Electric field in a solid cylinder with an offset hole

Click For Summary

Homework Help Overview

The problem involves finding the electric field within a cylindrical cavity located in a solid cylinder that has a uniform charge density. The original poster mentions the concept of superposition and expresses confusion regarding the integration process and the application of Gauss's Law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of superposition to determine the electric field, with some suggesting the need to consider the contributions from both the entire cylinder and the cavity. Questions arise regarding the reference frames for the electric fields and how to treat the smaller cylinder in relation to the larger one.

Discussion Status

Some participants have offered hints and suggestions for approaching the problem, while others are exploring different interpretations of the electric field contributions. There is an acknowledgment of confusion regarding the calculations and the reference frames, indicating an ongoing exploration of the topic.

Contextual Notes

Participants mention challenges related to combining fields from different cylindrical coordinate systems and the implications of the cavity's position relative to the larger cylinder. There is also a reference to the course context, which may influence the problem's complexity.

calandra
Messages
3
Reaction score
0
[SOLVED] Electric field in a solid cylinder with an offset hole

Homework Statement


question5.jpg

its teenie, but it says: Superposition: an infinite cylinder of radius R and uniform charge density (row) contains a cylindrical cavity of radius R/2 as shown. Find the electric field in the cavity.



Homework Equations



Gauss's Law I guess.



The Attempt at a Solution


My first thought was 'zero. duh'...but that's not exactly superposition. I know there has to be something about the whole cylinder minus the hole...but I'm not sure how to go about integrating with a common origin or...I've just gotten myself very confused. HELP!
 
Physics news on Phys.org
Here's a hint: A negative charge can cancel a positive charge. :wink:
 
Let [itex]E_1[/itex] the wanted electric field, [itex]E_2[/itex] the electric field produced by the cylinder [itex]R/2[/itex] alone and [itex]E_3[/itex] the electric field produced by the cylinder [itex]R[/itex]. Then

[tex]E_1+E_2=E_3\Rightarrow E_1=E_3-E_2[/tex]


The electric fields [itex]E_2, E_3[/itex] are easily obtained by Gauss law.
 
thank you thank you
however...easily obtained by gauss's law...easily...I'm having a very blonde afternoon...I can't seem to pick a reference frame that works for both E1 and E2. And do I treat E2 as a cylindrical shell with uniform charge distribution? Or as nothing...not that it really matters...its still zero right?
 
Come up with two solid uniform cylinders of charge which add up to give you the needed charge configuration.
 
Hey there, I take it you're in Phys 231/239 as well?
 
wow...lol.
Yup 239.
Have you finished it?
 
I'm done the assignment except for this question. I too was having some trouble combining the two fields (awkward using two separate cylindrical coordinate systems). I think I've figured it out now though.

I'm technically in 231, but it's just 239 renamed for us purple folk.
 


Doc Al said:
Come up with two solid uniform cylinders of charge which add up to give you the needed charge configuration.
Hi, I'm having trouble with the same problem. The center of the cavity is at a distance A from the center of the big cylinder. My calculations of superpositioning the two fields yields an electric field within the cavity equal to zero. Is this correct?
 
  • #10


irish_coffee said:
My calculations of superpositioning the two fields yields an electric field within the cavity equal to zero. Is this correct?
No, that's not correct. Show how you came to that conclusion.
 
  • #11


I know, I discovered it was nonsense. I now have calculated the vector product of the vector field produced by the big solid cylinder with charge density rho and the vector field produced by the small cylinder with charge density minus rho. Because this adds up to the situation we have here (a big cylinder with a cavity with density of zero). The direction of the resulting vector field in an arbitrary point P within the cavity is in the direction of the vector connecting the center of the big cylinder and the center of the cavity.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
2K
  • · Replies 6 ·
Replies
6
Views
13K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K