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Consider a coaxial cable which consists of an inner cylindrical conductor of radius R1, and a shell cylindrical conductor of radii R2 and R3. The 2 conductors are separated with a dielectric material of permittivity ε. Consider the length of the cable, ℓ, much larger than R3. The cable is connected to a voltage source V. Determine the polarization charge distribution per unit of length.

So I have no problem calculating the surface polarization charge distribution. Having:

$$D=\frac{\lambda}{2 \pi r}$$ $$E=\frac{\lambda}{2 \pi \epsilon r}$$ $$P=\frac{\lambda (\epsilon - \epsilon_0)}{2 \pi r \epsilon}$$ $$V=\frac{\lambda}{2 \pi \epsilon} \times \log{\frac{R_2}{R_1}}$$

Than we should have in r=R1:

$$\sigma= -\frac{V (\epsilon - \epsilon_0)}{R_1 \log{\frac{R_2}{R_1}} }$$

Now they as for the density by unit of length and the answer should be:

$$\lambda= -\frac{2 \pi V (\epsilon - \epsilon_0)}{\log{\frac{R_2}{R_1}} }$$

According to the answer, the relation between lambda and sigma should be:

$$\lambda = 2 \pi R_1 \sigma$$

However I don't understand how it can be? So my question is exactly how to get to that relation between lambda and sigma (if that relation is correct, of course).

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  • $\begingroup$ If $\sigma$ is the surface charge density - charge per unit area of surface - and the surface area of a cylinder of unit length and radius $R_1$ is the circumference of the cylinder times its length you can find the charge on a unit length of the cylinder. $\endgroup$ – Farcher Oct 29 '17 at 14:51
  • $\begingroup$ Hello Farcher! Thanks for being so quick at answering me. I would ask you if you could please elaborate more with mathematical expressions because I'm having trouble understanding ... $\endgroup$ – Granger Obliviate Oct 29 '17 at 15:43

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