Electric Field and Conductor Properties

AI Thread Summary
The discussion revolves around understanding electric flux in relation to a cylindrical insulator placed in an external electric field. It concludes that the net electric flux through the cylinder is zero, as there is no charge enclosed within it despite the presence of an external field. Additionally, it clarifies that the electric field at the surface of a conductor depends on its area, curvature, and total charge, confirming that option e is correct. The conversation emphasizes the importance of Gauss' Law in determining electric flux and highlights the relationship between surface charge density and electric field intensity. Overall, the key takeaway is the application of Gauss' Law to analyze electric fields and flux in conductors and insulators.
Soaring Crane
Messages
461
Reaction score
0

Homework Statement



A cylindrical piece of insulating material is placed in an external electric field, as shown in the figure. The net electric flux passing through the surface of the cylinder is

a.positive.

b.negative.

c.zero.

Figure : http://i131.photobucket.com/albums/p289/SoaringCrane/Tip23_fig11.gif


Homework Equations



Possibly: electric flux = E*A*cos theta

The Attempt at a Solution



I’m really unsure about this question. Would the net flux be a. positive? The flux on the ends would be 0 because the E and dA vectors are perpendicular to each other while the flux on the wall near the middle will be positive since E and dA are parallel to each other?


Homework Statement



The electric field at the surface of a conductor


a.is parallel to the surface.

b.depends only on the total charge on the conductor.

c.depends only on the area of the conductor.

d.depends only on the curvature of the surface.

e.depends on the area and curvature of the conductor and on its charge.



Homework Equations



This is really a conceptual question. Possibly electric flux = integral[E*dA] = Q_enclosed/epsilon_0

The Attempt at a Solution



Is the correct choice e. depends on the area and curvature of the conductor and on its charge? I know that the electric field intensity increases as the curvature increases. The electric field for an infinitely long sheet is based on the surface charge density, so area and charge (from Gauss’ law) are other factors?

Thanks.
 
Physics news on Phys.org
You are trying to work way too hard for the first problem. What IS Gauss' Law?
For the second one, yes it's conceptual. You know it depends on curvature and surface charge density. And you do know a) is false, right? (Why?) So everything is pointing at e) alright.
 
Last edited:
Gauss' law states that the net electric flux for a closed surface is equal to the total electric charge over epsilon_0.
 
Soaring Crane said:
Gauss' law states that the net electric flux for a closed surface is equal to the total electric charge over epsilon_0.

You mean "total charge enclosed in the surface". Right?
 
Last edited:
Yes, Q_enclosed/epsilon_0. What am I doing incorrectly for the first question?
 
You aren't doing anything wrong! But how much charge is enclosed in the cylinder?
 
It does not say how much charge is inside. (Is there even any charge inside??)
 
Exactly! Judging by the phrase "external electric field" and looking at the picture I would say there is no charge inside. Hence?
 
The net electric flux is 0.
 
  • #10
Yes. That would be correct.
 
Back
Top