A semicircle with radius R has uniform charge density Show

A semicircle with radius R has uniform charge density -. Show that at all points along the \"axis\" of the semicircle (the line through the center, perpendicular to the plane of the semicircl, as shown in the figure) the vectors of the electric field all point toward a common point in the plane of the semicircle. where is this point?

Solution

Length of the arc subtended by d@=Rd@

charge in arclength Rd@=dQ=QRd@/(R/2)=2Qd@/

electric field due to this charge at P,

         dE=kdQ/R^2=2kQd@/(R^2)

y-component of the electric field,

          dEy=dEcos@

                =2kQcos@d@/(R^2)

net y-componet of the electric field due to first half of the rod

      integral(dEy)=[2kQ/(R^2)]integral(cos@d@)

                            =[2kQ/(R^2)][sin@] from limits @=0 to @=?/2

                          = 2kQ/(R^2)

at an angle @ with respect to the the horizontal, charge in arc length is,

       Rd@=dQ =-QRd@/(?R/2) = -2Qd@/

the electric field due to this charge at P,

       dE =kdQ/R^2 =-2kQd@/(R^2)

the y-component of the electric field,

      dEy = -dEsin@ = 2kQsin@d@/(R^2)

net y-component of the electric filed   

      integral(dEy) =[2kQ/(R^2)]integral(sin@d@)

                           = [2kQ/(R^2)][-cos@] from limits @=0 to @=?/2,

                           = [2kQ/(R^2)]

Tus total y-component of the firld due to the entire rod

                      Ey= 2kQ/(R^2)+2kQ/(R^2)

                          = 4kQ/(R^2), vertically downward or pointed along y-axis