Solving an Electric Field Due to an Infinite Cylinder

AI Thread Summary
To find the electric field at r = 8.0 cm due to an infinitely long cylinder with a uniform volume charge density of 200 nC/m^3, Gauss's Law is the appropriate approach. The confusion arose from attempting to convert volume charge density to linear charge density and using the wrong geometry for calculations. The correct method involves using a cylindrical Gaussian surface with a radius of 8 cm to enclose the charge. The charge enclosed is calculated based on the cylinder's volume and density, leading to the correct application of Gauss's Law. The final solution confirms that the electric field is approximately 0.23 kN/C.
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Homework Statement


An infinitely long cylinder of radius 4.0 cm carries a uniform volume charge density
ρ = 200 nC/m^3 What is the electric field at r = 8.0 cm


Homework Equations


I'm confused as to how to do this problem, I've tried converting from volume charge density to simply charge density λ and then solving with the equation for the E field due to an infiinite line charge, but this doesn't give me the right answer.


The Attempt at a Solution


(200 x10 ^-9)((4/3)(pi)(.04^3)) = Q

λ = Q / L so λ = Q (as we're taking L to be 1 meter in the above equation)

Efield due to infinite line charge= 2kλ/(.08)

the correct answer is supposed to be .23kN/c
 
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It looks like there is enough symmetry to use Gauss's Law.
 


SammyS said:
It looks like there is enough symmetry to use Gauss's Law.

I know but I tried finding Q as I showed above, by taking ρ = 200nC/m^3 and multiplying it by the volume (4/3 pi (.04)^3) to get Q = 5.36E-11

I then use Gauss's law: E(4pi(.08)^2) = 5.36E-11/ε0

this however gives me E = 75.306

What am I doing wrong?
 


(4/3)πR3 is the volume of a sphere of radius, R.

You should be working with a cylinder.
 


Edit: never mind I found the answer, Thanks again for the help.
 
Last edited:


A cylinder of height 1 m is OK. A radius of 4 cm will give you the charge enclosed by a cylinder of height 1 m and radius of 8 cm, which is your Gaussian surface.
 
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