Polarization charge density

[Dassinia]

Hello,

Homework Statement

We use an accelerator to uniformly implant electrons in a block of lucite, its surface is 25 cm², thickness 12mm.
The current of the electron beam is 0,1 µA and the implantation takes 1s.
The electrons are "trapped" in a layer of uniform density of 2mm thickness whose center is at 6 mm of the block surface. The two large surfaces have same potential V=0.

- Find the polarization charge density ρ_p in each of the three regions of the block.
- Find the polarization charge density of surface σ_p over the two large surfaces of the block.

Homework Equations

ρ= ∇. D
D=ε_oE+P
P=ε_o X_eE
σ_p= P.n^

The Attempt at a Solution

I'd say that outside the layer where the electrons are ρ_p =0
And I am really really stuck ..
I have to return this before the end of tomorow, and I'm not getting any of this.. I've read the course notes a hundred times
Thanks !

 

[anathema]

Hello,

Thank you for your post. I can provide some guidance and help with your problem.

First, let's define some variables and equations that we will need to solve this problem. We have the electric displacement field D, which is related to the electric field E and polarization P by the equation D = ε0E + P. We also have the charge density ρ, which is related to D by the equation ∇ · D = ρ. And finally, we have the polarization charge density σp, which is related to P and the unit normal vector n by the equation σp = P · n.

Now, let's break down the problem into three regions: the region outside the layer where the electrons are, the region inside the layer, and the two large surfaces of the block.

In the region outside the layer, we can assume that there are no free charges or polarization charges, so ρp = 0. We also know that the electric field is perpendicular to the surface of the block, so we can use Gauss's law to find the electric field E = σ/ε0. Since the potential on the two large surfaces is V = 0, we can use the definition of potential to find the electric field E = -∇V = 0. Therefore, we have D = ε0E = 0, which means that there is no polarization in this region.

In the region inside the layer, we know that the electrons are uniformly implanted in a 2mm thick layer at a distance of 6mm from the surface. This means that the polarization P is also uniform in this region. Using the equation P = ε0χeE, where χe is the electric susceptibility of the material, we can find the polarization P. Then, using the equation D = ε0E + P, we can find the electric displacement field D. Finally, we can use the equation ∇ · D = ρ to find the charge density ρp.

For the two large surfaces of the block, we can use the definition of potential to find the electric field E = -∇V = 0. We also know that the polarization P is perpendicular to the surface, so we can use the equation σp = P · n to find the polarization charge density σp.

I hope this helps guide you in finding the solutions to your problem. If you need further assistance, please don't hesitate