Question about E-Fields and Potential of a Coaxial Cylinder

In summary: The Gauss theorem states that the electric field is constant in a cylindrical shell that encloses a charge.
  • #1
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Homework Statement


I have posted the given question and conditions in the attached image

Homework Equations


(Q_enclosed/ epsilon_0) = closed integral (E-field) dA
Q_encolsed = p*A
V(r)= -Int_(from origin to r) (E-field(r'))*dl'

The Attempt at a Solution


[/B]
a)
E=0------>For r<a

(p*v)/(epsilon_0) = pi*(a^2)*L*E
E= ((p*v)/(epsilon_0))*(1/(pi*(a^2)*L)------>For the solid inside's edge r=a

E= 0-------> Between soild inside's edge and the outer shell's edge (Q_out = -Q_in) a<r<b

(Q_enclosed)/(epsilon_0) = 2pi*b*L*E
E= ((Q_enclosed)/(epsilon_0))*(1/(2pi*b*L))-------For the outer shell's edge r=b

b)As for potential, V, I am very uncertain as to what I should do.
c) I think this portion will be much clearer once a) and b) are clearified.Thanks!
 

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  • #2
Hello artoribio, :welcome:

In your work for (a), is there an area where the E field is not zero ?
I agree with ##|\vec E | = 0## for 0 < r < a and also for b < r : you can form a gaussian cylinder that contains no charge.
What I don't understand is what you mean with
artoribio said:
E= 0-------> Between solid inside's edge and the outer shell's edge (Q_out = -Q_in) a<r<b
where is the gaussian surface ? And why do you start with E = 0 ?

PS the image is hard to read; at PF we prefer it if you type out the problem statement...
 
  • #3
I thought that the area would not have a zero E-field outside of the coaxial cylinder. I did not put an |⃗E| equation for b<r so I'm am not quite sure what you are referring to. As for your quoted portion, I was thinking that directly in between a and b the fields from the solid center and the shell would cancel out because Q_out + Q_in = 0.

I apologize for the image and will write out any future questions I might post. Thank you for the heads up. I had no idea!
 
  • #4
artoribio said:
directly in between a and b the fields from the solid center and the shell would cancel out because Q_out + Q_in = 0.
Ah ! So in between a positive charge and a negative charge there is no electric field :woot: ?
 
  • #5
I was thinking in terms of two positive nodes. Is that wrong? In the case of a positive and negative set it would just be straight lines between the center and and the shell considering, right?
 
  • #6
artoribio said:
I was thinking in terms of two positive nodes. Is that wrong?
I am afraid so. It even seems that that is a given in your exercise ??

I retype (I am tired of switching to the unsharp picture) and correct the

---------------------------------------
Problem statement
  • We have a coaxial cylindrical shell wth a solid center and a uniform charge density ## \rho ## C/m3
  • Inner solid cylinder has radius a. Outer shell has inner radius b.
  • Qout = - Qin
Find
  • ##\vec E## everywhere
  • ##V##
  • Plot both
-----------------------------------------

That wasn't so hard ! let's also do

2. Homework Equations
  1. ##\oint \vec E \cdot d\vec A = {\displaystyle Q_{\rm enclosed} \over \displaystyle \varepsilon_0 } ##
  2. ## Q_{\rm enclosed} = \rho A L \ \ ## Coulomb ## \ \ \ (\rho A ## has the wrong dimension ! ##)##
  3. ## V(r)= - \displaystyle \int_0^r E(r') \, dr' ##
-----------------------------------------

Explanation of

  • My adding density : ##\rho## is used for volume charge density (##\sigma## for area density, ##\lambda## for line density). You can use a length L = 1 for your calculations, though (see below). But writing ##Q = \rho A## is ugly.
  • ##\bf Q_{\rm out} = -Q_{\rm in}##: the charge on the inner cylinder induces an opposite charge on the inside of the outer cylinder shell. If that shell is grounded, the potential on the outside of that cylinder remains zero. And V = 0 outside both cylinders is then a solution for the electric field. Since the field is unique it is the solution.

Your equation #1 is Gauss theorem. It works usefully if there is a symmetry to exploit. In this case there is cylindrical symmetry, so (in cylinder coordinates) ##\ \ \vec E_z = 0 \ \ \&\ \ \vec E_\phi = 0 \ \ ## (can you explain why ?) and ##\ \ \vec E = E(r) \hat r \ \ ## (as you assumed and work out). Sensible gaussian surfaces are coaxial cylinder shells and you have a nonzero E field everywhere where such a cylinder encloses charge. Back to the drawing board for b < r therefore !
 
  • #7
I think that I understand your explanations. As for the symmetry of the cylinder or in general if the field lines are perpendicular to the surface then you can take the E values. So in this case E_z = 0 because it is parallel to the charge. That is what I gathered from a Khan Academy video. As for E_phi = 0 There are equal and opposite sets that cancel out in the z-direction and sum in the perpendicular direction of the surface of the cylinder. Do I need to find E for the areas 0, a, and b, or just r<a, a<r<b, and b<r?
 
  • #8
artoribio said:
So in this case E_z = 0 because it is parallel to the charge.
How can a vector be parallel to a charge ?
artoribio said:
As for E_phi = 0 There are equal and opposite sets that cancel out in the z-direction and sum in the perpendicular direction of the surface of the cylinder
How does that make ##E_\phi = 0 ## ?
artoribio said:
Do I need to find E for the areas 0, a, and b, or just r<a, a<r<b, and b<r?
just r<a, a<r<b, and b<r
 
  • #9
Maybe it was parallel to the surface?

I think that E_phi just ends up becoming E_perpendicular(to the surface) because the lateral portions cancel, no?
 

1. What is an E-Field and how is it related to potential in a coaxial cylinder?

An E-Field, or electric field, is a region in space where an electric charge experiences a force. In a coaxial cylinder setup, the E-Field is created between the inner and outer cylinders, with the potential difference between the two cylinders determining the strength of the field.

2. How does the potential of a coaxial cylinder affect the E-Field?

The potential difference between the inner and outer cylinders directly affects the strength of the E-Field. The greater the potential difference, the stronger the E-Field will be.

3. What factors determine the potential in a coaxial cylinder?

The potential in a coaxial cylinder is determined by the charge on the inner and outer cylinders, the distance between them, and the material properties of the cylinders and the surrounding medium.

4. Can the potential in a coaxial cylinder be calculated?

Yes, the potential can be calculated using the mathematical equation V = kQ/ln(b/a), where V is the potential, k is the Coulomb constant, Q is the charge on the inner cylinder, b is the radius of the outer cylinder, and a is the radius of the inner cylinder.

5. How can the E-Field and potential of a coaxial cylinder be used in practical applications?

The E-Field and potential of a coaxial cylinder have many practical applications, such as in capacitors, transmission lines, and medical equipment. They can also be used for electrostatic experiments and in industrial processes such as electroplating and electrostatic painting.

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