Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Gauss's Law anyone?

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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|
In the problem they drew an object in the shape of a can, why do we assume that that's the shape?
I meant to say "In the solution"

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
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?
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|
Oh, I apologize, I misread the question. Yes, left and right. So that's clear, then?
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|
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.
So the total flux is equal to the flux through the top of the can plus the flux through the bottom:|dw:1359856720464:dw|
Why is \[Q_{inside}=\rho A2a?\] I understand that the charge inside (Q) is \[\phi_{net}=\frac{Q_{inside}}{\epsilon_0}\]
\[\rho\] is the charge per unit volume of the plastic
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.
So A is the area of ......?
|dw:1359857331358:dw| A is the cross-sectional area of the can
are we calculating the charge inside of that can? in the area that I have highlighted? |dw:1359857447106:dw|
The only charge inside the can is in the volume of plastic
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?
I meant to say, why is this slab of plastic charged?
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.
I see. Thanks @Jemurray3 !!!
No problem
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question