- #1

PeteyCoco

- 38

- 1

## Homework Statement

Given a volume charge density function defined as follows:

[itex]\rho=\frac{dQ}{d\tau}= \begin{cases}z-z^{2} & 0<z<1\\

z+z^{2} & -1<z<0\\

0 & \text{everywhere else}

\end{cases}[/itex]

and is independent of x and y.

Determine the electric field everywhere (i.e. along the z axis) by an application of Gauss's Law in Differential form. Explain why the field is zero at [itex]z=\pm1[/itex]

## The Attempt at a Solution

I struggled with this one for a bit, because I couldn't visualize what the math I was doing meant physically. Here's where I got:

For each of the regions holding a charge, the field produced has no x and y components (they cancel by symmetry), so the differential form of Gauss's Law becomes

[itex] \nabla\bullet\bar{E}=\frac{\partia{lE_{z}}}{\partial{z}}= z -z^{2}[/itex] ,

(working with one part of the slab)

I integrated this to get

[itex]E_{z} = \frac{1}{2}z^{2} - \frac{1}{3}z^{3} + k, \text{k a constant to be determined}[/itex]

To find the constant, k, I looked at the outer edge of the surface at z = +1. I made a guess that the field here is zero because the field contribution by the thin charged sheets that make up the slab cancel (the field above a uniformly charged thin sheet of infinite dimensions is uniform). I'm going to prove this later hen I work through the problem with the integral form of Gauss's Law.

Anyway, knowing that we can find the particular solution to the differential equation above

[itex]E_{z} = \frac{1}{2}z^{2} - \frac{1}{3}z^{3} + \frac{1}{6}[/itex]

So that's where I've gotten. I stopped only because I have no idea if what I'm doing is even remotely correct. Some help would be appreciated.