Question

(1) Consider two point charges of magnitude q/2 separated by distance R and held in relative position by a massless rod. An identical massless rod with the same charges lies a distance R away, in the same plane, parallel, q and with centers aligned. I guess you better see the drawing! 2 By dimensional analysis, the force between these two objects must be some factor times kq2/R2, in other words: kq2 Find the factor C.

Answer 1

Force on the two charges are by the opposite two charges are as shown in fig. I marked the forces as different name easy to read F_2 & F_2 are equal & F_1 & F_y are also equal. The net force is F = F_2 + F_3 + (F_1 + F_4) cos theta sin theta = 2 F_2 + 2 F_1 sin theta F_1 cos theta & F_4 cos theta are equal & opposite so cancelled. F_2 = k(q/2)^2/R^2 = 1/4 kq^2/R^2 F_1 = k(q/2)^2/(squareroot R)^2 = 1/8 kq^2/R^2 Net. Force = F = 2 F_2 + 2 F_1 sin theta = 2/4 kq^2/R^2 + 2/8 kq^2/R^2 1/ squareroot 2 = (1/2 + squareroot 2/8) kq^2/R^1 = C = 1/2 + squareroot 2/8 C = 0.67677

C=0.66777

Thanks