Im not getting the right answers for thecomponents and im really confused in how
ID: 1735074 • Letter: I
Question
Im not getting the right answers for thecomponents and im really confused in how to find them. If you can show me step by step, that would begreat! Thank you for all your help its muchappreciated. In the figure below, two point charges q1 =6 muC and q2 =-6 muC interact with a third pointcharge Q = 10 muC. Find themagnitude and direction of the total (net) force on Q. Magnitude Im not getting the right answers for thecomponents and im really confused in how to find them. If you can show me step by step, that would begreat! Thank you for all your help its muchappreciated.Explanation / Answer
Given : q1 = 6x10-6 C q2 = - 6x10-6 C Q =10x10-6 C distance between q1 and Q : 0.50m distance between q2 and Q : 0.50m Required: Find the magnitude and direction of thetotal force exerted on Q by q1 and q2. Solution: a) Fq1 on Q = (kq1Q) /d2 = [ (9x109)(6x10-6)(10x10-6)] / [(0.5)2] = 2.16 N Fq2 onQ = (kq2Q) / d2 = [ (9x109)(-6x10-6)(10x10-6)] / [(0.5)2] = - 2.16 N = arcsin (0.3/0.5) =36.87o Fq1 on Q(x) = 2.16 cos 36.87o = 1.728 N Fq1 on Q(y) = - 2.16 sin 36.87o = -1.296 N Fq2 on Q(x) = - 2.16 cos 36.87o = -1.728 N Fq2 onQ(y) = - 2.16 sin36.87o = -1.296N Fnet = [ Fq1 on Q (x) +Fq2 on Q (x) ] i + [Fq1 on Q (y) + Fq2 on Q(y) ] j note:Fq1 on Q (x) and Fq2 on Q(x) will just cancel. Fnet (vector form) =0 i - 2.592 j Fnet =2.592 N , direction is 270o fromthe + x-axis (counterclockwise) or 90oclockwise from the + x-axis