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Chapter 19Electric Force and Electric Field
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Interaction of Electric Charge
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Charging an object A glass rod is rubbed with
silk
Electrons are transferredfrom the glass to the silk
Each electron adds a
negative charge to the silk
An equal positive charge isleft on the rod
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Charge
Unit:
C, Coulomb
+
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Electric ChargeElectric charge is one of the fundamental attributes of the
particles of which matter is made.
qproton e
qelectron e
e
1.6
10
19
C
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Electric Field
Like charges (++)
Electric dipole:
Opposite charges (+)
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Vectors are arrows
i
j
v1 2
i
j
v3 i j
v2
1.5i 0.5 j
v4 2i j
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What are these vectors?
i
jv1
2 i
v3 i 2 j
v2i 1.5 j
v4 3
i 2
j
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Magni tudeof a vector
= Length of the arrow
4
3
v 4 i 3 j
v 42 32 5
If v ai bj
v
v a2
b2
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What are the magnitudes?
i
jv1
2 i
v3 i 2 j
v2i 1.5 j
v4 3
i 2
j
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Magnitudes (solution)
i
jv1
2irv
1
2
v3 i 2 j
rv
3
(1)2 (2)2
12 22 5 2.24
v2i 1.5 j
rv2 1
2 1.52 1.80
v4 3i 2 j
r
v4 32
22
13 3.61
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Adding and subtracting vectors
u 2i 9
j
rv 5i 7.2 j
ru
rv 7i 16.2 j
ru
rv 3i 1.8 j
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Add and subtract
i
jv1 2i
v3 i 2 j
v2i 1.5 j
v4
3
i
2
j
Find:rv1
rv2 ,
rv1
rv2 ,
rv4
rv3
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v1 2i
v3 i 2 j
v4 3i 2 j
Find v1 v2 , v1 v2 , v4 v3
Solution
v2i 1.5 j
Very important!!!
rv1rv2
rv1 rv2
rv4
rv3
rv4
rv3
v1
v2 3i 1.5 j
r
v1
r
v2 (3)2
(1.5)2
32
1.52
3.35
rv4
rv3 (3i 2 j) (i 2 j)
3i 2 j i 2 j 2i 4 j
rv4
rv3 (2)
2 (4)2 22 42 4.47
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Notations
The following are all common notations of a vector:rv,v,v
The following are common notations of the magnitude
(i.e. the length of the arrow):
v,rv
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Vector Components
5
4
-3
i
j
v 4 i 3 j
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Terminology
v 4
i 3
j
Thex-component ofrv is 4, not 4 i .
They-component of
r
v is -3, not -3
j.
Usually written as:
vx 4,vy 3
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Decomposing a vector
v
Given v 7, 40o,
what are thex,ycomponents ofrv?
vx
vy
It means finding vx ,vy !
Hint: Once you know
one side of a right-
angle triangle and
one other angle, you
can find all the
lengths using cos,sin or tan.
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Trigonometry
cosb
h
sin ah
tana
b
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Solution
v 7
Given v 7, 40o,
what are thex,ycomponents ofrv?
vx v cos
7cos40o 5.36
vy v sin
7sin40o
4.50
v 5.36i 4.5j
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CheckOn the other hand, given
rv 5.36i 4.50 j,
you can deduce rv and .
v
vx
vy
v 5.362
4.52
7 (as expected)
tan
opposite
adjacent
vy
vx
4.5
5.36 0.8396
tan1(0.8396) 40o (as expected)
A l f
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Angles of a vector
x
Find the angles the four vectors make with the positivex-axis.
1 30o
2 180o 30o 150o
3 180o 30o 210oor 3 (180
o 30o) 150o
4 360o 30o 330oor 4 30
o
y
v1
v4
v2
v3
30
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Calculating the angles
v ai bj
tan1(b
a) , where
0oif a 0
180 oif a 0
Examples :
rv 2i 3j tan1(32
) 0o 56.3o
rv 2i 3j tan1(
3
2) 180o 123.7o
r
v 2
i 3
j tan1
(
3
2) 180o
236.3o
rv 2i 3j tan1(
3
2) 0o 56.3o( 360o 56.3o 303.7o)
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(-1) times a vector?
5
4
3 i
jv 4
i 3
j
i
j
What is - v?
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v 4i 3
j vv 4i 3
j
Points in the OPPOSITE direction!
What is - v?
5
4
3
v 4
i 3
j5
4
3v
v
4
i
3
j
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In General
If this is v
This is -v
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Adding Vectors Diagrammatically
u
vu v
v
You are allowed to move an arrow around as
long as you do not change its direction andlength.
Method for adding vectors:
1. Move the arrows until
the tail of one arrow is at
the tip of the other
arrow.
2. Trace out the resultant
arrow.
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Addition of vectors
You are allowed to move an arrow around as
long as you do not change its direction.
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Adding in a different order
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Order does not matter
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Subtracting Vectors Diagrammatically
u
v
u
v
v
u v
u v
What about uv?
v
u v u (v)
If we know (rv) we know
ru
rv
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Example
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Example
What isA B C?
A
C
B
D
A B C
AB
C
D
Addi t 1
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Adding vectors 1Add the three vectors to find the
total displacement.
s1 2.6 jrs2 4i
s3
3.1cos(45
o
)
i
3.1sin(45
o
)
j
2.19
i
2.19
j
stotal
s1
s2
s3 (4 2.19)i (2.6 2.19) j 6.19i 4.79j
or more precisely:r
stotal (6.19
i 4.79
j)km
Adding Vectors 2
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Adding Vectors 2
FindA 2
B.
A 2.8cos(60o
)
i 2.8sin(60o
)
j 1.40
i 2.42
jrB 1.9cos(60o)i 1.9sin(60o) j 0.95i 1.65j
2
B 2(0.95i 1.65 j) 1.90i 3.30 j
r
A 2r
B (1.40 1.90)
i (2.42 3.30)
j
0.50i 5.72 j
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Vector Notation of E field
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Vector Notation of E field
r: Unit vector pointing from the charge to the observer.
r 1
q
r
rE
q
40r2r
Charges produce electric field. The closer you are to the
charge, the stronger is the electric field.
Unit: V/m = N/C
0
8.85
10
12
C
2
N
1
m
2
: Permittivity
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Store k in your calculator
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Store kin your calculator
Type:
8.99E9 then STO
then ALPHA then K
then ENTER
If q=2C, r=1.3m, to
find the Efield, type:K*2/1.32
Electric Field (Magnitude)
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Electric Field (Magnitude)The magnitudeof the electric field produced by a single
point charge qis give by:
Eq
40
r2
Dont forget the absolute value!
Magnitude is alwayspositive.
Warnings
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WarningsDo not confuse the displacement vector rwithE.
r(blue arrow) points from the charge to the observer.rE(red arrow) is ALWAYS drawn with its tail at the obeserver,
and can point either away from OR toward the charge.
+Observerr E
r E
rec on o one
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rec on o onecharge)
If q 0,Eis in the SAME direction as r.
If q 0,r
Eis in the OPPOSITE direction as r.
rE
q
40r2r
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What is
r?
Example
q1 1nC
Observer atP.rrP1is the displacement vector from q1toP.
rP1is the corresponding unit vector.
rP1 1i 0.5 jrrP1 1
2 0.52 1.12
rP11i 0.5 j
1.12 0.89i 0.45 j
rP1
rP1rrP1
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Example (Continued)
The electric field vector is given by the red
arrow.
q1 1nCr
E1 q1
40rP12 rP1
109
40(1.12)2
(0.89i 0.45 j)
(6.38i 3.22j)V/m
f f
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The strategy in finding the electric field vector
1. Draw an arrow from the charge to the observer
2. Write down the vector rrand its magnitude rr
3. Calculate r
rrrr
4. CalculaterE q
40r2r
5. If there are more than one charge, repeat for each one
6.rEtotal
rE1
rE2L
rEN
Fi d th it t
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Find the unit vectors
A
B
C
D
E
q1
Find :r
rA1rrA1
rA1
then do the same
for the other point
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Warnings
Do not confuse the displacement vector rwithE.
rr(blue arrow) points from the charge to the observer, and is
used to calculate
r.rE(red arrow) is ALWAYS drawn with its tail at the obeserver.
q1 1nC
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Example - Two Charges
See supplementary notes
-1nC
+1nC
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Example
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Example
4 cm 3 cm
q1 q2P
q1 1 108C, q2 2 10
8C
Find the E field at point P
S l ti Pr1 r2
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Solution
4 cm 3 cm
q1 q2P
q1 1 108C, q2 2 10
8CE1
q1
40r12r1
q1
40r12i
(1 108 )
4(8.85 1012 )(0.04)2
i (56.2
i )kV/ m
E1E2
E2q2
40r22r2
q2
40r22
(i )
(2 10
8 )
4(8.85 1012 )(0.03)2(i ) (199.8i )kV/ m
1 r2
r1i , r2
i
EtotalE1E2 (56.2i 199.8i )kV/ m (256.0i )kV/ m
Example
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Example
x 7-x
q1 q2P
q1 1 108C, q2 2 10
8C
Find the point P such that E = 0.
7 cm
xamp q1 1 108C, q2 2 108C
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xampe
x 7-x
q1 q2P
q1 , q2
q1
40x2
q2
40 (7 x)2
q1
x2
q2
(7 x)2
(7 x)2
x2 q
2
q1 7 x
x q
2
q1 2
7 x 2x
x 71 2
16.9cm(rejected) or 2.9cm
Th diff b t
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The difference between
field vectors and field lines
Field vectors Field lines
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Properties of field lines
Field lines never cross each other
Field lines never terminate in vacuum
Field lines originate from positive charges andterminate 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
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Dipole
Field vectors Field lines
Similar to this
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Similar to this
You connects the field vectors tofind the field lines.
Electric Field and Electric Force
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Electric Field and Electric Force
Electric field can be used to calculate the electric forc
F q E
FandEare parallel when qis positive.
FandEare opposite when qis negative.
Example :
A chargeq 2Cin an electric fieldrE (2i 3j)N/Cwill feel a force:
rF (2C)(2i 3j)N/C (4i 6 j)N
Two point charges
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Two point charges
q1 q2r
E21E12
The E field (magnitude) at
point 2 due to charge 1:
E21q1
40r2
The force (magnitude) on
charge 2 due to charge 1:
F21 q2E21 q1q2
40r2
The E field (magnitude) at
point 1 due to charge 2:
E12q2
40r2
The force (magnitude) on
charge 1 due to charge 2:
F12 q1E12 q1q2
40r2
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Another Notation
Fq1q2
40r2
F kq1q2
r2
k1
40 8.99 109Nm2C2
Finding the Electric Force
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Finding the Electric Force
q1
q2
q3
There are two (equivalent) methods of
finding the force on a charge (say, q1).
Method 1 (usingEfield - recommended) :
Find the electric field at point 1 due to the other two chargesrE1
rE12
rE13 (Field at point 1 due to q2 and q3)
rF1 q1
rE1
Method 2 (using Coulomb's Law):
Find the forcevectorson charge 1 due to the other two charges using Coulomb's LawrF1
rF12
rF13 (Forces at point 1 due toq2 and q3)
Note that you must first write the forces as vectors,cannotadd the magnitude instearF1
rF12
rF13
n ng e orce on a
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n ng e orce on acharge
At pointP:rE (9.60i 3.16 j)N/C
P
-1nC
+1nCIf a charge q3 2nCis placed at point P,
the force on it will be given by :rF3 q3
rE (2nC)(9.60i 3.16 j)N/C
(19.20i 6.32 j) 109N
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