# Electric field for two coaxial, infinite thin, infinite long cylinders

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

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

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

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

I think the problem means that the surface density of the inner cylinder is $$\sigma$$ and the surface density of the external cylinder is $$\sigma'$$.
• 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. Jul 1, 2019 at 15:59