Question

A thin rod 39.2 cm long is charged uniformly with a positive charge density of 46.0 ?C/m. The rod is placed along the y-axis and is centered at the origin. A charge of +43.0 ?C is placed 51.2 cm from the midpoint of the rod on the positive x-axis. Calculate the electric field at a point on the x-axis, which is halfway between the point charge and the center of the rod. (Express your answer in terms of the unit vector x. For example, if the electric field is -10.5x N/C, then input -10.5 N/C.)

Answer 1

given
L = 39.2 cm = 0.392 m
lamda = 46 micro C/m
q = 43 micro C
r = 51.2/2 = 25.6 cm = 0.256 m

electric field on the perpendicular bisector,

E_rod = (2*k*lamda/r)*sin(theta)

= (2*k*lamda/r)*(L/2)/sqrt((L/2)^2 + r^2)

= (2*9*10^9*46*10^-6/0.256)*(0.392/2)/sqrt( (0.392/2)^2 + 0.256^2)

= 1.97*10^6 N/c (towards +x axis)

Eq = k*q/r^2

= 9*10^9*43*10^-6/0.256^2

= 5.90*10^6 N/c (towards -x axis)

so, Enet = 1.97*10^6 - 5.90*10^6

= 3.93*1066 N/C <<<<<<<<<------------Answer