Exercise:
A wire with length ell is uniformly charged constant linear charge density lambda. abcliste abc Show that the electric field at a po on the perpicular bisector of the wire is given by Er k_c fraclambda ellr fracsqrtell/^+r^ where r is the distance of the observation po from the wire. center includegraphicswidth.mm#image_path:longitudinal-setup-# center Verify that for r ll ell this ts to Er fracpi varepsilon_ fraclambdar abc Derive the formal expression for the electric field at a po on the line defined by the wire outside the wire. center includegraphicswidth.mm#image_path:longitudinal-setup-# center abcliste

Solution:
abcliste abc In order to calculate the electric field at the given po we have to divide the wire o infinitesimally small segments that can be treated as po charges textdq. The partial electric field produced by such a segment is see figure: center includegraphicswidthtextwidth#image_path:field-of-charged-wirsolution-# center textdE k_C fractextdqd^ k_C fractextdqr^+x^ k_C fraclambda textdxr^+x^ where in the last step we have used that the linear charge density is lambda textdq/textdx. vspacemm Because of the symmetry of the situation we only have to consider the y-component of the field the x-component is cancelled by the contribution of a segment on the right side of the wire. Using similar triangles we find fractextdE_ytextdE fracrd Longrightarrow textdE_y fracrd textdE fracrsqrtr^+x^ k_C fraclambda textdxr^+x^ k_C lambda r fracleftr^+x^right^/ textdx The total field corresponds to the egral along the wire: Er _-ell/^+ell/ textdE_y k_C lambda r _-ell/^+ell/ fracleftr^+x^right^/ textdx The antiderivative of fx fracleftr^+x^right^/ is given by the function Fx fracxr^leftr^+x^right^/ as can be verified by showing that F'x fx. It follows for the electric field Er k_C lambda r left fracxr^leftr^+x^right^/ right_-ell/^+ell/ k_C lambda fracellrleftr^+ell/^right^/ k_C fraclambda ellr fracsqrtr^+ell/^ quad square For ell gg r the first term in the square root can be neglected so E &approx k_C fraclambda ellr fracsqrtell/^ k_C fraclambda ellr fracell k_C fraclambdar fracpivarepsilon_ fraclambdar fracpivarepsilon_ fraclambdar quad square abc The partial electric field produced by an infinetisimal segment is see figure center includegraphicswidthtextwidth#image_path:field-of-charged-wiraxis-solution-# center textdE k_C fractextdqleft r-x right^ k_C fraclambdaleft r-x right^ textdx It follows for the total field Er k_C lambda _-ell/^+ell/ fracleft r-x right^ textdx k_C lambda left fracr-x right_-ell/^+ell/ k_C lambda left fracr-ell/ - fracr+ell/ right k_C lambda fracr+ell/ - r + ell/r-ell/r+ell/ k_C fraclambda ellr^-ell/^ For r gg ell the second term in the denominator can be neglected and we find E &approx k_C fraclambda ellr^ k_C fracQr^ where Q is the total charge on the wire. As expected far from a charged wire with finite length the field is approximately equal to the field of a po charge with the same charge. abcliste