# 16 C - TuHS Physics

```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)
B
FBA
+16 C
A
+153 C
75 cm
C
FCA
Find the force on A:
1. Calculate the forces
2. Figure out direction
+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
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
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
```