Gauss's Law problem

  • Thread starter chillaxin
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A long straight conducting rod (or wire) carries a linear charge density of +2.0uC/m. This rod is totally enclosed within a thin cylindrical shell of radius R, which carries a linear charge density of -2.0uC/m.
A) Construct a Gaussian cylindrical surface between the rod and the shell to derive then electric field in the inner space as a function of the distance from the center of the rod.
B) Construct a Gaussian cylindrical surface outside both the rod and the shell to calculate the electric field outside the shell.

This is what i have so far.

E=q/4piEor^2
E=+2.0uC/m / 4pi8.85x10^-12(-2uC/m)^2
E=4.5x10^9Nm^2/C
 

Answers and Replies

  • #2
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Anybody?!?!?!?
 
  • #3
OlderDan
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chillaxin said:
A long straight conducting rod (or wire) carries a linear charge density of +2.0uC/m. This rod is totally enclosed within a thin cylindrical shell of radius R, which carries a linear charge density of -2.0uC/m.
A) Construct a Gaussian cylindrical surface between the rod and the shell to derive then electric field in the inner space as a function of the distance from the center of the rod.
B) Construct a Gaussian cylindrical surface outside both the rod and the shell to calculate the electric field outside the shell.

This is what i have so far.

E=q/4piEor^2
E=+2.0uC/m / 4pi8.85x10^-12(-2uC/m)^2
E=4.5x10^9Nm^2/C
The field is certainly not constant in the region between the rod and the cylinder. Are these anwers to multiple parts? Just the first part?
 
Last edited:
  • #4
Unless I'm mistaken, the total charge inclosed in the whole system is zero. If the enclosed charge is zero, the electric field is zero. Thus from what I can draw, the answer to B is zero. The answer to A requires using the enclosed charge to be the positive portion and then solving for E.
 

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