# 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!

• cylinders have a radius and height, so the energy per unit length is actual the energy of a cylindrical capacitor with unit height. May 24, 2017 at 7:38

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: $$\frac{d(ka^2 \ln(\frac{b}{a}))}{da} =0$$ $$\Longrightarrow 2a\ln(\frac{b}{a})-a=0$$ $$\Longrightarrow \ln(\frac{b}{a}) = \frac{1}{2}$$ $$\Longrightarrow a=\frac{b}{\sqrt{e}}$$