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Lecture 04
The Electric Field
Chapter 22 - HRW
Week of February
17 05
Electric Field
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Physics 2049 News
WebAssign was due today
 Another one is posted for
Friday
 You should be reading chapter
22; The Electric Field.

 This
is a very important concept.
 It is a little “mathy”

There will be a QUIZ on Friday.
 Material
from chapters 21-22.
 Studying Works!
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This is WAR
Ming the
merciless
this guy is
MEAN!



You are fighting the enemy on the
planet Mongo.
The evil emperor Ming’s forces are
behind a strange green haze.
You aim your blaster and fire …
but ……
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Nothing Happens! The Green
thing is a Force Field!
The Force may not be with you ….
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Side View
The
FORCE FIELD
Force
Big!
|Force|
o
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Position
Electric Field
5
Properties of a FORCE
FIELD
It is a property of the position in
space.
 There is a cause but that
cause may not be known.
 The force on an object is
usually proportional to some
property of an object which is
placed into the field.

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EXAMPLE: The
Gravitational Field That
We Live In.
M
m
mg
Mg
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The gravitational field:
g




The gravitational field strength is
defined as the Force per unit
mass that the field creates on an
object
This becomes
g=(F/m)=(mg/m)=g
The field strength is a VECTOR.
For this case, the gravitational
field is constant.
 magnitude=g
(9.8 m/s)
 direction= down
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Final Comment on
Gravitational Field:
Force
mg
g

g
unit _ mass m
Even though we know what is causing
the force, we really don’t usually think
about it.
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Newton’s Law of
Gravitation
m
R
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MEarth
10
The Calculation
mM
FG
2
REarth
F
M
g G
2
m
REarth
g  6.67 x10
11
24
6 x10

6 2
(6.4 x10 )

g  9.77 m / s
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
11
Not quite correct ….
Earth and the Moon (in
background), seen from space)
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More better …
Moon
Fmoon
m
mg
FEarth
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MEarth
13
To be more precise …

g is caused by
 Earth
(MAJOR)
 moon (small)
 Sun (smaller yet)
 Mongo (extremely teeny tiny)

g is therefore a function of
position on the Earth and even
on the time of the year or day.
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The Electric Field E

In a SIMILAR WAY
 We
DEFINE the ELECTRIC FIELD
STRENGTH AS BEING THE FORCE
PER UNIT CHARGE.
 Place a charge q at a point in space.
 Measure (or sense) the force on the
charge – F
 Calculate the Electric Field by
dividing the Force by the charge,
F  qE
F  Newtons 
E 
q  Coulomb 
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Electric Field Near a Charge
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Two (+) Charges
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Two Opposite Charges
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A First Calculation
Q
A Charge
r
q
The spot where we want
to know the Electric Field
Place a “test charge
at the point and
measure the Force on it.
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Doing it
Q
qQ
F  k 2 runit
r
F
Q
E   k 2 runit
q
r
F
q
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A Charge
r
The spot where we want
to know the Electric Field
Electric Field
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GeneralqQ
F  k 2 runit
r
F
Q
E   k 2 runit
q
r
General
E  E j  
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Fj
q
 k
Electric Field
Qj
r
2
j
r j ,unit
22
Continuous Charge Distribution
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ymmetry
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Let’s Do it Real Time
Concept – Charge per
unit length m
dq= m ds
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The math
ds  rd
Ey  0
Why?
0
dq
E x  (2)  k 2 cos( )
r
0
0
rd
E x  (2)  k 2 cos( )
r
0
0
2k
2k
Ex 
cos( )d 
sin(  0 )

r 0
r
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A Harder Problem setup
dE
 dE
y

r
x
A line of charge
m=charge/length
dx
L
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Ex  k
L
2

mdx cos( )
(r  x )
2
L

2
r
cos( ) 
(r  x )
2
L/2
E x  2k
2

0
E x  2krm
2
rmdx
2
2 3/ 2
(r  x )
L/2

0
dx
2
2 3/ 2
(r  x )
(standard integral)
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Completing the Math
Doing the integratio n :
kLm
Ex 
2


L
2
r r   
 4
In the limit of a VERY long line :
L
2


L
2
r   
 4
klm
2km
Ex 

r
L
r 
2
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Electric Field
1/r dependence
29
Dare we project this??
Point Charge goes as 1/r2
 Infinite line of charge goes as
1/r1
 Could it be possible that the
field of an infinite plane of
charge could go as 1/r0? A
constant??

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The Geometry
Define surface charge density
s=charge/unit-area
(z2+r2)1/2
dq=sdA
dA=2prdr
dq=s x dA = 2psrdr
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dE z  k
(z2+r2)1/2

dq cos( ) k 2prsdr
z

z2  r2
z2  r2 z2  r2

 
R
E z  2pksz 
0
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
Electric Field
z
rdr
2
r

1/ 2

2 3/ 2
32
Final Result
(z2+r2)1/2
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 s 
z
1 
E z  
2
2
2

z R
 0 
When R  ,




s
Ez 
2 0
Electric Field
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Look at the “Field Lines”
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What did we learn in
this chapter??

We introduced the concept of
the Electric FIELD.
 We
may not know what causes
the field. (The evil Emperor
Ming)
 If we know where all the charges
are we can CALCULATE E.
 E is a VECTOR.
 The equation for E is the same
as for the force on a charge from
Coulomb’s Law but divided by
the “q of the test charge”.
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What else did we learn
in this chapter?
We introduced continuous
distributions of charge rather
than individual discrete
charges.
 Instead of adding the individual
charges we must INTEGRATE
the (dq)s.
 There are three kinds of
continuously distributed
charges.

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Kinds of continuously
distributed charges

Line of charge
m
or sometimes l = the charge per unit
length.
 dq=mds (ds= differential of length
along the line)

Area
s
= charge per unit area
 dq=sdA
 dA = dxdy (rectangular coordinates)
 dA= 2prdr for elemental ring of charge

Volume
 r=charge
per unit volume
 dq=rdV
or 4pr2dr or some other
expressions we will look at later.
 dV=dxdydz
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The Sphere
dq
r
thk=dr
dq=rdV=r x surface area x thickness
=r x 4pr2 x dr
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Summary
qQ
F  k 2 runit
r
F
Q
E   k 2 runit
q
r
General
Fj
Qj
E   E j     k 2 r j ,unit
q
rj
E  k
rdV (r )
r
2
 k
sdA(r )
r
2
 k
mds(r )
r
2
(Note: I left off the unit vectors in the last
equation set, but be aware that they should
be there.)
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To be remembered …



If the ELECTRIC FIELD at a point
is E, then
E=F/q (This is the definition!)
Using some advanced
mathematics we can derive from
this equation, the fact that:
F  qE
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Example:
The electric field in a region of space
is given by the expression :
E  3x 2
What force would a 0.5 coulomb
charge experience if it is placed
at the coordinate (2,8)?
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Solution
This is an easy one!
E  3x  3  4  12( N / C )
The y coordinate doesn' t matter.
The force  qE  0.5 C 12 (N/C)
or
F  6 Newtons
2
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In the Figure, particle 1 of charge q1 = -9.00q and
particle 2 of charge q2 = +2.00q are fixed to an x
axis.
(a) As a multiple of distance L, at what coordinate
on the axis is the net electric field of the particles
zero?
[1.89] L
(b) Plot the strength of the electric field as a
function of position (z).
q1 = -9q
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q2=+2q
Electric Field
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Let’s do it backwards…
Take an arbitrary position x which
for the heck of it has a coordinate
that is greater th an L. Then ...
 2q
9q 
E ( x)  k 
 2 
2
x 
 ( x  L)
Let x  αL

2
9 
E ( x)  qk 
 2 2 
2
 (L  L)  L 
qk  2
9 
E ( x)  2 
 2 
2
L  (  1)  
We can look at this as a function
of  and just plot the quantity in
brackets. Let' s use EXCEL for this.
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EXCEL
a
First Term
Second Term
Sum
-3
0.13
-1.00
-0.88
-2.9
0.13
-1.07
-0.94
-2.8
0.14
-1.15
-1.01
-2.7
0.15
-1.23
-1.09
-2.6
0.15
-1.33
-1.18
-2.5
0.16
-1.44
-1.28
-2.4
0.17
-1.56
-1.39
-2.3
0.18
-1.70
-1.52
-2.2
0.20
-1.86
-1.66
-2.1
0.21
-2.04
-1.83
-2
0.22
-2.25
-2.03
-1.9
0.24
-2.49
-2.26
-1.8
0.26
-2.78
-2.52
ETC ….
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Bracket
100.00
50.00
??
-2
-1.8
-1.6
-1.4
-1.2
-1
-0.8
-0.6
-0.4
0.00
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
-50.00
-100.00
alpha
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alpha=1.89
46
The mystery solved!!!
For x  L
 2q
9q 
E ( x)  k 
 2 
2
x 
 ( x  L)
For x  L
  2q
9q 
E ( x)  k 
 2 
2
x 
 ( x  L)
For x  0
  2q
9q 
E ( x)  k 
 2 
2
x 
 ( x  L)
.
Week of February
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BE CAREFULL!
Electric Field
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In the Figure, the four particles are fixed in
place and have charges q1 = q2 = +5e, q3 =
+3e, and q4 = -12e. Distance d = 9.0 mm. What
is the magnitude of the net electric field at point
P due to the particles?
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Figure 22-34 shows two charged particles on an
x axis, q = -3.20 10-19 C at x = -4.20 m and q =
+3.20 10-19 C at x = +4.20 m.
(a) What is the magnitude of the net electric field
produced at point P at y = -5.60 m?
[7.05e-11] N/C
(b) What is its direction?
[180]° (counterclockwise from the positive x
axis)
Week of February
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Figure 22-40 shows two parallel nonconducting rings
arranged with their central axes along a common line.
Ring 1 has uniform charge q1 and radius R; ring 2 has
uniform charge q2 and the same radius R. The rings
are separated by a distance d = 3.00R. The net
electric field at point P on the common line, at
distance R from ring 1, is zero. What is the ratio
q1/q2?
[0.506]
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In the Figure, eight charged particles form a square
array; charge q = +e and distance d = 1.8 cm. What
are the magnitude and direction of the net electric
field at the center?
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