If we substitute \(x\) as \({L \over 2}\), and \(v\) as \(c\) in the expression previously derived, we can solve for \(L\).
Hence,
\[c = \sqrt{ {\sigma R q \over m \epsilon_0} \left[ \sinh^{-1} \left( {L \over 2R} \right) + \sinh^{-1} \left( {L \over 2R} \right) - \sinh^{-1} \left( {L \over R}\right)\right]}\]
We know that,
\[\sinh^{-1} x \pm \sinh^{-1} y = \sinh^{-1}(x\sqrt{1+y^2} \pm y \sqrt{1+x^2})\]
We can then approximate \(L>>R\).
Hence, we get
\[L = 4R\sinh\left({c^2 m \epsilon_0 \over \sigma R q}\right) = 2R\left[e^{c^2 m \epsilon_0 \over \sigma R q} - e^{-c^2 m \epsilon_0 \over \sigma R q}\right]\]
To get an idea of how enormous this quantity is, consider:
\(R=1\)m; \(m=1\)g; \(\sigma=10\text{Cm}^{-2}\); \(q=1\)C.
These are already extreme numbers, since 1 C of charge in 1g is very difficult, along with the high charge density, \(\sigma\).
Substituting,
\[{c^2 m \epsilon_0 \over \sigma R q}\approx 566.4\]
\[\implies e^{c^2 m \epsilon_0 \over \sigma R q} \approx 9.67 \times 10^{245}\]
\[\text{and, } e^{-c^2 m \epsilon_0 \over \sigma R q} \to 0\]
\[L = 2R e^{566.4}\]
\[L \approx 2 \times 10^{245}\text{m}\]
Clearly, \(L>>R\).
For comparison, the size of the observable universe is approximately \(8.8 \times 10^{26}\)m. This implies our cylinder ought to be \(2.2 \times 10 ^{219}\) times bigger than the observable universe.
Note that if we had considered relativistic effects, as the particle approaches the speed of light, its mass increases. This implies that the length of the cylinder would be much larger, and in fact, tend to infinity.