Applying Guass' Law to Cylindrical Symmetry

In summary, the individual is asking for clarification on whether there would be an electric field at a radial distance of R/2 for a long, thin wall metal tube with uniform charge per unit length. They understand that there would be no net flux using a gaussian surface smaller than the radius of the tube and that the electric field would be zero at the center of the tube due to symmetry. However, they are unsure if there would be an electric field at R/2 and are seeking an explanation. The responder explains that the symmetry of the charge distribution and Gauss's law allow us to conclude that the field is zero for r < R.
  • #1
raytrace
9
0
OK, having some trouble wrapping my head around this so would appreciate some clarification.

Let us say I had a long, thin wall metal tube of radius R with a uniform charge per unit length. Would there be some magnitude of E of the electric field at a radial distance of R/2?

I understand that there would be no net flux using a gaussian surface smaller than the radius of the tube. I also understand that at the center of the tube, E would be zero. However, I would think that at the radius of R/2 there would be some electric field there.

Could someone explain to me as to why or why not there is E at R/2?
 
Physics news on Phys.org
  • #2
raytrace said:
I understand that there would be no net flux using a gaussian surface smaller than the radius of the tube. I also understand that at the center of the tube, E would be zero. However, I would think that at the radius of R/2 there would be some electric field there.
Since you realize that there's no charge enclosed in a gaussian surface at that radius, why would you think there would be a non-zero E?
 
  • #3
Doc Al said:
Since you realize that there's no charge enclosed in a gaussian surface at that radius, why would you think there would be a non-zero E?

Well, a net zero flux from a gaussian surface just means that you have an equal number of electric field lines going in as going out. Just because the flux is a net zero doesn't mean that there doesn't exist an electric field, right?
 
  • #4
The flux is a net zero does mean that there doesn't exist an electric field!
 
  • #5
raytrace said:
Well, a net zero flux from a gaussian surface just means that you have an equal number of electric field lines going in as going out. Just because the flux is a net zero doesn't mean that there doesn't exist an electric field, right?
Right, but the symmetry of the charge distribution allows you to make a much stronger statement. In this case, due to the cylindrical symmetry of the charge distribution, you know that at any radius the field must be radial and uniform. It's that symmetry combined with Gauss's law that allows you to conclude that the field is zero for r < R.

ThomasYoung said:
The flux is a net zero does mean that there doesn't exist an electric field!
In general you are correct. But in this case it does mean that there is no electric field for r < R.
 

1. What is Guass' Law and how does it apply to cylindrical symmetry?

Guass' Law is a fundamental law in electromagnetism that relates the electric field to the charge distribution. When applied to cylindrical symmetry, it states that the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space.

2. What is cylindrical symmetry and why is it important in applying Guass' Law?

Cylindrical symmetry is a type of symmetry where an object or system has the same properties when rotated around a central axis. In the context of Guass' Law, cylindrical symmetry allows us to simplify the calculation of the electric field by reducing it to a one-dimensional problem along the axis of symmetry.

3. How do you determine the direction of the electric field when using Guass' Law for cylindrical symmetry?

The direction of the electric field can be determined by using the right-hand rule. If the charge distribution is positively charged, the electric field will point away from the surface, and if the charge distribution is negatively charged, the electric field will point towards the surface.

4. Can Guass' Law be applied to non-uniform charge distributions in cylindrical symmetry?

Yes, Guass' Law can be applied to non-uniform charge distributions as long as the charge distribution is symmetrical around the axis of rotation. In these cases, the electric field will still be determined by the enclosed charge and the distance from the axis.

5. How is Guass' Law used in practical applications involving cylindrical symmetry?

Guass' Law is used in a variety of practical applications such as calculating the electric field inside a cylindrical capacitor, determining the electric field of a long current-carrying wire, and analyzing the electric field inside a hollow cylindrical conductor. It is also used in the design of devices such as MRI machines and particle accelerators.

Similar threads

I Electrostatic force exerted on an electron inside a nucleus
    • Classical Physics
Replies
19
Views
865
B Determining whether the non-integral form of Gauss' law applies
    • Classical Physics
Replies
1
Views
682
Using Gauss' law to find the induced surface charge density ##\sigma##
    • Advanced Physics Homework Help
Replies
2
Views
722
I Why isn't the Lagrangian invariant under ##\theta## rotations?
    • Classical Physics
Replies
5
Views
737
I How Gauss’ Law is applied to cylinders
    • Classical Physics
Replies
20
Views
1K
I Proving Gauss' Law using a Cubical Surface
    • Classical Physics
Replies
5
Views
2K
Electric Field Inside Cylindrical Capacitor
    • Introductory Physics Homework Help
Replies
2
Views
1K
I Atmospheric Density vs Altitude in an O'Neil Cylinder
    • Classical Physics
Replies
1
Views
595
I Trying to derive Gauss' law using a cylindrical surface
    • Electromagnetism
Replies
2
Views
647
I How to use geometrical symmetries -- general advice? (Vector potential)
    • Electromagnetism
Replies
2
Views
308

Hot Threads

Recent Insights

Back
Top