Calculating Electric Field of Cylindrical Rod and Shell

AI Thread Summary
The discussion revolves around calculating the electric field E(r) between an infinitely long positively charged cylindrical rod and a surrounding negatively charged cylindrical shell. Using Gauss's law, the electric field due to the rod is derived as E1 = λ/(ε(2πr)), while the shell contributes no electric field within the Gaussian surface since it encloses no negative charge. The confusion arises from the assumption that the shell's electric field affects the Gaussian surface, which is incorrect. The final answer for the electric field between the rod and shell is E = -λ/(ε(2πr)). Understanding Gauss's law is crucial for correctly applying it in this scenario.
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Homework Statement



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.

Homework Equations


Gauss's law


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\lambda
E1=\lambda/(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\lambdaL

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

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

-\lambda/(e(2(pi)r))

That was my answer but I am not sure if this is right or not
 
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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.
 
kuruman said:
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?
 
No, it does not.
 
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