## JenniferSmart1 Group Title Gauss's Law anyone? one year ago one year ago

1. JenniferSmart1

A very large (infinite), uniformly charged slab of plastic of thickness 2a occupies the region between the z=-a plane and the z=+a plane. |dw:1359854942958:dw|

2. JenniferSmart1

In the problem they drew an object in the shape of a can, why do we assume that that's the shape?

3. JenniferSmart1

I meant to say "In the solution"

4. JenniferSmart1

It further says Find the Electric field everywhere due to this charge configuration. The charge per unit volume is $\rho$ The solution is $\vec{E}=E_z\hat{k}=$ and then it gives us a piecewise solution I can write them all out if you want me to

5. Jemurray3

The crux of the argument is that the only possible orientation of the electric field is toward or away from the plate. That is, there can be no component other than either up or down. Do you understand why that is?

6. JenniferSmart1

I don't quite understand it. Wouldn't the electric field point away from that slab of plastic? To the right and left|dw:1359856455623:dw|

7. Jemurray3

Oh, I apologize, I misread the question. Yes, left and right. So that's clear, then?

8. JenniferSmart1

uhm yes. but why do we assume that the shape is the following? It's supposed to look like a symmetrically shaped can. |dw:1359856604911:dw|

9. Jemurray3

What you're trying to do is to use the symmetry argument to determine the electric field strength. A "can" shape is a good choice because the only contribution to the electric flux is through the top and bottom of the can. The sides don't contribute, because the electric field is tangential to them.

10. Jemurray3

So the total flux is equal to the flux through the top of the can plus the flux through the bottom:|dw:1359856720464:dw|

11. JenniferSmart1

Why is $Q_{inside}=\rho A2a?$ I understand that the charge inside (Q) is $\phi_{net}=\frac{Q_{inside}}{\epsilon_0}$

12. JenniferSmart1

$\rho$ is the charge per unit volume of the plastic

13. Jemurray3

rho is equal to the charge per unit volume. Therefore, the total charge is equal to rho times the volume of plastic contained inside the can. 2a* A is equal to the volume contained in the can.

14. JenniferSmart1

So A is the area of ......?

15. Jemurray3

|dw:1359857331358:dw| A is the cross-sectional area of the can

16. JenniferSmart1

are we calculating the charge inside of that can? in the area that I have highlighted? |dw:1359857447106:dw|

17. Jemurray3

The only charge inside the can is in the volume of plastic

18. JenniferSmart1

yeah that makes sense, haha.....sorry about the silly questions. one last question: Why is plastic charged? Wouldn't it make more sense to charge a metal?

19. JenniferSmart1

I meant to say, why is this slab of plastic charged?

20. Jemurray3

You can put charge on whatever you'd like, be it plastic or metal. However, the behavior of charge on metals is very different from that of plastics. Specifically, you can't have uniform charging of metals -- all the charge flows to the outer surfaces.

21. JenniferSmart1

I see. Thanks @Jemurray3 !!!

22. Jemurray3

No problem

23. JenniferSmart1

To recap: We have an infinitely large uniformly charged slab of plastic of thickness 2a. We used the shape of a can to determine the charge inside that plastic. Aha! and that charge should be consistent throughout that infinitely large slab of plastic.