At the begginning of class, we were asked to open the new pink bookleet titled Electric and Magnetic Fields and turn to page 13 which we went over the question. The answer for this question is:
F1 on B = (k)(qA)(qB)/R^2
= (9x10^9)(4.5x10^-6)(-8.2x10^6)/(0.040m^2)
= 208 N {L}
DISTANCE BETWEEN THE OTHER TWO CHARGES IS:
_/ means sqaure root
C = _/(0.040m^2)+(0.030m^2)
= 0.050m
" means inverse
Oi = tan"~0~(0.030m/0.040m)
= 37 degrees
F2 Fc on B = (k)(qC)(qB)/(R^2)
= (9x10^9)(8.2x10^-6 c)(6.0x10^-6 c)/(0.040m^2)
= 177 N at 217 degrees from positive x-axis
~0~ means theta
dg means degrees
COMPONENTS OF F1 are:
F2 x = (F2)(cos~0~)
= 177 N(Cos~0)(~217dg)
= 142 N {L}
F2 y = (F2)(Cos~0~)(Sin~0~)
= (177 N)(Sin~0~217dg)
= 106 N {Down}
COMPONENTS OF NET (Resultant Force) FORCES ARE
Fnet x = -208 N - 142 N
= -350 N or 350 N {L}
Fnet y = 106 N
Then...............
Fnet = _/(350 N^2)+(106 N^2)
= 366 N or 350 N {L}
~0~ 2 = Tan" (106 N)/(350 N)
= 17 dg below the negative x-axis
Fnet = (3.7x10^2) at 197dg from positive x-axis
Then in class we were asked to read pages 14, 15, 17, 18, 19, 20. The things that was important were the things in bold. The pages also went on a previous lessons about work. Remeber that potentional energy increases if the height of an object increases creating a bigger fall. Then we were asked to do questions in the booklet pages 42, 43 but we didn't go over them. Tommorow we will I think.
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