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## Homework Statement

A semi-infinite (infinite in y and z, bounded in x) slab of charges carries a charge per unit volume ##\rho##. Electric potential due to this slab is a function of horizontal distance, x from the center of the slab. It is linear for ## x \lt -1m## & ## x \gt 1m##, and is given by ##V(x) = \frac 15 2 x^2 - \frac 25 2## for ##-1m \leq x \leq 1m##. Find the x-component of the electric field. Use Gauss' law to find the charge density ##\rho## of the slab.

Vector Potential Plot:

## Homework Equations

Gauss' Law:

##\frac {q_{encl}} {\epsilon _0} = \oint \vec E \cdot d \vec A ##

##\rho = \frac Q V##

##\vec E = -\nabla V##

## The Attempt at a Solution

Using the equation for the electric potential from the plot I have ##V(x) = 15x + V_0## for ##x## being outside of the bounds. (##x \lt -1m## and ##x \gt 1m##)

I get the electric field to be:

##\vec E = -\nabla V##

##\vec E = -15 \hat x##

Then for the electric potential outside of the bounds, ##V(x) = \frac 15 2 x^2 - \frac 25 2##

Here I get the electric field to be:

##\vec E = -15x \hat x##

But I'm not sure that it makes sense to have two different values for the electric field? Shouldn't I just get one function?

Then when I solve for charge density I get:

##\rho = \frac {q_{encl}} V##

##Volume = A x##

##\rho = \frac {[\oint \vec E \cdot d \vec A] \epsilon _0 } {A x} ##

##\rho = \frac {\epsilon _0} x E_x##

Which doesn't really make sense because I get the charge density falling off at ##\frac 1 x## outside of the bounds, and I get it to be constant within the slab. Which should be the other way around (it should be constant outside of the bounds of ##x##). I'm not sure if I attempted this problem wrong, any help would be appreciated, thanks!