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Gauss's law, cylinder of length L

by lidl
Tags: cylinder, gauss, length
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lidl
#1
Feb13-10, 01:17 PM
P: 10
http://img404.imageshack.us/img404/4148/gauss3j.png (infinitly long)

[itex]\Phi=\Phi_{0}(1-{r \over R}) {C \over m^2}[/itex]

What's the charge of the cylinder of length L?

Again, almost everything was wrong...
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lidl
#2
Feb13-10, 06:44 PM
P: 10
anyone? :(
lidl
#3
Feb14-10, 12:24 PM
P: 10
Here's link to my paper: http://img202.imageshack.us/img202/5889/physics.jpg

JaWiB
#4
Feb14-10, 12:42 PM
P: 289
Gauss's law, cylinder of length L

You need to be much clearer in the problem statement and your work.

Is the cylinder hollow? What do r, R, [tex]\Phi[/tex], and [tex]\Phi_0[/tex] represent?

Also, it's probably OK to write notes in your homework like "the electric field points radially outward so the flux through the top and bottom of the gaussian cylinder is zero" rather than relying on vague drawings
lidl
#5
Feb14-10, 01:04 PM
P: 10
This is from the test, as you can see, and everything I knew was written down...
lidl
#6
Feb14-10, 02:34 PM
P: 10
please delete this post
lidl
#7
Feb14-10, 02:37 PM
P: 10
ok, now I got the following:

Q=integral of /rho(1-r/R) times rdr d/phi dl

Which is the answer for the question of what's the charge of the cylinder L.

How to calculate it?


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