-1
$\begingroup$

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),

Zangwill_3.3(b)-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.

$\endgroup$

closed as off-topic by Kyle Kanos, M. Enns, ZeroTheHero, Jon Custer, Emilio Pisanty Jan 21 at 14:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, M. Enns, ZeroTheHero, Jon Custer, Emilio Pisanty
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

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 $

$\endgroup$
  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.