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Non Linear Arrays of charges Contents: •2-D Arrays •Example •Whiteboards Basic Concept FCB A B FAB C •Three equal positive charges A, B, C •The force on B is the sum of the forces of repulsion from A and C Basic Concept FCB FAB A B FAB+ FCB C It’s Vector Time Kiddies!! •Three equal positive charges A, B, C •The force on B is the sum of the forces of repulsion from A and C Non Linear Arrays +16 C A +153 C 75 cm C Find the force on A: +312 C 190 cm B FBA +16 C A +153 C 75 cm C FCA Find the force on A: +312 C 190 cm 1. Calculate the forces 2. Figure out direction (angle) 3. Add forces (as vectors) B FBA +16 C A +153 C 75 cm C FCA Find the force on A: 1. Calculate the forces 2. Figure out direction 3. Add forces (as vectors) +312 C 190 cm FBA = kqBqA = k(312 C)(16 C) = 10.76 N r2 (1.92 + .752) FCA = kqCqA = k(153 C)(16 C) = 19.56 N r2 (.752 + .752) B FCA = 19.56 N FBA = 10.76 N +16 C 2. Figure out direction 3. Add forces (as vectors) A +153 C 75 cm C B 190 cm +312 C B C Figuring out direction: C = 45o B = Tan-1(.75/1.9) = 21.54o = 158.46o (trig angle) (180-21.54) Find the Trig angle – ACW from x axis 90o 27o 0o Or 360o 180o This is the trig angle 270o 17o 51o 15o T = 270 – 15 = 255o T = 270 + 51 = 321o T = 360 – 17 = 343o OR T = -17o So - do you remember how to add vectors?? What can we do to AM vectors? FBA = 10.76 N 21.54o FCA = 19.56 N y x 45o A AM to VC: FCA = 19.56cos(45o) x FCA = 13.83 x + 19.56sin(45o) y + 13.83 y FBA = 10.76cos(158.46o) x FBA = -10.01 x + 10.76sin(158.46o) y + 3.95 y FCA = 19.56 N y FBA = 10.76 N 21.54o x 45o A VC + VC = VC FCA = 13.83 x + 13.83 y FBA = -10.00 x + 3.95 y FCA + FBA = +3.83 x + 17.78 y FCA = 19.56 N y FBA = 10.76 N 21.54o VC to AM: FCA + FBA = +3.83 x x 45o A + 17.78 y Magnitude = (3.832 + 17.782) = 18 N 17.78 N Angle = Tan-1(17.78/3.83) = 78o 3.83 N Whiteboards: Non-Linear Charge Arrays 1|2 C +162 C Find the force on A, and the angle it makes with the horizontal. .80 m +15 C A 2.1 m FBA = 15.66 N, FCA= 34.13 N F = (15.662 + 34.132) = 37.55 N = 38 N Angle = Tan-1(34.13/15.66) = 65o 38 N, 65o Below x axis, to the left of y +512 C B 15.66 A 34.13 C +180 C Find the force on C, and the angle it makes with the horizontal. .92 m +150 C A 1.9 m FAC= 286.8 N, FBC = 188.8 N ABC = Tan-1(.92/1.9) = 25.84o FAC = 0 N x + 286.8 N y FBC = -188.8cos(25.84o) x + 188.8sin(25.84o)y Ftotal = -170. x + 369 y 410 N, 65o above x axis (to the left of y) +520 C B