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
[tex]\vec{E} d(2 \pi \vec{r} L) = \frac{\sigma L}{\epsilon_{0}}[/tex]
hence [tex]\vec{E} = \frac{\sigma}{2 \pi r \epsilon_{0}} \hat{r}[/tex]
is this fine??
thank you in advance for the help!
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
[tex]\vec{E} d(2 \pi \vec{r} L) = \frac{\sigma L}{\epsilon_{0}}[/tex]
hence [tex]\vec{E} = \frac{\sigma}{2 \pi r \epsilon_{0}} \hat{r}[/tex]
is this fine??
thank you in advance for the help!