Why does Coulomb's Law have a pi in the denominator while Gauss' Law does not?

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SUMMARY

The discussion centers on the difference in the denominators of the electric field equations derived from Coulomb's Law and Gauss' Law for a uniform straight infinite line charge. Specifically, the Gauss' Law solution yields a denominator of 2rε, while Coulomb's Law results in a denominator of 2πrε. This discrepancy arises from the geometric considerations inherent in each law's application, particularly the integration over the cylindrical symmetry in Coulomb's Law, which introduces the factor of π.

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Old Guy
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I'm sure I'm missing something simple here, but to the point:

I calculate the electric field a distance r from a uniform straight infinite line charge using Gauss' Law and get an answer; I do the same calculation using Coulomb's law and get the same answer but a pi remains in the denominator (that was not there in the Gauss' Law solution). I don't know how to enter the equation here, but in both answers the numerator is the linear charge density. The Gauss' Law denonminator is 2r times epsilon; Coulomb's law denominator is 2(pi)r times epsilon. Help!
 
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oops!

Sorry, all; I just realized I posted this in the wrong forum.
 

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