Electric Field and Charge Density

[jesuslovesu]

[SOLVED] Electric Field and Charge Density

Oops, nevermind I guess I just use div(E) = rho/e0

Homework Statement

A layer of charge fills the space between x = -a and x = a. The layer has a charge density $$\rho (x)$$. The electric field intensity everywhere inside the charge distribution is given by $$E(x) = \hat{x} Kx^3$$ where K is a constant[/tex]

Homework Equations

The Attempt at a Solution

I asked my professor about this and he said the $$\rho(x)$$ should be a volume charge density. So basically it's an infinite slab (in the y and z dir) Having some difficulty in finding the charge density.

I am assuming the charge density is NOT constant everywhere, correct?
I recognize that this requires a Gauss's Law formulation. Similar to an infinite plane if I am not mistaken.
$$E(A) = Qin/e0$$
$$E(2A) = \rho (x) * A * (2a)/e0$$
$$E = \rho (x) * a/e0$$
Can I just plug in E and rearrange to get $$\rho (x)$$ ? Am I handling the ends correctly? I am basically following the same procedure for finding the E of an infinite plane except I am using 2a as the thickness.

 

[member 553083]

SOLUTION:Yes, you can use Gauss's Law to solve this problem. First, we need to calculate the electric field at a given point x inside the charge distribution: E(x) = \hat{x} Kx^3Then, we can calculate the total charge inside the charge distribution by integrating the charge density over the region between x=-a and x=a: Q_{total} = \int_{-a}^{a} \rho(x) dxFinally, we can use Gauss's Law to relate the electric field and charge density: E(x) = \frac{Q_{total}}{A\epsilon_0}where A is the area of the charge distribution (in this case it is 2a). We can then rearrange this equation to solve for the charge density: \rho(x) = \frac{E(x)\epsilon_0}{A} = \frac{\hat{x}Kx^3\epsilon_0}{2a}