Question

1. Physics is used in many magic tricks. The diagram below shows a balancing rod that has a charge of +q and +2q on the ends. Directly beneath these charges charges, each +Q attached to the floor provide an upward force rod's charges. The magician's assistant must also lay down on this apparatus (indicated by the mass "W" in the diagram) are two other a) Where should this mass of weight W be placed (measured from the left side of the rod) to perfectly balance the rod? b) As you well know, magicians often prove to the audience that there are not strings supporting the levitating assistant. What value should h have so that the rod exerts no vertical force on the bearing (assume it is still perfectly balanced)? L Rod +q Bearing +2q +Q W +Q

Please show all the steps in the solution, thank you.

Answer 1

h2 Ans:a) The rod is perfectly balanced , hence that about the bearing is zero. F = KQq (electric ne force on tq due to tQ) F = K (29)) (electric force on +2q due to Q) | for 2ka | Znet about the bearing = -5,*4 + Fz ç - Wa(2-4)=0 WZ 2+{w + {( F2-F1) ** Ź + zi ( 52 - F.) = {(1+ (2 keqe - Kqy)) 2kQ2 h2 b) We want FN=0 {Fy = F, F2- W- FN = 0 . W= Ft Ez = Koal + 2kQg h 3kQq or h= 3kaq - search eine 62 3K 04 9. w 1 3Q9 97,