# Find the electric potential of an uniformly charged disk using spherical coordinates

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!

• They are both surface integrals. It might have been clearer if the author had been more explicit that $\mathrm{d}s'=r'\mathrm{d}\phi ' \mathrm{d}r'$ – garyp Oct 17 '16 at 12:47
• Why do we use $\rho$ in the second formula?? Does it mean $\phi$ is a volume integral? – WeiShan Ng Oct 17 '16 at 12:59
• 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? – garyp Oct 17 '16 at 13:41
• 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? – garyp Oct 17 '16 at 14:57
• 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. – WeiShan Ng Oct 17 '16 at 16:00

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.
• 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. – Lelouch Oct 17 '16 at 15:02
• 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 ? – Lelouch Oct 17 '16 at 15:04
• 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" – garyp Oct 17 '16 at 15:05