Cylindrical symmetry, Gauss's Law

Click For Summary

Homework Help Overview

The problem involves calculating the electric field inside a semiconducting nanowire with a specific volume charge density described by ρ(r)=ρ0(r/R). The context is rooted in the application of Gauss's Law and cylindrical symmetry.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of Gauss's Law and the integration of charge density to find the electric field. Questions arise about the correct interpretation of the integral and the calculation of enclosed charge (Qenc).

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the integral and the relationship between charge and electric field. Some guidance has been provided regarding the need to consider the charge distribution correctly, but no consensus has been reached on the correct approach.

Contextual Notes

There is a mention of confusion regarding the use of epsilon naught and the factors involved in the calculations, indicating potential misunderstandings in the application of Gauss's Law.

mawgs
Messages
5
Reaction score
0

Homework Statement


[/B]
A semiconducting nanowire has a volume charge density ρ(r)=ρ0(r/R) where R is the radius of the nanowire. How would you calculate the electric field inside the wire?

Homework Equations



Gauss's Law

The Attempt at a Solution


[/B]
I know that by symmetry the E field only points radially out. Using Gauss's law and finding the dq by integrating in terms of dr, a small ring in the gaussian cylinder. So would you integrate ρ0(r/R)2piLdr from 0 to r?
 
Last edited:
Physics news on Phys.org
mawgs said:
would you integrate ρ0(r/R)2piLdr from 0 to r?
That is a reasonable integral to perform, but what exactly do you think it will give you?
 
Qenc?
 
mawgs said:
Qenc?
Yes. So what is the next step to find the field?
 
haruspex said:
Yes. So what is the next step to find the field?

Set the Qenc over E0 equal to E2πrL. When I did this, I got ρ0r/E0R. I know this is wrong because the homework is online and gives instant feedback. When I integrated, I got .5(2πρ0Lr)/R. I'm not sure where I went wrong.
 
mawgs said:
Set the Qenc over E0 equal to E2πrL.
Sorry, what's E0? Sounds like a field. Why would you divide the charge by a field?
mawgs said:
I know this is wrong
You haven't taken into account the charge between r and R.
 
haruspex said:
Sorry, what's E0? Sounds like a field. Why would you divide the charge by a field?

You haven't taken into account the charge between r and R.

Sorry, I meant epsilon naught.
 
mawgs said:
Sorry, I meant epsilon naught.
Ok. But I think you lost a factor of 2 in there somewhere.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
11
Views
2K
Replies
6
Views
1K
Replies
9
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K