Infinite Line Charges

Assume that the rod remains fixed in place. Let $x$ be the distance from a point on the rod to the end of the rod that is farthest from the charged particle. Let $D$ be the distance from that same end to the charged particle. Let the rod length be $L$ .

Infinitesimal rod charge:

$dQ = \lambda \, dx$

Infinitesimal potential energy contribution:

$d U = - \frac{dQ \, q}{4 \pi \epsilon_0 \, (D - x) } = - \frac{\lambda \, q \, dx}{4 \pi \epsilon_0 \, (D - x) }$

Total potential energy:

$U = - \frac{\lambda \, q}{4 \pi \epsilon_0}\int_0^L \frac{dx}{ (D - x) } = \frac{\lambda \, q}{4 \pi \epsilon_0} \ell n \Big (\frac{D - L}{D} \Big)$

Difference in potential energy from $(D = 2 L)$ to $(D = \frac{3}{2} L)$ :

$\Delta U = \frac{\lambda \, q}{4 \pi \epsilon_0} \Big [ \ell n \Big (\frac{2 L - L}{2 L} \Big ) - \ell n \Big (\frac{3/2 \, L - L}{3/2 \, L} \Big ) \Big ] \\ = \frac{\lambda \, q}{4 \pi \epsilon_0} \ell n \Big (\frac{3}{2} \Big )$

Setting the change in potential energy equal to the particle kinetic energy:

$\Delta U = \frac{1}{2} m v^2 \\ \frac{\lambda \, q}{4 \pi \epsilon_0} \ell n \Big (\frac{3}{2} \Big ) = \frac{1}{2} m v^2 \\ v^2 = \frac{\lambda \, q}{2 \, \pi \epsilon_0 m} \ell n \Big (\frac{3}{2} \Big )$