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

3. 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

4. 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!!

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

6. 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)

7. 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) = 10.76 N r 2 (1.9 2 +.75 2 ) F CA = kq C q A = k( 153  C)(16  C) = 19.56 N r 2 (.75 2 +.75 2 )

8. TOC ABC F CA = 19.56 N F BA = 10.76 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) = 21.54 o = 158.46 o (trig angle) (180-21.54) BB CC

9. 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 = 270 + 51 = 321 o  T = 360 – 17 = 343 o OR  T = -17 o

10. A F CA = 19.56 N F BA = 10.76 N 45 o 21.54 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 + 19.56sin(45 o ) y F CA = 13.83 x + 13.83 y F BA = 10.76cos(158.46 o ) x + 10.76sin( 158.46 o ) y F BA = -10.01 x + 3.95 y x y

11. A F CA = 19.56 N F BA = 10.76 N 45 o 21.54 o VC + VC = VC F CA = 13.83 x + 13.83 y F BA = -10.00 x + 3.95 y F CA + F BA = +3.83 x+ 17.78 y x y

12. A F CA = 19.56 N F BA = 10.76 N 45 o 21.54 o VC to AM: F CA + F BA = +3.83 x+ 17.78 y x y 3.83 N 17.78 N Magnitude =  (3.83 2 + 17.78 2 ) = 18 N Angle = Tan -1 (17.78/3.83) = 78 o

13. Whiteboards: Non-Linear Charge Arrays 11 | 22 TOC

14. 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 = 15.66 N, F CA = 34.13 N F =  (15.66 2 + 34.13 2 ) = 37.55 N = 38 N 38 N, 65 o Below x axis, to the left of y A 34.13 15.66 Angle = Tan -1 (34.13/15.66) = 65 o

15. 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 = 286.8 N, F BC = 188.8 N  ABC = Tan -1 (.92/1.9) = 25.84 o F AC = 0 N x+ 286.8 N y F BC = -188.8cos(25.84 o ) x+ 188.8sin(25.84 o )y F total = -170. x+ 369 y 410 N, 65 o above x axis (to the left of y)