerns the magnitude. The sign and thus direction of the vector depends entirely on the geometry of your system and how you define your coordinates. The key is that the magnitude of the centripetal force is equal to the magnitude of the Coulomb force.

In vector form it would be $$\frac{1}{4 \pi \epsilon_{0}}\frac{e \cdot -e}{R^{2}}(-\mathbf{\hat r}) = m_{e}\frac{v^{2}}{R}(\mathbf{\hat r})$$

The vectors are antiparallel, so one force has the opposite sign of the other, hence the minus sign on the left. The two negatives cancel, and both sides have the same sign.

Same for the second part. The Lorentz force term is negative, so it becomes

$$\left( \frac{1}{4 \pi \epsilon_{0}}\frac{e \cdot -e}{R^{2}} - evB\right) (-\mathbf{\hat r})$$

which is positive.

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  • $\begingroup$ Your explanation is helpful. $\endgroup$
    – Physkid
    Commented Sep 1, 2016 at 13:00
  • $\begingroup$ One last question from this thread: The electric force is in the direction towards the charge -e. Why then is the radial direction given by you $-\hat{r}$? $\endgroup$
    – Physkid
    Commented Sep 1, 2016 at 13:11
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    $\begingroup$ We're dealing with the orbiting electron, so the force is radially inward, from the electron to the central positive charge. We defined our coordinate system such that outward is positive (centripetal) and inward is negative (Coulomb). Hence the negative sign. $\endgroup$
    – Kyle Arean-Raines
    Commented Sep 1, 2016 at 13:24
  • $\begingroup$ As electrons cannot realistically orbit anything (the bound state is not a classical orbit), where does that leave your derivation? $\endgroup$
    – Gert
    Commented Sep 1, 2016 at 13:30
  • $\begingroup$ This is clearly a classical mechanics problem... $\endgroup$
    – Kyle Arean-Raines
    Commented Sep 1, 2016 at 13:31
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The total force exerted on an electron is : $ -\left(\frac{e^2}{4\pi\epsilon_0R^2}+evB \right)\vec{e}_r $ and the acceleration is $-\frac{v^2}{R}\vec{e}_r$.

So by using the 2nd Newton's law, you get the expression (projected on the radial direction) : $ \frac{e^2}{4\pi\epsilon_0R^2}+evB =m\frac{v^2}{R}. $

Regards

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