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I'm currently having trouble understanding why the electric field inside a coaxial conductor (assuming that the charge is evenly distributed on the inner and outer conductor and opposite signed) is the same as a line charge.

I know that the field is just depended on the enclosed charge if you derive the equation with Gauß' law, but in my head it doesn't make sense since the inner charge in the coaxial conductor is "pushing" the test charge outwards and the outer charge is "pulling" the charge, so the forces should be superpositioned but thats not correct since the formula is:

$$ E = \frac{1}{2\pi\epsilon}\cdot \frac{Q}{r \cdot l} $$

with \$ 2\pi r l \$ as the surface of the integral of the enclosed charge Q.

Can anyone explain why the outer charge doesn't have an influence on the electric field? Also maybe in comparision to dot charges where I have to superposition the different electric fields of 2 opposite charges.

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2 Answers 2

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but in my head it doesn't make sense since the inner charge in the coaxial conductor is "pushing" the test charge outwards and the outer charge is "pulling" the charge

Your intuition here definitely makes sense. Suppose that the inner conductor is positively charged, the outer conductor is negatively charged, and we have a positive test charge somewhere in between the inner conductor and the outer conductor. Since the outer conductor is negatively charged and the test charge is positive, it seems like the outer conductor should attract the test charge towards itself, right?

Below is a very crude diagram I made representing the situation. The orange circles are the inner and outer conductors, and the blue dot is the test charge.

A cross-section of a coaxial cable, with a test charge located in between the inner conductor and the outer conductor, on the left side of the inner conductor

Intuitively, it seems like the positive test charge should be pulled to the left, since it's attracted to the negatively charged outer conductor. So why doesn't that happen?

Well, notice that only a small portion of the outer conductor is on the left side of the test charge. Most of the outer conductor is to the right of it. So, on the one hand, the test charge is closer to the left-hand part of the outer conductor than to the right-hand part, and for that reason, we would expect the left-hand part to pull on it more strongly. On the other hand, the right-hand part of the outer conductor (meaning the portion of the outer conductor which is to the right of the test charge) is much larger than the left-hand part, and for that reason, we would expect the right-hand part to pull on it more strongly.

If we were to do the math, we would find out that these two effects cancel each other out perfectly: both the left-hand part and the right-hand part of the outer conductor pull on the test charge with exactly equal force, and so the overall force from the outer conductor is exactly zero!

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    \$\begingroup\$ I had to read your explanation a few times, but you are correct. The outer sleeve is in fact a uniformly charged hollow cylinder, which can be shown to have zero electric field inside, except for some fringe effects at the ends. All of the electric field inside the cable is due to the charge on the inner conductor. \$\endgroup\$
    – Bart
    Commented Sep 11, 2023 at 12:28
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The problem must be addressed using Maxwell's equations. With them we study the electromagnetic field in a multi-connected system. I also added the study of field propagation on a two-wire line. enter image description here enter image description here

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