I'm working on a problem taken from Zangwill's Modern Electrodynamics, where I'm asked to derive the well known result of the electric field $\mathbf{\vec{E}(\vec{r})}$ both inside and outside a uniformly charged sphere of radius $R$ by superposing the fields from a collection of uniformly charged disks.

I'm having a hard time understanding how to get from the volume charge density $\rho$ of the sphere to the surface charge density $\sigma$ of one disk.
According to the solution provided (see the picture below for the notation),


a disk of radius $r = R\sin{\theta}$ has a surface charge density $\mathrm{d}\sigma = \rho R \sin{\theta}\mathrm{d}\theta$ .

Could someone show me, step by step, how to get to this result?
Thanks in advance.


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The charge of the disk is $dQ=\rho d\tau =\rho \pi {{r}^{2}}(R\sin (\theta )d\theta )$ and the surfacic charge is $\sigma =dQ/S=\rho (\pi {{r}^{2}})(R\sin (\theta )d\theta )/\pi {{r}^{2}}=\rho R\sin (\theta )d\theta $

The main point is to see that the height of the disk is $R\sin (\theta )d\theta $ and not $Rd\theta $

  • $\begingroup$ Fracticelli, There is an important detail that's been overlooked, $\sigma$ is not equal to $dQ/S$. $dQ$ is the charge enclosed within a cylinder (disk with height) of base area $\pi R^2\sin(\theta) $ and of height $Rd\theta$ whereas $\sigma \times S$ is the charge enclosed within a two dimensional disk with an area of $\pi R^2 \sin(\theta)$. $dQ$ is enclosed within a volume, $\sigma \times dS$ is enclosed within a surface, thus $\sigma \ne dQ/S$. $\endgroup$ – Hilbert Aug 30 at 17:30

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