When do you use R or r for Gauss' Law

AI Thread Summary
The discussion focuses on the correct usage of 'r' and 'R' in the equation E A = 1/4piEo * Qenc when applying Gauss's law. It clarifies that 'R' should be used on the right side when the Gaussian surface is outside the charge distribution (r > R), as it represents the total charge enclosed. Conversely, when the Gaussian surface is inside or on the surface (r ≤ R), 'r' can be used on both sides, leading to simplifications through cancellation. The physical reasoning behind this distinction relates to the charge enclosed by the Gaussian surface, which varies based on its position relative to the charge distribution. Understanding this concept is crucial for accurately applying Gauss's law in electrostatics.
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I've seen many examples of spheres, cylinders, and planes and I'm trying to understand when to use r or R in the equation E A = 1/4piEo * Qenc. I've seen examples where the r's will cancel giving a simplified answer and then others where you have something like R^2 / r^2.

So my question is how do you know what type of r to write on the left side of the equation AND the right side of the equation. Thank you.
 
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I thought you treated r the same on both sides of the eqn if the Gaussian surface was outside of the original surface, but that doesn't seem correct based on the examples I've seen in Griffiths.
 
I think I've noticed a pattern but I still don't understand the physical meaning behind it. It seems if the Gaussian is written outside of the original surface (r>R) then you use R on the right side of the equation. If (r< or equal to R) then you can use r on both sides of the equation, which will result in some cancellation when you bring over A to the right side and calculate E. But physically I would like to understand why this is. Any help would be greatly appreciated, thanks.
 
I'll attempt to answer my own question. The right side of the equation is Q enclosed which means the charge inside of the Gaussian surface.. so if your Gaussian is outside the surface, the charge enclosed is the total charge of the surface, but if your Gaussian is within the original surface, the charge enclosed is only equal to the charge within that Gaussian and not the total... as long as the charge density is uniform and we have charge inside. Right?
 
I'm referring to rewriting Q enclosed in terms of the surface charge (sigma) and volume charge density (ro)..
 
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