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

[yevi]

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)??

 

[yevi]

The answer should be:
E=2 $$\pi$$ k$$\rho$$$$\frac{r^2-a^2}{r}$$$$\hat{r}$$

So I did something wrong...

 

[malty]

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.

 

[yevi]

what do you mean closed?
The tube is hollow...

If I can't use gauss, what should i use?

 

[yevi]

I still don't understand why I can't use gauss here.

 

[yevi]

Anyone? :)