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Problem: calculate the Electric Field for two coaxial (along z-Axis), infinite thin and infinite long charged cylinders with $r_1 < r_2$ and surface charge density $-\sigma \frac{r_1}{r_2}$ respectively $\sigma>0$.

Solution: So from Gauss follows that $$\int_{\partial V} \vec{E}\cdot d\vec{F}=\int_{\partial V} Edf = E\cdot M = 4\pi Q_{in} \quad \Rightarrow \quad \vec{E}=4\pi\frac{Q_{in }}{M} \tag{1}$$

wheres $M$ is the surface area of our Gauss-Cylinder $$M=2\pi r h \tag{2}$$

We have $$Q_{in}=\begin{cases}0, & r<r_1\\ 2\pi r_1 h\cdot \sigma', & r_1\leq r < r_2 \\ 0, &r_2 \leq r\end{cases} \tag{3}$$

Whereas $\sigma'=\sigma \frac{r_1}{r_2}$

Now we get, using (1):

$$\vec{E}=\begin{cases}0, & r < r_1\\ 4\pi\frac{r_1}{r}\sigma', &r_1\leq r < r_2 \\ 0, &r_2\leq r \end{cases} \tag{4}$$

For the potential,

Question:

So first of all, I'm confused by their use of the symbol $\sigma$. Where I used $\sigma'$, they just used $\sigma$. Their official solution is:

$$Q_{in}=\begin{cases}0, & r<r_1\\ 2\pi r_1 h\cdot \sigma, & r_1\leq r < r_2 \\ 0, &r_2 \leq r\end{cases} \tag{5}$$

$$\vec{E}=\begin{cases}0, & r < r_1\\ \frac{\sigma}{r}r_1, &r_1\leq r < r_2 \\ 0, &r_2\leq r \end{cases} \tag{6}$$

But I just can't see how they got their solution.

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I think the problem means that the surface density of the inner cylinder is $\sigma$ and the surface density of the external cylinder is $\sigma'$.

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  • $\begingroup$ hmm, I think they really did mean that the surface density is $\pm \sigma \frac{r_1}{r_2}$. The $\sigma > 0$ and the fact that we have a capacitor should indicate the $\pm$ - at least that's how I read it and their solution does match up with that (outer electric field vanishes). But then I'm still wondering what $\sigma$ should stand for. $\endgroup$
    – handy
    Jul 1, 2019 at 15:59
  • $\begingroup$ If the surface density is the same on the two surfaces, the electric field outide is not zero as the outer surface is larger (i.e. the charge on it when you apply Gauss' law is bigger and you have a total charg which is not zero) while, if it's as I say, the total charge is zero. $\endgroup$
    – Rhino
    Jul 1, 2019 at 21:32
  • $\begingroup$ I meant that the charge is the same, that implies that the density varies of course. $\endgroup$
    – handy
    Jul 3, 2019 at 12:05
  • $\begingroup$ Were you able to get what they got? $\endgroup$
    – handy
    Jul 3, 2019 at 12:28
  • $\begingroup$ Yes, if it's as I say I get the official solution $\endgroup$
    – Rhino
    Jul 3, 2019 at 17:34

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