Gauss's Law 2 concentric cylinders in electrostatic equilibrium

In summary: Have fun.In summary, the question concerns two long, concentric, metallic, hollow cylinders in a vacuum with different radii and charge densities. At electrostatic equilibrium, the total charge on the inner surface of the larger cylinder is -2λL, while the outer surface has no charge. The electric field in the cavity between the cylinders is E = λ/πrεo, where r is the outer radius of the smaller cylinder. The question also addresses the use of linear density and the Gaussian surface.
  • #1
shale
4
0

Homework Statement



Consider the cross sections of two, very long, concentric, metallic, hollow cylinders placed in a vacuum. The small cylinder has inner radius A and outer radius B while the larger cylinder has inner radius 2A and outer radius 2B. Initially the small and big hollow cylinders have charge densities of +2λ and -2λ respectively.

1) AT electrostatic equilibrium what is the total charge of the inner surface of the big cylinder if the cylinders both hav e length L?

2) what is the electric field in tha cavity between the two cylinders? se r as the distance measured from the center of the spheres.

Homework Equations



relevant equations would be the flux which is EA=Qenco

The Attempt at a Solution



for number 1

since this is at electrostatic equilibrium, the net charge AT the CYLINDER is ZERO. therefore Qin=0... but inner cylinder has +2λL, and so the inner surface of the larger cylinder would be -2λL... and for the outside surface... (will it be +2λL for the net charge on the conductor to be zero.?? or there will be no charge because the said "net" charge in that conductor should be -2λL??) I'm confused...

and is it correct for me to multiply the charge densities to L??


for number 2

I set up an equation like this...

EA= E2πrL = ([2λ(πr2L)] - [-2λ(πR2L)]) /εo


we cancel and it becomes like this

E = [λπ(r2+R2)] /εor


where r is the outer radius of the smaller cylinder and R is inner radius of larger one.. i think I'm wasting my effort because all this is wrong :-ss please help.
 
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  • #2
Welcome to PF. :smile:

shale said:

Homework Statement



Consider the cross sections of two, very long, concentric, metallic, hollow cylinders placed in a vacuum. The small cylinder has inner radius A and outer radius B while the larger cylinder has inner radius 2A and outer radius 2B. Initially the small and big hollow cylinders have charge densities of +2λ and -2λ respectively.

1) AT electrostatic equilibrium what is the total charge of the inner surface of the big cylinder if the cylinders both hav e length L?

2) what is the electric field in tha cavity between the two cylinders? se r as the distance measured from the center of the spheres.

Homework Equations



relevant equations would be the flux which is EA=Qenco

The Attempt at a Solution



for number 1

since this is at electrostatic equilibrium, the net charge AT the CYLINDER is ZERO. therefore Qin=0... but inner cylinder has +2λL, and so the inner surface of the larger cylinder would be -2λL...
Yes.

and for the outside surface... (will it be +2λL for the net charge on the conductor to be zero.??
No, because the net charge on the outer conductor is NOT zero.

... or there will be no charge because the said "net" charge in that conductor should be -2λL??)
Yes. There is zero charge on the outer surface of the outer cylinder, for exactly the reason you said here.

i'm confused...

and is it correct for me to multiply the charge densities to L??

Yes, if you want to get the total charge in a length L of a cylinder.

for number 2

I set up an equation like this...

EA= E2πrL = ([2λ(πr2L)] - [-2λ(πR2L)]) /εo

Not quite, but you're not far off.

Think about the surface you are defining, and the fact that you need to use the charge enclosed by that surface here.

  • Does the surface contain the inner cylinder?
  • Does the surface contain the outer cylinder?

Based on the answers to those 2 questions, what is the charge enclosed within the surface?

we cancel and it becomes like this

E = [λπ(r2+R2)] /εor


where r is the outer radius of the smaller cylinder and R is inner radius of larger one.. i think I'm wasting my effort because all this is wrong :-ss please help.
 
  • #3
Redbelly98 said:
Not quite, but you're not far off.

Think about the surface you are defining, and the fact that you need to use the charge enclosed by that surface here.

  • Does the surface contain the inner cylinder?
  • Does the surface contain the outer cylinder?


do mean the gaussian surface? ---- 1) Yes, and 2) No..

Redbelly98 said:
Based on the answers to those 2 questions, what is the charge enclosed within the surface? ?

so total charge enclosed is +2λL


should i still use EA=Qenco...? if so then...

EA= E2πrL = 2λL/εo

(I think I was wrong in my first attempt because i multiplied it to the volume and not the area, λ is always linear density right? because in this question it's not clearly stated)

E = λ/πrεo


is this one correct now? or is there a different approach on how this is solved? :-ss

THANKS FOR YOUR REPLY! I'm really glad someone responded.!
 
  • #4
shale said:
do mean the gaussian surface?
Yes.
(I think I was wrong in my first attempt because i multiplied it to the volume and not the area, λ is always linear density right? because in this question it's not clearly stated)
I think so, but I didn't really think about that and just assumed that it was the linear density. You're right, it should be clearly stated.

E = λ/πrεo

is this one correct now? or is there a different approach on how this is solved? :-ss
Looks good! :smile:

THANKS FOR YOUR REPLY! I'm really glad someone responded.!

You're welcome.
 

Related to Gauss's Law 2 concentric cylinders in electrostatic equilibrium

1. What is 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. It is named after the German mathematician and physicist Carl Friedrich Gauss.

2. What is electrostatic equilibrium?

Electrostatic equilibrium refers to a state in which the electric charges within a system are at rest and the electric field at every point within the system is constant. This means that the net force on any charge within the system is zero.

3. How does Gauss's Law apply to 2 concentric cylinders in electrostatic equilibrium?

In this scenario, Gauss's Law states that the electric flux through a closed surface between the two cylinders is equal to the charge enclosed within that surface divided by the permittivity of free space. This can be represented mathematically as Φ = Q/ε0, where Φ is the electric flux, Q is the charge, and ε0 is the permittivity of free space.

4. What is the significance of using concentric cylinders in this example?

Using concentric cylinders allows us to create a symmetrical system in which the electric field is constant at every point within the system. This simplifies the calculation of the electric flux and makes it easier to apply Gauss's Law.

5. How does the electric flux change if the distance between the concentric cylinders is increased?

If the distance between the cylinders is increased, the electric flux through the closed surface will decrease, as the charge enclosed within the surface is spread out over a larger area. This is because the electric field strength decreases with distance, and therefore the electric flux decreases as well.

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