PHY1220 Week 1 Discussion: Review on Vectors; Coulomb’s Law
1. You are given vectors A and B in terms of magnitude and direction shown below. (i) Draw and label the x and y
components of each vector. (ii) Then determine the numerical values of the x and y components of each vector.
Ay
y
y
A
Bx
Ay
5
1300
x
Ax
x
450
By
4
By
B
Ax  A cos 130 o  (5) cos 130 o  3.2
Bx  (4) cos(360 o  45o )  2.82
Ay  A sin 130o  (5) sin 130o  3.8
B y  (4) sin(360 o  45o )  2.82




2. Vector A and B are as in problem 1. Draw vectors  A and  B , find their magnitudes and directions. Report
the direction by an angle counterclockwise from the +x-axis.


Sol: vectors  A and  B are shown as the thick blue arrows in the drawings above.

 A : magnitude = 5 (magnitude is always positive), direction: 130+180 = 310o

 B : magnitude = 4 (magnitude is always positive), direction: 180-45 = 135o
3. You are given the following vectors in terms of the x and y components. For each vector, (i) draw and label the x
and y components. (ii) Then draw and label the vector. (iii) Then determine the magnitude and the direction of
each vector. Report the direction by an angle counterclockwise from the +x-axis.
Ax = 3, Ay = 2
Bx = 0, By = 2
Cx = -3, Cy = -4
y
y

A
θ
Ay
Ax
y

B
By θ
θ
x
Cx
x
Cy
A  Ax2  Ay2  (3) 2  (2) 2  3.6
A 
2
  tan  y   tan 1    33o
 3
 Ax 
1
Dx = -1, Dy = 2
B  (0) 2  (2) 2  2
direction :
  90
o

C
C  (3) 2  (4) 2  5
4
  53 o
  tan  1 

3


  53o  180  233o
Ex = 3, Ey = -4
y
x
α
Fx = -2, Fy = 0
y
y

D
Dy
θ
α
θ
Ex
x
D  (1) 2  (2) 2  2.23
  tan 1  21   63 o
  180  63o  117o
x
α
Dx

E
 θ
F
Fx
Ey
E  (3) 2  (4) 2  5
4
o
  53
 3 o
o
  tan  1 
  360  53  307
F  (0) 2  (2) 2  2
  180 o
x
    

4. Vectors A , B , C , D , E , and F are the same as in problem 3. Find the following vectors using the graphical
method, then calculate the magnitude and direction of each vector using the component method.
  
  
  
H ED
G  A D
RCD
y
y
y


D
G
θ

A
θ
θ
x

E
D
H

Rα
 
D C
x
α

D
Rx  Cx  Dx  3  1  4
H x  E x  Dx  3  (1)  2
H y  E y  D y  (4)  (2)  2 R y  C y  D y  (4)  2  2
G x  Ax  Dx  3  (1)  2
G y  Ay  D y  2  2  4
G  G x2  G y2
H  H x2  H y2
R  Rx2  R y2
 (2) 2  (4) 2  4.5
 (2) 2  (2) 2  2.83
 (4) 2  (2) 2  4.47
 Ry 
  tan 1  
 Rx 
 Gy
 Gx
  tan 1 
x
 Hy
 Hx
  tan 1 

4
  tan 1    63 o
2




2
o
 tan 1 
  27
4
2
o
 tan 1 
  45
2


  27 o 180  207o
  360  45o  315o


5. A vector A has a magnitude of 40.0m and points in a direction 20.0o below the +x-axis. A second vector B , has a
magnitude of 75.0 m and points in a direction 50.0o above the +x-axis.
    
(a) Sketch the vectors A , B , C  A  B . (graphical method).
y
y
B
  
C  A B
50o
Θ1
x
o
20
B
x
A
A

(b) Using the component method, find the magnitudes and directions of the vectors C . Report the direction by an angle
counterclockwise from the +x-axis.
Sol: Ax  (40 m) cos(360 o  20 o )  37 .6m
Ay  (40m) sin( 360o  20o )  13.7m
By  (75m) sin 50o  57.5m
Bx  (75 m) cos 50 o  48 .2m



The x component of the vector C  A  B is： C x  Ax  Bx  37 .6m  48 .2m  85 .8m
C y  Ay  B y  13 .7m  (57 .5m)  43 .8m
Similarly, the y component is

C  Cx2  C y2  (85.8m) 2  (43.8m) 2 = 96.3m
The magnitude of vector C is

 Cy 
43 .8m 
o
  tan 1 
  27
 85 .8m 
 Cx 
1  tan 1 
Direction of C is
The Electric (electrostatic) force between two point charges:
Like charges repel; unlike charges attract.
Coulomb’s Law: F  k
q1 q2
r2
(magnitude of electric forces between two point charges)
6. Two point charges q1 and q2 are fixed at the two bottom corners of a square,
as shown in the figure. The length of each side of the square is 0.5 m.
+y
A third charge q = -0.5μC is placed at corner A,
(a) Determine the magnitude and direction of the
electric force F1 on charge q by charge q1.
Draw vector F1 on the figure.
Sol: “−“ charge q is attracted by + charge q1
F1 
F2
A
q = −0.5μC
+x
0.5m
k q q1
2
1
r
(8.99 109 N  m 2 / C 2 )(0.5 106 C )(5 106 C )

 0.09 N
(0.5m) 2
r1
F1
q1
+5.00μC
r2
q2
−2.60μC
(b) Determine the magnitude and direction of the electric force F2 on charge q by charge q2. Draw vector F2 on
the figure.
Sol: the straight-line distance between q and q2 is: 𝑟2 = √2 × 0.5𝑚 = 0.71𝑚
“−“ charge q is repelled by “-“ charge q2
F2 
k q q2
r22
(8.99 109 N  m 2 / C 2 )(0.5 106 C )(2.6  106 C )
2
 0.023N
(0.71m) 2
(c) Draw the NET electric force vector Fnet  F1  F2 using the graphical method on the figure.
+y
F2
135o
+x
F1 Fnet
F2
F1
r2
(d) Determine the magnitude and direction of the NET electric force Fnet on charge q by charges q1 and q2.
Sol: F1x  (0.09 N ) cos(270o )  0
F1 y  (0.09N )sin(270o )  0.09 N
F2 y  (0.023N )sin(135o )  0.0163N
F2 x  (0.023 N ) cos(135o )  0.0163 N
The x component of the vector Fnet  F1  F2 is： Fnet , x  F1x  F2 x  0  (0.0163)  0.0163N
Similarly, the y component is
Fnet , y  F1 y  F2 y  0.09  (0.0163)  0.0737 N
The magnitude of vector Fnet is
2
2
2
2
Fnet  Fnet
, x  Fnet , y  (0.0163N )  (0.0737 N )  0.075 N
1
Direction of Fnet is
 Fnet , y
tan 1 
F
 net , x

1  0.0737 N 
o
  tan 
  77

0.0163
N



180o + 77o = 257o counterclockwise from the +x-axis.
Fnet is in 3rd quadrant