# Coaxial cable has radius a of copper core and radius b of copper shield. Between there is an insulat

Coaxial cable has radius a of copper core and radius b of copper shield. Between there is an insulator with specific resistivity $\zeta$. 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: $dR=\zeta \frac{dr}{2\pi rdr}$ The solution sheet says: $dR=\zeta \frac{dr}{2\pi rL}$
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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Layne Bailey
The area of your element is $2\pi rL$ and the thickness of your element is $dr$