Electric field in two concentric conducting cylinders

In summary: You will want to integrate E from ra to rb. Then, you can use V = -∫E dr. (See http://www.math.union.edu/~barbanej/IMP13S02/PhysicsNotes/Energy.pdf]this[/URL] for a derivation and a discussion of the relationship between E and V.)In summary, two concentric cylindrical conducting shells with surface charge density +σ and radii ra and rb respectively, separated by a vacuum, have an electric field that is zero inside the shells. For r between ra and rb, the electric field is given by E = σra / rε0. The electric field outside the shells is zero as well, since the outer shell has no net charge.
  • #1
Ryaners
50
2

Homework Statement


Two concentric cylindrical conducting shells of length L are separated by a vacuum. The inner shell has surface charge density +σ and radius ra. The outer shell has radius rb. Using Gauss’ Law, as a function of radius r find: The direction and magnitude of electric field inside and outside the shells. Be sure to clearly state the Gaussian surfaces that you are using!
Find an expression for the voltage between the shells.

Homework Equations


Electric flux = ∫ EdA = qencl / ε0
σ = q/A

The Attempt at a Solution


I'm using a cylindrical Gaussian surface. I understand that when r < ra the field E = 0. Here's what I've gotten so far for when r is between ra and rb:
E = qencl / ε0⋅Agauss = qencl / ε0⋅2πrl
⇒ E = σ2πral / ε0⋅2πrl
⇒ E = σra / rε0

That doesn't seem very... right. I don't think there should be a dependence on ra for starters. Would the expression for the flux through a Gaussian surface with r > rb be the same also, as there isn't any non-induced charge on it? Also, for the last part of the question I don't understand why there would be a voltage between the shells when the charge on the outer shell is just induced by the inner cylinder, ie the net charge on the outer cylinder is 0. Isn't it? Sorry if I'm not making much sense, I'm almost through my exams and my brain is extra melty. :frown:
 
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  • #2
Ryaners said:
⇒ E = σra / rε0

That doesn't seem very... right. I don't think there should be a dependence on ra for starters.
I don't see anything wrong. Does your result give the correct expression for E at the surface of the inner cylinder?
See item 3 here http://www.math.union.edu/~barbanej/IMP13S02/PhysicsNotes/Gauss's Law III.pdf

Would the expression for the flux through a Gaussian surface with r > rb be the same also, as there isn't any non-induced charge on it?
Yes. (The outer cylinder has no net charge.)

Also, for the last part of the question I don't understand why there would be a voltage between the shells.
If there is an electric field between the cylinders, then there will be a potential difference between the cylinders. How do you get ΔV from E?
 
  • #3
TSny said:
I don't see anything wrong. Does your result give the correct expression for E at the surface of the inner cylinder?
See item 3 here http://www.math.union.edu/~barbanej/IMP13S02/PhysicsNotes/Gauss's[/PLAIN] [Broken] Law III.pdf

Yes. (The outer cylinder has no net charge.)

If there is an electric field between the cylinders, then there will be a potential difference between the cylinders. How do you get ΔV from E?

Ok, for the first part: when r = ra that leaves E = σ/ε0 which is correct - thank you for pointing out the sanity check :) Thanks for the link too - very helpful!

To get V from E, you multiply E by r (to get electric potential energy) and then divide by q (to get electric potential) - is that right? So since the outer cylinder is at zero potential, the potential difference is just the electric potential of the inner cylinder, which would be:
Er/q = V = σr / ε0q
= σr / ε0σ(2πrl )
= 1 / 2πlε0
Does that seem reasonable? The dimensional analysis seems to check out!
 
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  • #4
The electric field between the cylinders is not uniform, it varies with r. So, you will need to use some calculus.
 

1. What is an electric field in two concentric conducting cylinders?

The electric field in two concentric conducting cylinders refers to the distribution of electric charges and their associated forces between two cylindrical conductors that share the same axis. The inner cylinder has a positive charge, while the outer cylinder has a negative charge, creating an electric field between them.

2. How is the electric field calculated between two concentric conducting cylinders?

The electric field between two concentric conducting cylinders can be calculated using the formula E = λ/2πεr, where λ is the charge density, ε is the permittivity of the medium between the cylinders, and r is the distance between the two cylinders. This formula assumes that the cylinders are infinitely long and have a uniform charge distribution.

3. How does the electric field change as the distance between the cylinders increases?

As the distance between the cylinders increases, the electric field decreases. This is because the electric field is inversely proportional to the distance between the cylinders, according to the formula E = λ/2πεr. Therefore, as the distance increases, the denominator of the formula increases, resulting in a smaller electric field.

4. What factors affect the strength of the electric field between two concentric conducting cylinders?

The strength of the electric field between two concentric conducting cylinders is affected by the charge density, the permittivity of the medium between the cylinders, and the distance between the cylinders. Additionally, the presence of any other nearby charges or conductors can also affect the strength of the electric field.

5. What are some real-life applications of the electric field in two concentric conducting cylinders?

The electric field in two concentric conducting cylinders has several real-life applications, including in capacitors, which use two concentric cylinders to store electric charge, and in particle accelerators, where the electric field is used to accelerate charged particles. It is also used in various electrical and electronic devices, such as motors and generators, to control the flow of electric current.

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