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Electric potential
A circular disk of radius $b$ carries a uniform surface charge density $\sigma$.
What is the electric potential of an arbitrary point along the z-axis from the disk ?

I understand how we get the first formula but what I couldn't figure out is the second formula. How does the surface charge density change to volume charge density?

Thanks for the help!

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  • $\begingroup$ They are both surface integrals. It might have been clearer if the author had been more explicit that $\mathrm{d}s'=r'\mathrm{d}\phi ' \mathrm{d}r'$ $\endgroup$ – garyp Oct 17 '16 at 12:47
  • $\begingroup$ Why do we use $\rho$ in the second formula?? Does it mean $\phi$ is a volume integral? $\endgroup$ – WeiShan Ng Oct 17 '16 at 12:59
  • $\begingroup$ I didn't even see that. It may be a careless mistake by the author. That would be in keeping with the careless mistakes in the first set of formulas: what happened to $b$? On the other hand, there might be something that you left out. What happened to the factor of 2? Is there more to this problem that we should know about? $\endgroup$ – garyp Oct 17 '16 at 13:41
  • $\begingroup$ Note that the units in the last equation are incorrect. I think it's a mistake, and that the author intended $\sigma$. But there is still the question of the factor of 2, so I'm not sure that there's not something else going on. Is there any text between those two expressions? $\endgroup$ – garyp Oct 17 '16 at 14:57
  • $\begingroup$ It's from a brief lecture note. And no, there isn't any text between these two expressions. That's why I'm so confused. $\endgroup$ – WeiShan Ng Oct 17 '16 at 16:00
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I cannot account for the missing factor of 2 in the last formula. However, the rest appears to be fine. The author showed the integral over a surface, because the object was planar, and then generalised with a volume charge density. To give an informal proof why $\rho$ will work here as well, we start with the mathematical definition : $\int_{-\infty}^{\infty}\int_0^{2\pi}\int_0^\pi\rho.r^2dr.d\phi .d\theta$ = q(in all space). Here, The disc is planar, so the azimuthal angle will have non zero value of $\rho$ only for some particular value of the angle, as all other planes contain no charge. So the given integral will reduce to :

$\int_{-\infty}^{\infty} \int_0^{2\pi} \sigma . r.dr. d\theta$ = q(on disc) since the azimuthal $\phi$ is effectively constant in this case(all other values of $\phi$ yield zero charge, since $\rho$ becomes zero for them.

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  • $\begingroup$ You may be on the right track, but your integrals have an incorrect volume element. And since it's a disk, why not use cylindrical coordinates? Doing this analysis in spherical coordinates leads to difficulties. (By the way, what you call the azimuthal angle is not that. The azimuthal angle is the other one.) $\endgroup$ – garyp Oct 17 '16 at 14:53
  • $\begingroup$ the angle in the plane is the polar angle, that sweeps over the disc. The azimuth determines the plane. There is charge on only one plane. And since $\phi \in [0,\pi)$, there will not be another plane with charge on it. Cylindrical is welcome, but i used the convention in the OP's question for analysis. $\endgroup$ – Lelouch Oct 17 '16 at 15:02
  • $\begingroup$ I am calling $\phi$ the azimuth by the way. I dont know if its the other way round, but my calculation seems fine. Can you specify the mistake i made in more detail ? $\endgroup$ – Lelouch Oct 17 '16 at 15:04
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    $\begingroup$ I guess it depends on how you orient your coordinate system. Typically the azimuthal angle runs from $0$ to $2\pi$ and the polar or altitude angle runs from $0$ to $\pi$. From "north" to "south" $\endgroup$ – garyp Oct 17 '16 at 15:05
  • $\begingroup$ Yes, I believe the OP will understand what i meant anyway. Any doubts are welcome. $\endgroup$ – Lelouch Oct 17 '16 at 15:06

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