Non Linear Arrays of charges Contents: 2-D Arrays Example Whiteboards

Basic Concept TOC Three equal positive charges A, B, C The force on B is the sum of the forces of repulsion from A and C ABC F CB F AB

Basic Concept TOC Three equal positive charges A, B, C The force on B is the sum of the forces of repulsion from A and C ABC F CB F AB F AB + F CB It’s Vector Time Kiddies!!

Non Linear Arrays TOC ABC +16  C +153  C +312  C 190 cm75 cm Find the force on A:

TOC ABC F CA F BA +16  C +153  C +312  C 190 cm75 cm Find the force on A: 1.Calculate the forces 2.Figure out direction (angle) 3.Add forces (as vectors)

TOC ABC F CA F BA +16  C +153  C +312  C 190 cm75 cm Find the force on A: 1.Calculate the forces 2.Figure out direction 3.Add forces (as vectors) F BA = kq B q A = k( 312  C)(16  C) = N r 2 ( ) F CA = kq C q A = k( 153  C)(16  C) = N r 2 ( )

TOC ABC F CA = N F BA = N +16  C +153  C +312  C 190 cm75 cm 2. Figure out direction 3. Add forces (as vectors) Figuring out direction:  C = 45 o  B = Tan -1 (.75/1.9) = o = o (trig angle) ( ) BB CC

TOC Find the Trig angle – ACW from x axis 0 o Or 360 o 90 o 180 o 270 o 27 o 51 o 15 o 17 o This is the trig angle  T = 270 – 15 = 255 o  T = = 321 o  T = 360 – 17 = 343 o OR  T = -17 o

A F CA = N F BA = N 45 o o So - do you remember how to add vectors?? What can we do to AM vectors? AM to VC: F CA = 19.56cos(45 o ) x sin(45 o ) y F CA = x y F BA = 10.76cos( o ) x sin( o ) y F BA = x y x y

A F CA = N F BA = N 45 o o VC + VC = VC F CA = x y F BA = x y F CA + F BA = x y x y

A F CA = N F BA = N 45 o o VC to AM: F CA + F BA = x y x y 3.83 N N Magnitude =  ( ) = 18 N Angle = Tan -1 (17.78/3.83) = 78 o

Whiteboards: Non-Linear Charge Arrays 11 | 22 TOC

W ABC +162  C +15  C +512  C 2.1 m.80 m Find the force on A, and the angle it makes with the horizontal. F BA = N, F CA = N F =  ( ) = N = 38 N 38 N, 65 o Below x axis, to the left of y A Angle = Tan -1 (34.13/15.66) = 65 o

W ABC +180  C +150  C +520  C 1.9 m.92 m Find the force on C, and the angle it makes with the horizontal. F AC = N, F BC = N  ABC = Tan -1 (.92/1.9) = o F AC = 0 N x N y F BC = cos(25.84 o ) x sin(25.84 o )y F total = x+ 369 y 410 N, 65 o above x axis (to the left of y)