# E-field and charge density outside two coaxial cylinders

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1. Feb 18, 2015

I am working on a problem of electrostatics, and I am having troubles in trying to figure out one part of it.

1. The problem statement, all variables and given/known data

It consists of an inner solid cylinder of radius $a$ with a voltage $V_A$, and an outer coaxial cylindrical shell of inner radius $b$ and outer radius $c$ charged with a voltage $V_B$. Find $V(\mathbf{r})$, $\mathbf{E}(\mathbf{r})$ everywhere, and the surface change density at the surfaces of the conductors.

2. Relevant equations
Maxwell's equations

3. The attempt at a solution
I have calculated the potential field $V$ as well as the $\mathbf{E}$ field for $r < a$, $a < r < b$, and $b < r < c$. I also calculated the surface charge density for the inner cylinder and the inner layer of the cylindrical shell. But can't figure out anything for $c \leq r$.

I tried using Gauss's law, taking the origin of potential at infinity ($V(\infty) = 0$), finding the $\mathbf{E}(\mathbf{r})$ field for $c<r$ and calculating $\displaystyle V_B = \int_{\infty}^{c} \mathrm{d}V = -\int_{\infty}^{c} \mathbf{E}(\mathbf{r} )\mathrm{d}\mathbf{r}$ in order to find the value of surface charge density at $r = c$. Also, I tried using that outside the shell we can assume the space is dielectric, so as $-\nabla ^{2} V = 0$. None of this yields reasonable values, or at least I am doing something wrong in the procedure.

Procedure used:

• Using a Gauss cylinder of radius $r>c$, $Q_{in} = Q = 2\pi c L \sigma_{c}$, where $L$ is the length of the cylinder.
• Through Gauss's law, $\Phi = E(r) Area = E \cdot 2 \pi r L = Q /\varepsilon_0 = \dfrac{2\pi c L \sigma_{c}}{\varepsilon_0}$ $\Rightarrow$ $\mathbf{E}(r) = \dfrac{\sigma_{c}}{\varepsilon_0}\dfrac{c}{r}\hat{r}$.
• Finding $\sigma_c$: $\displaystyle \left. V_B = \int_{\infty}^{c} \mathrm{d}V = -\int_{\infty}^{c} \mathbf{E} \mathrm{d}\mathbf{r} = - \dfrac{\sigma_c}{\varepsilon_0} c \ln r \, \right|_{\, \infty}^{\, c} = \infty$, which doesn't make sense.

Last edited: Feb 18, 2015
2. Feb 19, 2015

### BvU

There is something with the potential from an infinitely long line charge that makes it impossible to choose V = 0 at infinity. You bumped into that. But you can still use Gauss law for a a cylindrical surface around the outer cylinder as you did. Just don't go as far as infinity.