Exercise:
A short wire with charge q is placed in the radial field produced by a cylindrical rod carrying a charge Q see figure. The orientation of the wire is along the field lines of the radial field i.e. perpicular to the rod. The charge q is much smaller than Q so that it does not affect the charge distribution in the rod. center includegraphicswidthcm#image_path:perpicular-wirsetup-# center abcliste abc Derive a formal expression for the force on the wire. abc Calculate the force acting on a wire with length lO and charge qO in the field of a rod with charge QO and length LO for a distance dO. abcliste

Solution:
abcliste abc The electric field due to the charged rod at a distance x from the rod is given by Ex fracpivarepsilon_fraclambdax fracQpivarepsilon_ L x The force on a short segment of the wire with charge textdq at distance x from the rod is see figure textdF Ex textdq center includegraphicswidthcm#image_path:perpicular-wirsolution-# center For a uniform charge distribution in the wire the charge of the segment can be written as textdq fracqell textdx where textdx is the length of the segment. vspacemm Since the force on a segment deps on the distance from the rod we have to egrate the force along the wire: F _d-ell/^d+ell/ Ex fracqell textdx _d-ell/^d+ell/ fracQpivarepsilon_ L x fracqell textdx fracQ qpivarepsilon_ L ell _d-ell/^d+ell/ fracx textdx The antiderivative of /x is ln x hence F fracQ qpivarepsilon_ L ell left ln x right_d-ell/^d+ell/ fracQ qpivarepsilon_ L ell left lnleftd+fracellright - lnleftd-fracellright right fracQ qpivarepsilon_ L ell lnleftfracd+fracelld-fracellright FF abc The force on the wire is F FF fracQtimesqpitimesncepstimesLtimesl &quad quad timeslnleftfractimesd+ltimesd-lright F approx resultFP- abcliste