- #1
stunner5000pt
- 1,447
- 5
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!
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!