Finding the Electric Field of a non-uniform Semi-Infinite Slab of Charge

[euphtone06]

Homework Statement

Non uniform slab of charge density $$\rhov = \rho_{0}cos(pi*x/(2d))$$
extends infinitely in the y and z planes is present between -d < x < d.
Find the Electric Field everywhere.

Homework Equations

$$\int\epsilon_{0}{E^\rightarrow}\bullet{ds^\rightarrow}= \int\rho_{v}dv$$

The Attempt at a Solution

So I have an understanding of how Gauss' Integral Law is supposed to function but this semi-infinite slab is confusing me.

First I will attempt to solve the rhs of Gauss' integral law
$$\int\rho_{v}dv = \int_{0}^{L}\int_{0}^{W}\int_{-d}^{d}\rho_{0}cos(\pi x/(2d))dxdydz = \int_{0}^{L}\int_{0}^{W}\frac{2d\rho_{0}sin(\pi x/(2d))}{\pi} |^{d}_{-d}dydz = \frac{4d\rho_{0}WL}{\pi}$$

Now I believe I did that correct the lhs has me much more confused.

Since it is infinite about the y and z axis wouldn't it observe some of the same principles as a infinite sheet charge?

I don't believe we can define ds as the surface area is infinite. So that forces the use of the integral. Which is where I believe I am going wrong. There is only charge at the +x and -x axis outside of d and -d. How do I set up the Lhs?
$$\int_{0}^{L}\int_{0}^{W}\epsilon_{0}E a_{x}dydz = \epsilon_{0}EWL$$
This cannot be correct because when applying Gauss' Differential Law there is no way to get the surface charge density without a x variable in which to differentiate.

Thats where I am stuck how is the LHS of Gauss' Integral Law to be implemented such that when applying Gauss' Differential Law I receive the surface charge density $$\rho_{0}cos(pi*x/(2d))$$

Thanks for the help

 

[anathema]

!

Dear fellow scientist,

Thank you for reaching out for help with this problem. I can see that you have a good understanding of Gauss' Integral Law and have made a solid attempt at solving the problem. Let me try to guide you towards the correct solution.

First, let's consider the electric field at a point P outside the slab, at a distance r from the center of the slab along the x-axis. Since the slab extends infinitely in the y and z planes, we can assume that the electric field will have the same magnitude and direction at all points on a circle of radius r centered at the origin. This is similar to the electric field of an infinite line charge, where the electric field is uniform along a circle of constant radius.

Now, let's consider a small element of charge on the surface of the slab, located at a distance x from the origin along the x-axis. The electric field due to this element of charge at point P can be calculated using Coulomb's Law:

dE = \frac{1}{4\pi\epsilon_{0}}\frac{\rho_{0}cos(\pi x/(2d))dx}{r^{2}}

Note that we have used the fact that the charge density is constant along the y and z directions, so we only need to consider the x-component of the electric field. Also, since we are considering a small element of charge, we can treat it as a point charge.

Now, we can integrate this expression over the entire surface of the slab to get the total electric field at point P:

E = \frac{1}{4\pi\epsilon_{0}}\int_{-d}^{d}\frac{\rho_{0}cos(\pi x/(2d))dx}{r^{2}}

This integral can be evaluated to get the final expression for the electric field at point P:

E = \frac{\rho_{0}}{2\epsilon_{0}r^{2}}\left(sin(\pi d/(2r)) - sin(-\pi d/(2r))\right)

= \frac{\rho_{0}}{\epsilon_{0}r^{2}}sin(\pi d/(2r))

= \frac{2\rho_{0}}{\epsilon_{0}\pi r}cos(\pi x/(2d))

Note that this expression is valid for all points outside the slab, including points on the x-axis. For points inside the slab,

 

[baba89]

!

Thank you for your question. It seems like you have a good understanding of the problem and Gauss' Integral Law. Let me try to clarify the concept for you.

First, you are correct in your approach to solving the right-hand side of Gauss' Integral Law. Your integral for the charge density is correct and gives you the total charge enclosed within the semi-infinite slab.

Now, for the left-hand side of the equation, we need to consider the electric field at a point P due to the entire semi-infinite slab. This electric field will be a vector quantity, with components in the x, y, and z directions. We can use the principle of superposition to find the electric field at point P, by considering the electric field due to each small element of charge within the slab.

For simplicity, let's consider a small section of the slab at a distance x from the origin, with width dx and charge density \rho_{0}cos(pi*x/(2d)).

Using Gauss' Differential Law, we can write:

\oint\epsilon_{0}{E^\rightarrow}\bullet{ds^\rightarrow}= \int\rho_{v}dv

Since we are considering a small section of the slab, the electric field will be constant and perpendicular to the surface of the slab. This means that the left-hand side of the equation can be simplified to:

\epsilon_{0}E\int ds = \epsilon_{0}E(2d)dydz = 2\epsilon_{0}Edydz

Now, for the right-hand side of the equation, we can substitute the expression for charge density that we found earlier:

\int\rho_{v}dv = \int_{-d}^{d}\rho_{0}cos(\pi x/(2d))dx = \frac{4d\rho_{0}}{\pi}\int_{-d}^{d}cos(\pi x/(2d))dx

Integrating this, we get:

\frac{4d\rho_{0}}{\pi}\left[\frac{2d}{\pi}sin(\pi x/(2d))\right]_{-d}^{d}

Simplifying this, we get:

\frac{4d\rho_{0}}{\pi}\left[\frac{2d}{\pi}(0-0)\right] = 0

Therefore, we have:

2\epsilon_{