Find the electric field of a cylindrical charge

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SUMMARY

The discussion focuses on calculating the electric field (E) of a cylindrical charge using Gauss's Law. The derived formula is E = Q/(2*pi*e*p) for p ≥ a, where Q is the charge per unit length, e is the permittivity, and p is the radial distance from the axis of the cylinder. The user clarifies that the charge Q is uniformly distributed along the axis of the cylinder, leading to the conclusion that Q can be expressed as Q = y*h, where h is the length of the cylinder. This aligns with established results found in academic resources.

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  • Knowledge of cylindrical coordinate systems
  • Basic concepts of charge distribution
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zelscore
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Homework Statement
Figure 2.29 depicts a long cylindrical charge with a radius a placed in free space
of permittivity e. A charge Q per unit length is uniformly distributed within a
circular cylinder. Determine the electric field.
Hint: Construct a cylindrical Gaussian surface surrounding the charge and apply
Gauss’s law.

Fig 2.29 is basically just a cylinder on a z-axis and where the radius is described as p=a, with the letter Q inside the cylinder
Relevant Equations
From wikipedia https://en.wikipedia.org/wiki/Gaussian_surface#Cylindrical_surface
Q = yh, where y is the charge density, and h is the length of the cylinder
Flux of electric field E is the sum of all surface flux
Gauss law: Flux = Q/e
I begin by calculating the flux to be the flux of the cylinders lateral surface, which equals E*2*pi*p*h (p is the radius)
The other two surfaces have E ortogonal to dA, so their flux is 0.
Using Gauss law together with the calculated flux above, I get
Flux = Q/e
Flux = E*2*pi*p*h

Solve for E
E*2*pi*p*h = Q/e
E = Q/2*pi*e*p*h
Q = yh, so
E = y/2*pi*e*p for p >= a which is the same result as wikipedia gets, and this website too http://www.ncert.nic.in/html/learni...city/electrostatics/ef_cylinder_of_charge.htm

using Q instead of y:
E = Q/2*pi*e*p*h for p >= a

HOWEVER, the book I use get
E = Q/2*pi*e*p for p >= a

Which leads me to believe that, in Q = yh, and where h is the length of our "long" cylinder, h can be set to 1 and thus Q = y?

This is my first time posting so if I didn't follow the rules somehow please do tell.
 
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zelscore said:
Homework Statement: Figure 2.29 depicts a long cylindrical charge with a radius a placed in free space
of permittivity e. A charge Q per unit length is uniformly distributed within a
circular cylinder. Determine the electric field.
I'm confused. Is there one cylinder or two? If one, is the charge on the surface of the cylinder, along the axis of the cylinder, or both?
 
haruspex said:
I'm confused. Is there one cylinder or two? If one, is the charge on the surface of the cylinder, along the axis of the cylinder, or both?
There is one cylinder. The charge Q is inside the cylinder, on the axis of the cylinder (is how I interpreted the question, and solved it). That is why I had Q = yh and not Q = y*(volume of cylinder). Because I assumed the charge is concentrated on the line in the cylinder
 

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