# Surface charge density from volume charge density [closed]

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?

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

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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 $$Rd\theta$$
• 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 $\sigma \ne dQ/S$. – Hilbert Aug 30 at 17:30