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Gauss law 2

by hover
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hover
#1
Feb6-10, 03:59 PM
P: 344
1. The problem statement, all variables and given/known data

An infinitely long conducting cylindrical rod with a positive charge lambda per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of -2 \lambda and radius r_1, as shown in the figure.

What is E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the distance r from the axis of the cylindrical rod?
Express your answer in terms of lambda, r, and epsilon_0, the permittivity of free space.

2. Relevant equations
Gauss's law


3. The attempt at a solution

Basically I put a Gaussian surface just larger than the rod but smaller than the shell. First I calculate the electric field from the rod.

E(2L(pi)r)=Q/e
E=Q/(e(2L(pi)r))
Q=L[tex]\lambda[/tex]
E1=[tex]\lambda[/tex]/(e(2(pi)r))

Thats field one. Now I do the same thing to calculate the second field

E=q/(e(2L(pi)r))

where q = -2[tex]\lambda[/tex]L

E2= -2[tex]\lambda[/tex]/(e(2(pi)r))

Now I should add the fields to find the field inside the shell and outside the rod and I get

-[tex]\lambda[/tex]/(e(2(pi)r))

That was my answer but im not sure if this is right or not
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kuruman
#2
Feb6-10, 04:45 PM
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What second field? Your gaussian surface encloses no negative charge. What you call "field one" is the answer. It seems you are unsure about how Gauss's Law works. It might be a good idea to review it.
hover
#3
Feb6-10, 05:00 PM
P: 344
Quote Quote by kuruman View Post
What second field? Your gaussian surface encloses no negative charge. What you call "field one" is the answer. It seems you are unsure about how Gauss's Law works. It might be a good idea to review it.
But doesn't the electric field from the shell also emit an electric field that passes through the same gaussian surface?

kuruman
#4
Feb6-10, 05:42 PM
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Gauss law 2

No, it does not.


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