Cylindrical symmetry, Gauss's Law

AI Thread Summary
To calculate the electric field inside a semiconducting nanowire with a volume charge density of ρ(r)=ρ0(r/R), Gauss's Law is applied, recognizing that the electric field points radially outward due to symmetry. The discussion involves integrating the charge density to find the enclosed charge (Qenc) within a Gaussian cylinder. There is confusion regarding the correct application of the formula, particularly in how to relate Qenc to the electric field and the permittivity of free space (epsilon naught). A participant notes a potential error in the calculations, suggesting a missing factor of 2 in the integration process. The conversation emphasizes the importance of accurately accounting for the charge distribution when applying Gauss's Law.
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Homework Statement


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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


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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?
 
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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.
 
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