Cylindrical symmetry, Gauss's Law

In summary, the conversation discusses how to calculate the electric field inside a semiconducting nanowire with a volume charge density of ρ(r)=ρ0(r/R). The suggested method is to use Gauss's Law and integrate ρ0(r/R)2πLdr from 0 to r, but the student is having trouble with the calculations. They suggest dividing Qenc by E0, but this is incorrect and they need to take into account the charge between r and R.
  • #1
mawgs
5
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?
 
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  • #2
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?
 
  • #3
Qenc?
 
  • #4
mawgs said:
Qenc?
Yes. So what is the next step to find the field?
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
mawgs said:
Sorry, I meant epsilon naught.
Ok. But I think you lost a factor of 2 in there somewhere.
 

1. What is cylindrical symmetry?

Cylindrical symmetry is a type of symmetry in which an object or system has the same appearance or behavior when viewed from any direction along its central axis. This means that the object's properties, such as shape or electric field, remain constant as you rotate it around the axis.

2. How is cylindrical symmetry related to Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. When dealing with a system that has cylindrical symmetry, the electric field and charge distribution are also symmetric about the axis, allowing for simplified calculations using Gauss's Law.

3. Can Gauss's Law be applied to objects with cylindrical symmetry?

Yes, Gauss's Law can be applied to objects with cylindrical symmetry. This includes objects such as cylinders, cones, and even infinite sheets or wires with cylindrical symmetry. As long as the electric field and charge distribution are symmetric about the axis, Gauss's Law can be used to calculate the electric flux and charge enclosed.

4. What is the formula for the electric field of a charged cylinder?

The formula for the electric field of a charged cylinder is given by E = λ/2πε0r, where λ is the linear charge density of the cylinder, ε0 is the permittivity of free space, and r is the distance from the axis of the cylinder. This formula can be derived using Gauss's Law and cylindrical symmetry.

5. Why is Gauss's Law useful in calculating electric fields for objects with cylindrical symmetry?

Gauss's Law is useful in calculating electric fields for objects with cylindrical symmetry because it allows for simplified calculations by taking advantage of the symmetry of the system. Instead of having to integrate over the entire surface, Gauss's Law only requires the calculation of the electric flux through a closed surface. This makes it a powerful tool for solving complex problems in electromagnetism.

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