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Coaxial cable has radius a of copper core and radius b of copper shield. Between there is an insulator with specific resistivity ζ. What is the resistance of this cable with length L between the core and the shield?

First, I tried to solve this like this: $dR=\zeta \frac{l}{S}$

In our case the length is dr, and therefore I suppose that the area of this ring is 2πrdr: $dR=\zeta \frac{dr}{2πrdr}$ The solution sheet says: $dR=\zeta \frac{dr}{2πrL}$

I know that something is wrong with my equation, because dr goes away and then I cannot integrate from a to b. But why is in the solution L instead of dr? As I understand problem instruction L= b-a. And therefore L=dr which doesn't make sense to me.

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  • $\begingroup$ @Farcher I imagine that the area of my element is 2πrdr, not 2πrL. What am I missing? $\endgroup$ – continuity Apr 18 '18 at 11:03
  • $\begingroup$ The element for integration is like a “pipe” of length $L$, inner radius $r$ and outer radius $r+dr$. $\endgroup$ – Farcher Apr 18 '18 at 11:18
  • $\begingroup$ @Farcher Got it now! I thought that b-a=L from problem instruction and it didn't make any sense to me. $\endgroup$ – continuity Apr 18 '18 at 11:27
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The area of your element is $2\pi r L$ and the thickness of your element is $dr$.

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