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Chapter 21
Electric Charge and Electric Field
Interaction of Electric Charge
Charging an object
• A glass rod is rubbed with silk
• Electrons are transferred from
the glass to the silk
• Each electron adds a negative
charge to the silk
• An equal positive charge is left
on the rod
Charge
+
Unit:
C, Coulomb
−
Electric Charge
Electric charge is one of the fundamental attributes of the particles of
which matter is made.
q proton  e
qelectron  e
e  1.6  10 19 C

Electric Field
Like charges (++)
Electric dipole:
Opposite charges (+−)
Vectors are arrows
ĵ
v1  2iˆ  ĵ
iˆ
v2  1.5iˆ  0.5 ĵ
v3   iˆ  ĵ
v4  2iˆ  ĵ
What are these vectors?
v1  2iˆ
ĵ
iˆ
v2  iˆ  1.5 ĵ
v4  3iˆ  2 ĵ
v3   iˆ  2 ĵ
Magnitude of a vector
= Length of the arrow
v  4 iˆ  3 ĵ
3
v  4 3 5
2
2
r
If v  aiˆ  bĵ
v
2
2
 v  a b
4
What are the magnitudes?
v1  2iˆ
ĵ
iˆ
v2  iˆ  1.5 ĵ
v4  3iˆ  2 ĵ
v3   iˆ  2 ĵ
Magnitudes (solution)
r
ˆ
v1  2i  v1  2
ĵ
iˆ
v2  iˆ  1.5 ĵ
r
 v2  12  1.5 2  1.80
v3  iˆ  2 ĵ
r
 v3  (1)2  (2)2
 12  2 2  5  2.24
v4  3iˆ  2 ĵ
r
 v4  32  2 2
 13  3.61
Adding and subtracting vectors
u  2iˆ  9 ĵ
r
v  5iˆ  7.2 ĵ
r r
u  v  7iˆ  16.2 ĵ
r r
u  v  3iˆ  1.8 ĵ
Add and subtract
v1  2iˆ
Find:
r r r r r r
v1  v2 , v1  v2 , v4  v3
ĵ
iˆ
v2  iˆ  1.5 ĵ
v4  3iˆ  2 ĵ
v3   iˆ  2 ĵ
Solution
v1  2iˆ
v  iˆ  1.5 ĵ
r r r r r r
Find v1  v2 , v1  v2 , v4  v3
2
v3   iˆ  2 ĵ
v4  3iˆ  2 ĵ
Very important!!!
r r
r
r
v1  v2  v1  v2
r r
r
r
v4  v3  v4  v 3
r r
v1  v 2  3iˆ  1.5 ˆj
r r
v1  v 2  (3) 2  (1.5) 2  32  1.5 2  3.35
r
r
v 4  v 3  (3iˆ  2 ˆj )  ( iˆ  2 ˆj )
 3iˆ  2 ˆj  iˆ  2 ˆj  2iˆ  4 ˆj
r
r
v 4  v 3  (2) 2  (4) 2  2 2  4 2  4.47
Notations
The following are all common notations of a vector:
r
v, v,v
The following are common notations of the magnitude
(i.e. the length of the arrow):
r
v, v
Vector Components
ĵ
4
5
-3
ˆi
ˆ
v  4 i  3 ĵ
Terminology
v  4 iˆ  3 ĵ
r
The x-component of v is 4, not 4iˆ.
r
The y-component of v is -3, not -3 ĵ.
Usually written as:
vx  4, vy  3
Decomposing a vector
r
Given v  7,   40o,
r
what are the x, y components of v?
Hint: Once you know
one side of a rightangle triangle and one
other angle, you can
find all the lengths
using cos, sin or tan.

v

vy
vx
It means finding vx ,vy !
A quick reminder
h



hcos

hsin 
Trigonometry
b
cos  
h
a
sin  
h
a
tan  
b
Solution
r
Given v  7,   40o,
r
what are the x, y components of v?
r
vy  v sin 
v 7
 7sin 40  4.50
o

r
vx  v cos
 7 cos 40  5.36
o
r
 v  5.36iˆ  4.5 ˆj
Check
v
r
On the other hand, given v  5.36iˆ  4.50 ĵ,
r
you can deduce v and  .
vy

v  5.36  4.5  7 (as expected)
2
2
opposite vy 4.5
tan  
 
 0.8396
adjacent vx 5.36
   tan 1 (0.8396)  40o (as expected)
vx
Angles of a vector
Find the angles the four vectors make with the positive x-axis.
y
r
v2
r
v1
1  30 o
30°

2  180 o  30 o  150 o
r
v3

3  180 o  30 o  210 o or 3  (180 o  30 o)  150 o

4  360  30  330 or 4  30
o
o
o
o

x
r
v4


Calculating the angles
r
v  aiˆ  bˆj
o

0
if a  0
b
   tan 1 ( )   ,where    o
a
180 if a  0
Examples :
3
r
v  2iˆ  3 ˆj    tan 1 ( )  0 o  56.3o
2
3
r
v  2iˆ  3 ˆj    tan 1 ( )  180 o  123.7 o
2
3
r
v  2iˆ  3 ˆj    tan 1 ( )  180 o  236.3o
2
3
r
v  2iˆ  3 ˆj    tan 1 ( )  0 o  56.3o( 360 o  56.3o  303.7 o)
2
Write down the following three vectors in i j notation.
Find the sum of these vectors also.
v2  4 sin 50oiˆ  4 cos 50o ĵ
 3.1iˆ  2.6 ĵ
5
4
50o
10o
4.5
v3  4.5 cos10oiˆ  4.5sin10o ĵ
 4.4iˆ  0.8 ĵ
60o
v1  5 cos 60oiˆ  5sin 60o ĵ
 2.5iˆ  4.3 ĵ
vtotal  1.2iˆ  0.9 ĵ
(-1) times a vector?
ĵ
v  4 iˆ  3 ĵ
5
3
4
v
What is - v ?
ˆi
ˆ
i
 ĵ
v
What is - v ?
v
ˆ
v  4 i  3 ĵ   v  4 iˆ  3 ĵ
Points in the OPPOSITE direction!
4
3
v
 v  4 iˆ  3 ĵ
5
5
ˆ
v  4 i  3 ĵ
4
3
In General
r
If this is v
r
This is -v
Adding Vectors Diagrammatically
You are allowed to move an arrow around as long
as you do not change its direction and length.
Method for adding vectors:
v
r
uv
u
v
1. Move the arrows until the
tail of one arrow is at the
tip of the other arrow.
2. Trace out the resultant
arrow.
Addition of vectors
You are allowed to move an arrow around as long
as you do not change its direction.
Adding in a different order
Order does not matter
Subtracting Vectors Diagrammatically
r r
What about u  v ?
v
r
uv
u
v
r r
r
u  v  u  ( v)
r
r r
If we know ( v) we know u  v
v
r
v
u
r
uv
Example
Example
r r r
What is A  B  C?
C
B
r r
r
A  B  C  D
A
D
r r
A BC
Adding vectors 1
Add the three vectors to find the
total displacement.
r
s1  2.6 ˆj
r
s2  4 iˆ
r
s3  3.1cos(45 o)iˆ  3.1sin( 45 o) ˆj  2.19iˆ  2.19 ˆj
r
r r r
stotal  s1  s2  s3  (4  2.19)iˆ  (2.6  2.19) ˆj  6.19iˆ  4.79 ˆj
or more precisely :
r
stotal  (6.19iˆ  4.79 ˆj )km
Adding Vectors 2
r
r
Find A  2 B.
r
A  2.8cos(60 o)iˆ  2.8sin( 60 o) ˆj  1.40iˆ  2.42 ˆj
r
B  1.9cos(60 o)iˆ  1.9sin( 60 o) ˆj  0.95iˆ  1.65 ˆj
r
2B  2(0.95iˆ  1.65 ˆj )  1.90iˆ  3.30 ˆj
r
r
 A  2B  (1.40  1.90)iˆ  (2.42  3.30) ˆj
 0.50iˆ  5.72 ˆj
Electric field is a vector
The direction of the electric field is given below:
Vector Notation of E field
Charges produce electric field. The closer you are to the charge, the
stronger is the electric field.
Unit: V/m = N/C
r
E
q
4 0 r
2
rˆ
 0  8.85  1012 C 2 N 1m2 : Permittivity
r̂ : Unit vector pointing from the charge to the observer.

r̂  1
r̂
q
Another Notation
r
E
k

q
q
kq
ˆ
ˆ
ˆ
2 r 
2 r 
2 r
4 0 r
4 0 r
r
1
4 0
1
 8.99  10 Nm C
9
2
2
Store k in your calculator
Type:
“8.99E9” then “STO”
then “ALPHA” then “K”
then “ENTER”
If q=2C, r=1.3m, to find
the E field, type:
“K*2/1.32”
Electric Field (Magnitude)
The magnitude of the electric field produced by a single point
charge q is give by:
E 
q
4 0 r
2
Don’t forget the absolute value!
Magnitude is always positive.
Warnings
r
Do not confuse the displacement vector r̂ with E.
r̂ (blue arrow) points from the charge to the observer.
r
E (red arrow) is ALWAYS drawn with its tail at the obeserver,
and can point either away from OR toward the charge.
+
r̂
−
r̂
Observer
E
E
Direction of E (one charge)
r
If q  0, E is in the SAME direction as r̂.
r
If q  0, E is in the OPPOSITE direction as r̂.

r
E
q
4 0 r
2
rˆ
Example
Observer at P.
r
rP1 is the displacement vector from q1 to P.
rˆP1 is the corresponding unit vector.
What is r̂?
q1  1nC
r
rP1
rˆP1  r
rP1


r
rP1  1iˆ  0.5 ˆj
r
rP1  12  0.5 2  1.12
1iˆ  0.5 ˆj
rˆP1 
 0.89iˆ  0.45 ˆj
1.12

Example (Continued)
r
E1 
q1
ˆ
2 rP1
4 0 rP1
q1  1nC
10 9
ˆ
ˆ

2 (0.89i  0.45 j )

4 0 (1.12)
 (6.38iˆ  3.22 ˆj )V /m
The electric field vector is given by the red arrow.
The strategy in finding the electric field vector
1. Draw an arrow from the charge to the observer
r
r
2. Write down the vector r and its magnitude r
r
r
3. Calculate r̂  r
r
r
q
4. Calculate E 
r̂
2
4 0 r
5. If there are more than one charge, repeat for each one
r
r
r
r
6. Etotal  E1  E2  L  EN
Find the unit vectors
Find :
r
rA1
r
rA1
rˆA1
A
B
C
then do the same
for the other points.


q1

D


E

Warnings
r
r
Do not confuse the displacement vector r with E.
r
r (blue arrow) points from the charge to the observer, and is
used to calculate r̂.
r
E (red arrow) is ALWAYS drawn with its tail at the obeserver.
q1  1nC

Example - Two Charges
See supplementary notes
More about r:
Sometimes it is easier to find r̂ using angles, for example:
r̂   cos iˆ  sin  ĵ

r̂  cos iˆ  sin  ĵ


r̂   sin  iˆ  cos ĵ
Example
Find the E field at point P
P
q1
4 cm
8
q2
3 cm
8
q1  1  10 C, q2  2  10 C
Solution
q1
P
r̂1
E1
E2
r̂1  iˆ, r̂2   iˆ
E1 
q1
4 0 r1
r̂ 
2 1
q1
4 0 r1
ˆ
i
2
r̂2
4 cm
q2
3 cm
q1  1  108 C, q2  2  108 C
(1  10 8 )
ˆ  (56.2iˆ )kV / m

i
4 (8.85  10 12 )(0.04)2
q2
q2
ˆ)
E2 
r̂

(
i
2
4 0 r2 2
4 0 r2 2
(2  10 8 )
ˆ )  (199.8iˆ )kV / m

(
i
4 (8.85  10 12 )(0.03)2
r
r
r
 Etotal  E1  E2  (56.2iˆ  199.8iˆ )kV / m  (256.0iˆ )kV / m
Example
Find the point P such that E = 0.
7 cm
q1
P
7-x
x
q1  1  108 C, q2  2  108 C
q2
Example
8
8
q1  1  10 C, q2  2  10 C
q1
q2
P
x
q1
q2
q1
q2

 2 
2
2
4 0 x
4 0 (7  x)
x
(7  x)2
(7  x)2 q2
7x
q2




 2
2
x
q1
x
q1
 7  x   2x
7
x
 16.9cm (rejected) or 2.9cm
1 2
7-x
The difference between field
vectors and field lines
Field vectors
Field lines
Properties of field lines
•Field lines never cross each other
•Field lines never terminate in vacuum
•Field lines originate from positive charges and
terminate at negative charge
•Field lines may go off to infinity
•The tangent of a field line gives the direction of the
E field at that particular point
Dipole
Field vectors
Field lines
Similar to this
You connects the field vectors to
find the field lines.
Electric Field and Electric Force
Electric field can be used to calculate the electric force:
r
r
F  qE
F and E are parallel when q is positive.
F and E are opposite when q is negative.
Example :

r
A charge q  2C in an electric field E  (2iˆ  3 ˆj )N /C will feel a force :
r
F  (2C)(2iˆ  3 ˆj )N /C  (4 iˆ  6 ˆj )N
Two point charges
q1
E12
The E field (magnitude) at
point 1 due to charge 2:
E12 
q2
4 0 r 2
The force (magnitude) on
charge 1 due to charge 2:
F12  q1E12
q1q2

4 0 r 2
r
q2
E21
The E field (magnitude) at
point 2 due to charge 1:
E21 
q1
4 0 r 2
The force (magnitude) on
charge 2 due to charge 1:
F21  q2 E21
q1q2

4 0 r 2
Coulomb’s Law
The mutual force due to two point charges has
magnitude:
q1q2
F
4 0 r 2
Another Notation
q1q2
F
4 0 r 2
k
1
4 0
q1q2
F k 2
r
 8.99  10 9 Nm 2C 2
Finding the Electric Force
There are two (equivalent) methods of finding
the force on a charge (say, q1).
q3
Method 1 (using E field - recommended) :
Find the electric field at point 1 due to the other two charges
r
r
r

E1  E12  E13 (Field at point 1 due to q2 and q3 )
q1
r
r
 F1  q1 E1
q2
Method 2 (using Coulomb' s Law) :
 two charges using Coulomb'

Find the force vectors on charge 1 due to the other
r
r
r
F1  F12  F13 (Forces at point 1 due to q2 and q3 )
Note that you must first write the forces as vectors,
r
r
r
F1  F12  F13
s Law
cannot add the magnitude instead
Finding the force on a charge
If we put a charge q3  2nC at point P, then the force is :
r
r
F  q3 E P  (2nC)(9.60iˆ  3.16 ˆj )N /C
 (19.20iˆ  6.32 ˆj )  10 9 N
r
F  19.2 2  6.32 2  20.21 10 9 N
+1nC
At point P :
r
E  (9.60iˆ  3.16 ˆj )N /C
-1nC
P

Motion of a charge in an E field
r
F  qE
r
r
 ma  qE
r q r
a E
m
r
Once you know E, you can find the acceleration.
From the acceleration you can deduce the motion
of the charge.
(perhaps using the equations of motion from Phys 270)
Line of Charge
a
dy
Linear charge density:
Q Q
 
l 2a
dQ   dy
x
x
cos   
r
y
sin   
r
y
tan  
x
r
y
r  x 2  y2

P

dEy
A rod of length l  2a, total charge Q.
Find E field at point P.
a
dE x
r
dE
x
x 2  y2
y
x 2  y2
Solution
r
dE 
dQ
dQ x ˆ y
ˆ
(cos  i  sin  ĵ) 
( 3 i  3 ĵ)
2
4 0 r
4 0 r
r
 dy
x
y
ˆ

( 2
i 2
ĵ)
2 3/2
2 3/2
4 0 (x  y )
(x  y )
 a
x
 Ex   dEx 
dy
2
2 3/2

4 0  a (x  y )
a

y
 Ey   dEy  
dy
2
2 3/2

4 0  a (x  y )
Solution (Cont.)
 a x
 a
x
Ex   dEx 
dy 
dy
3
2
2 3/2


4 0  a r
4 0  a (x  y )
x
Use: r 
, y  x tan 
cos 
 dy  x sec 2  d
  max
  max
max
max


3
2
 Ex 
cos  sec  d 
cos  d


4 0 x 
4 0 x 

(Q / 2a)
a
Q
1

sin  max 
(
)
2 0 x
2 0 x
4 0 x x 2  a 2
x 2  a2
Solution (Cont.)
 a
y
Ey   dEy  
dy
2
2 3/2

4 0  a (x  y )
 Ey  0
Because the integrand is an odd function.
Reminder:
f (x)   f (x) : odd function
f (x)   f (x) : even function
An odd function
Solution (Cont.)
Ex 
Q
1
, Ey  0
4 0 x x  a
r
Q
1
E
iˆ
4 0 x x 2  a 2
Special case:
2
2
As a    x 2  a 2  a
r
Q 1 ˆ
 ˆ
E
i
i
4 0 xa
2 0 x
You will see this again in Gauss' law.