Infinitely Long Cylindrical Surface Problem

AI Thread Summary
The discussion centers on applying Gauss' law to find the electric field of an infinitely long cylindrical surface with a uniform charge density. The problem specifies a cylinder of radius R = 4.00×10^-2 m and a charge density of 1.00×10^-2 C/m³, with the goal of calculating the electric field at a radius of r = 2.00×10^-2 m. Participants emphasize using Gauss' law, suggesting that the electric field can be derived from the charge enclosed within a Gaussian surface. The formula E = λ/(2πε₀r) is highlighted as relevant for an infinite cylinder, where λ represents the linear charge density. Ultimately, the discussion aims to clarify the application of Gauss' law to solve for the electric field in this specific scenario.
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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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