By using the method of images and symmetry, we get
\[\sigma = \frac{qR}{2 \pi (R^2 + l^2) ^{3\over2}} (\sigma q < 0)\]
From here on, we do the same steps as in Q 10.
\[F = {q^2 R^2 \over 16 \pi \epsilon_0}\left[ {1 \over (R^2 + x^2)^2} - { 1 \over (R^2 + (L-x)^2)^2}\right]\]
\[\implies v = \sqrt{ {q^2 \over 16 \pi \epsilon_0m R} \left[ \frac{1}{ {R \over x} + {x\over R} } + \frac{1}{ {R \over L} + {L\over R} } - \frac{1}{ {R \over L-x} + {L-x\over R} }+ \tan^{-1}\left( {x \over R} \right) + \tan^{-1}\left( {L \over R} \right) - \tan^{-1}\left( {L-x \over R} \right) \right] }\]
Here, we note that the motion is not oscillatory.
Velocity at end of the cylinder,
\[v\big|_{x=L} = \sqrt{ {q^2 \over 8 \pi \epsilon_0m R} \left[ \frac{1}{ {R \over L} + {L\over R} } + \tan^{-1}\left( {L \over R} \right) \right] }\]
If we let \(L\to\infty\),
\[v\big|_{x=\infty} = {q \over 4 \sqrt{m R \epsilon_0} }\]
Which is a constant finite terminal velocity.
For some estimation,
\(q=1\)C; \(m=1\)kg; \(R=1\)m
\[v\big|_{x=\infty} \approx 10^5 \text{ms}^{-1}\]