# Electric field and charge density outside two coaxial cylinders

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

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 Gaussian 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.

This is a variant of a standard question in electrostatics: finding the potential of an infinite line charge.

Let's consider the simpler case first, and then apply that to the more complex problem you asked.

The electric field of an infinite line charge is proportional to 1/r, where r is the radial distance from the axis of the line charge. As you found in your last line, integrals of this field produce natural logs, which diverge at either zero or infinity. This implies that you cannot choose either zero or infinity as the reference point of potential, because the line integral from either point to some finite radius diverges.

The textbook problem is usually got round by choosing the reference point at some arbitrary position, and then integrating from there to the location of interest; thus the natural log never diverges unless the point of interest is at zero or infinity. For instance, see this: How to find electric scalar potential of infinite wire with Poisson/Laplace equation?.

Taking this back to your question, then, I think the problem is ill-posed. I see three ways to resolve this:

1) The outer shell is grounded. This would force the electric field for c < r to be zero, but contradicts the problem statement.

2) The net charge of the system is specified to be zero, so the electric field is zero outside by Gauss's law. This seems most likely, as it naturally avoids the diverging natural log.

3) The reference point of potential is set at some finite, non-zero r. This seems unlikely, if the problem does not state anything about reference point.

• To complement this answer: one can only set the potential to be 0 at infinity ** if ** there is no charge at infinity. This is not the case for an infinitely long cylinder as in this problem; hence the apparent non-sensical answer. – ZeroTheHero Dec 25 '16 at 3:03