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bitrex
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I'm attempting to teach myself some electrostatics, so that I'll be prepared when I head back to school full time.
Determine the electric field inside the inner conductor, between the conductors, and outside the conductors of a coaxial cable when the inner conductor is negatively charged with a linear charge density of [tex]-\lambda[/tex] and the outer conductor is positively charged with a linear charge density of [tex]\lambda[/tex].
[tex]\iint \vec{E} \cdot dS = \frac{q_0}{\epsilon_0}[/tex]
[tex]\lambda = \frac{q}{l} = 2\pi r \sigma[/tex]
Using cylindrical symmetry, I can dispense with the integral and say that [tex] \vec{E}(2\pi r l) = \frac{2 \pi r l \sigma}{4\pi \epsilon_0}[/tex]. I construct my first Gaussian cylinder inside both the positively charged conductor and the negatively charged conductor. There is no charge enclosed by this Gaussian surface, so the electric field must be zero. I construct my second Gaussian surface outside the negatively charged conductor, but inside the positively charged conductor. The positively charged conductor doesn't contribute anything since it is outside the surface, but the negatively charged conductor, using the equation above, contributes an electric field of [tex]-\frac{\sigma}{4 \pi \epsilon_0}[/tex]. I construct a third Gaussian surface outside both the positively charged conductor and the negatively charged conductor. The net field here is the sum of both the positive flux pointing outward from the positive conductor, and the negative flux pointing inward towards the negative conductor, so the field is 0 outside both conductors.
Does that line of reasoning look correct?
edit: I'm getting confused in this problem by the two variants of Gauss' law I'm seeing - one I'm looking at has the right hand side as being [tex]4\pi k_e q_s[/tex]. That's why I'm getting mixed up with the 4 pi. I'll have to do this over.
Homework Statement
Determine the electric field inside the inner conductor, between the conductors, and outside the conductors of a coaxial cable when the inner conductor is negatively charged with a linear charge density of [tex]-\lambda[/tex] and the outer conductor is positively charged with a linear charge density of [tex]\lambda[/tex].
Homework Equations
[tex]\iint \vec{E} \cdot dS = \frac{q_0}{\epsilon_0}[/tex]
[tex]\lambda = \frac{q}{l} = 2\pi r \sigma[/tex]
The Attempt at a Solution
Using cylindrical symmetry, I can dispense with the integral and say that [tex] \vec{E}(2\pi r l) = \frac{2 \pi r l \sigma}{4\pi \epsilon_0}[/tex]. I construct my first Gaussian cylinder inside both the positively charged conductor and the negatively charged conductor. There is no charge enclosed by this Gaussian surface, so the electric field must be zero. I construct my second Gaussian surface outside the negatively charged conductor, but inside the positively charged conductor. The positively charged conductor doesn't contribute anything since it is outside the surface, but the negatively charged conductor, using the equation above, contributes an electric field of [tex]-\frac{\sigma}{4 \pi \epsilon_0}[/tex]. I construct a third Gaussian surface outside both the positively charged conductor and the negatively charged conductor. The net field here is the sum of both the positive flux pointing outward from the positive conductor, and the negative flux pointing inward towards the negative conductor, so the field is 0 outside both conductors.
Does that line of reasoning look correct?
edit: I'm getting confused in this problem by the two variants of Gauss' law I'm seeing - one I'm looking at has the right hand side as being [tex]4\pi k_e q_s[/tex]. That's why I'm getting mixed up with the 4 pi. I'll have to do this over.
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