Gauss' Law With an Infinite Cylinder of Charge

AI Thread Summary
The discussion revolves around applying Gauss' Law to determine the potential difference between a point above an infinite cylinder of charge and a point on its surface. The user initially struggles with understanding the professor's solution, particularly the term "E(s)*2pi*L*S." They clarify that the 2pi and L arise from integrating over the cylinder's surface, while S represents the distance from the point to the cylinder. The user realizes that the area A is calculated as 2pi times the length times the radius, and S is effectively treated as the radius in this context. This understanding resolves their confusion about the application of Gauss' Law in this scenario.
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[SOLVED] Gauss' Law With an Infinite Cylinder of Charge

Homework Statement



The test question was to find the potential difference between a point S above a cylinder of charger per length lambda, and a point on the surface of the cylinder having radius R and infinite length.

Homework Equations



Stupid me tried to grind through it during the test. But looking at the prof's solution:

http://www.phys.washington.edu/users/schick/321A/321-07exam1soln.pdf

Here's the test if you need to know the question how he worded it:

http://www.phys.washington.edu/users/schick/321A/321-07exam1.pdf

Problem #1.

I don't get how he got "E(s)*2pi*L*S

I get the 2pi and L come from integrating over phi and the length of the cylinder, but I can't figure out where that S came from.Is it because you'd need a cosine(theta) for the direction that the cylinder is from S? Where cosine is S/distance? Then wouldn't he have to add in a bunch of other stuff? I'm just not seeing something here and I feel embarrassed.
 
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\int E \cdot dA = EAcos(\Theta)

Where \Theta is the angle between the surface of the cylinder and the electric field.

Now, simplify this last expression. What is A? What is \Theta?
 
Ohhhhhh I get it. A is 2pi*Length*radius, and he's just using S as the radius, since it shouldn't matter.

Gotcha, thanks a bunch.
 
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