Gauss' Law and a uniform surface charge

stunner5000pt
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Use Gauss' Law to find the field inside and outside a long hollow cylindrical tube which carries a uniform surface charge sigma.

It has been a few months since i did this so i may be a bit rusty

As i can recall if there is a point inside a holow cylindrical tube there is no enclosed charge, hence the electric field is zero inside the tube for all points inside

for the outside
For a radial ditance of r from the tube of length L
\vec{E} d(2 \pi \vec{r} L) = \frac{\sigma L}{\epsilon_{0}}

hence \vec{E} = \frac{\sigma}{2 \pi r \epsilon_{0}} \hat{r}

is this fine??

thank you in advance for the help!
 
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