Calculating Electric Field in a Charged Tube Using Gauss's Law

AI Thread Summary
The discussion revolves around calculating the electric field within a hollow charged tube using Gauss's Law. The user initially attempts to apply Gauss's Law but questions the validity of their approach, particularly regarding the closed nature of the tube. Clarification reveals that the tube is hollow, leading to confusion about whether Gauss's Law is applicable. Participants emphasize the importance of the tube's geometry in applying Gauss's Law correctly. The conversation highlights the need for a proper understanding of the conditions under which Gauss's Law can be used effectively.
yevi
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A long tube charge with charged with uniform spatial density \rho.
The inner radius of the tube is: a
The outer radius of the tube is: b

Need to find the electric field in: a<r<b

My approach is Gauss:

E*S=4 \pi kq

The S is the Gaussean Surface it should be 2 \pi r^2 ??

and q should be \rho*(r^2-a^2)??
 
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The answer should be:
E=2 \pi k\rho\frac{r^2-a^2}{r}\hat{r}

So I did something wrong...
 
yevi said:
The answer should be:
E=2 \pi k\rho\frac{r^2-a^2}{r}\hat{r}

So I did something wrong...

Is the tube closed? because if it isn't then Gauss law can't be used.
 
what do you mean closed?
The tube is hollow...

If I can't use gauss, what should i use?
 
I still don't understand why I can't use gauss here.
 
Anyone? :)
 
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