Infinitely Long Cylindrical Surface Problem

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SUMMARY

The discussion centers on applying Gauss' law to determine the electric field generated by a uniformly charged infinitely long cylinder with a radius of 4.00×10-2 m and a charge density of 1.00×10-2 C/m3. The electric field at a distance of 2.00×10-2 m from the center is calculated using the formula E = λ/(2πε0r), where λ represents the linear charge density. Participants emphasize the importance of visualizing a cylindrical Gaussian surface to simplify the integration process and confirm that the electric field remains constant at all points around the cylinder.

PREREQUISITES
  • Understanding of Gauss' law and its application in electrostatics
  • Familiarity with electric field calculations for cylindrical geometries
  • Knowledge of charge density and linear charge density concepts
  • Basic calculus for integration in physics
NEXT STEPS
  • Study the derivation of Gauss' law in electrostatics
  • Learn about electric fields of charged cylinders and their properties
  • Explore the concept of linear charge density and its implications
  • Practice solving problems involving Gauss' law with different geometries
USEFUL FOR

Physics students, electrical engineers, and anyone studying electromagnetism who seeks to deepen their understanding of electric fields generated by charged cylindrical surfaces.

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Homework Statement


Here is my problem. I don't fully understand Gauss' law so any assistance there would be greatly appreciated
Charge is distributed uniformly throughout the volume of an infinitely long cylinder of radius R = 4.00×10-2 m. The charge density is 1.00×10-2 C/ m3. What is the electric field at r = 2.00×10-2 m?
(in N/C)


Homework Equations





The Attempt at a Solution


I understand that gauss' law is the integral EdA and that the E for a cylinder is lamda/2pi(E0)r , but I don't understand it for an infinite cylinder
 
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nevermind figured it out
 
You mention Gauss' law, so maybe you are supposed to use it to find the answer rather than that formula. You must imagine a cylinder that has the point you are interested in on its surface. Just the given charged cylinder will do nicely in this case. You figure out how much charge is inside - maybe use L for the length and let it tend to infinity in your final answer (most likely it will cancel out before then). Looks like E will be the same at all points around the cylinder so that integral is very easy. Solve for E. I would expect the answer to be just that formula lamda/2pi(E0)r which is for an infinitely long line of charge.

Of course you will put the numbers into find a numerical answer.
 

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