Find the electric field of a cylindrical charge

AI Thread Summary
The discussion focuses on calculating the electric field of a cylindrical charge using Gauss's law. The user calculates the electric flux through the lateral surface of the cylinder, concluding that the electric field E can be expressed as E = Q/(2πeph) for p ≥ a, where Q is the charge and h is the cylinder's length. There is some confusion regarding whether the charge is distributed along the surface or the axis of the cylinder, but it is clarified that the charge Q is concentrated along the axis. The user also notes a discrepancy between their result and a textbook's formula, leading to a consideration that Q could be simplified to y when h is set to 1. The conversation emphasizes the importance of correctly interpreting the charge distribution in cylindrical geometries.
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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