Field between charged cylinders The question is the following:

"For an air-filled cylindrical capacitor, with inner radius a and outer radius b, show 
  that the electric field between the cylinders is
  $$E = E_{in} \frac{a}{r}$$
  where $E_{in}$ is the field strength at the inner cylinder and r is the distance from its axis.
  A breakdown occurs in air if the electric field strength is larger than a particular
  value $E_0$. This determines the maximum value of $E_{in}$. For a fixed outer radius b, find the 
  inner radius a for which the energy per unit length in the electric field is a maximum."

I managed to do the first part, but then I am stuck. I introduced $\sigma$ as a variable for the charge density on the inner cylinder, but I seem unable to find an expression which cancels it and only leaves me with an expression of b in terms of a. I am also unsure about what "energy per unit length" is referring to. The answer is supposed to b: $ a = b/ \sqrt(e) $.
Many thanks! 
 A: Cylindrical capacitors have an inner radius, outer radius and also a length. So energy per unit length actually refers to the energy in a cylindrical capacitor with a unit length.

Now, the field at  distance $r$ from the axis of the capacitor is  $E = E_{in} \frac{a}{r}$. Thus the energy density per volume $\rho$ is given by:

$$\rho = \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2}\epsilon_0 E_{in}^2\frac{a^2}{r^2}$$

Thus, the total energy $\phi$ id given by integrating $\rho$ over the whole volume enclosed by the capacitor:

$$\phi = \int_{Volume} \rho dv = \int_{0}^{l}\int_{a}^{b} \frac{1}{2}\epsilon_0 E^2 * 2\pi rdr* dl $$
$$= \pi\epsilon_0 E_{in}^2a^2 \int_{a}^{b}\frac{dr}{r} \int_{0}^{l} dl$$
$$= \pi\epsilon_0 E_{in}^2a^2 \ln(\frac{b}{a})l$$

Thus, the energy per unit length is given by $\frac{\phi}{l}$
$$=\pi\epsilon_0 E_{in}^2a^2 \ln(\frac{b}{a})$$
$$=ka^2 \ln(\frac{b}{a})$$
Where $k=\pi\epsilon_0 E_{in}^2$.

 To maximise the energy per unit length, we need to differentiate it with respect to $a$ and equate to zero to get:
$$\frac{d(ka^2 \ln(\frac{b}{a}))}{da} =0$$
$$\Longrightarrow 2a\ln(\frac{b}{a})-a=0$$
$$\Longrightarrow \ln(\frac{b}{a}) = \frac{1}{2} $$
$$\Longrightarrow a=\frac{b}{\sqrt{e}}$$

As required by the question.
