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The Electric Field
January 13, 2006
Calendar
 Quiz Today
 Continuation with Coulomb’s Law and
the concept of the Electric Field
 MLK Holiday on Monday
 Next week we continue with same
topics – see the schedule on the
website
 Quiz on Friday
Last Time
 We found two kinds of charge.
 Like charges repel, unlike charges
attract.
 Discussed induction … a bit
 Fundamental unit of charge is the
COULOMB.
 Coulomb’s Law
 Assignment: read text about
induction.
One example of induction
Polarize
Ground
Remove Ground
Positive !
Coulomb’s Law
The Unit of Charge is called
THE COULOMB
1
q1q2
F
runit
2
40 r
1
40
 k  9 x109 Nm 2 / C 2
Smallest Charge: e ( a positive number)
1.6 x 10-19 Coul.
electron charge = -e
Proton charge = +e
What would be the magnitude of the electrostatic force
between two 0.500 C charges separated by the following
distances, if such point charges existed (they do not) and this
configuration could be set up?
(a) 1.25 m
__________N
(b) 1.25 km
__________N
Three point charges are located at the corners of an
equilateral triangle as shown in Figure P23.7. Calculate the
resultant electric force on the 7.00-μC charge.
Two small beads having positive charges 3q and q are fixed at
the opposite ends of a horizontal, insulating rod, extending
from the origin to the point x = d. As shown in Figure P23.10,
a third small charged bead is free to slide on the rod. At what
position is the third bead in equilibrium? Can it be in stable
equilibrium?
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 ……
Nothing Happens! The Green thing is a Force
Field!
The Force may not be with you ….
Side View
The
FORCE FIELD
Force
Big!
|Force|
o
Position
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.
EXAMPLE: The Gravitational
Field That We Live In.
M
m
mg
Mg
Mysterious Force
F
Electric Field
 If a charge Q is in an electric field E
then it will experience a force F.
 The Electric Field is defined as the
force per unit charge at the point.
 Electric fields are caused by charges
and consequently we can use
Coulombs law to calculate it.
 For multiple charges, add the fields as
VECTORS.
Two Charges
F  1  qq0
q
E     k 2 runit  k 2 runit
q0  q0  r
r
Doing it
Q
qQ
F  k
runit
2
r
F
Q
E 
 k 2 runit
q
r
F
q
A Charge
r
The spot where we want
to know the Electric Field
GeneralqQ
F  k 2 runit
r
F
Q
E
 k 2 runit
q
r
General
E  E j  
Fj
q
 k
Qj
r
2
j
r j ,unit
Force  Field
Two Charges
What is the Electric Field at Point P?
The two S’s
•Superposition
•Symmetry
What is the electric field at the
center of the square array?
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= 2rdr for elemental ring of charge
 Volume
 r=charge per unit volume
 dq=rdV
 dV=dxdydz or 4r2dr or some other expressions we
will look at later.
Continuous Charge Distribution
ymmetry
Let’s Do it Real Time
Concept – Charge per
unit length m
dq= m ds
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
2k
Ex 
r
0
2k
0 cos( )d  r sin(  0 )
A Harder Problem
setup
dE

dEy

r
x
A line of charge
m=charge/length
L
dx
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
(r 2  x 2 )3 / 2
(standard integral)
Completing the Math
Doing the integratio n :
kLm
Ex 
2


L
2
r r   
 4 
In the limit of a VERY long line :
L
 L2 
r   
 4 
kLm
2km
Ex 

r
L
r 
2
2
1/r dependence
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??
The Geometry
Define surface charge density
s=charge/unit-area
(z2+r2)1/2
dq=sdA
dA=2rdr
dq=s x dA = 2srdr
dq cos( ) k 2rsdr
z
dE z  k

2
2
z r
z2  r2 z2  r2

(z2+r2)1/2



R

E z  2ksz 
0
z
rdr
2
r

2 3/ 2

1/ 2
Final Result
(z2+r2)1/2
 s 
z
Ez  
 2 

1 
z 2  R2
0 

When R  ,
s
Ez 
2 0




Look at the “Field Lines”
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”.
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.
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= 2rdr for elemental ring of charge
 Volume
 r=charge per unit volume
 dq=rdV
 dV=dxdydz or 4r2dr or some other expressions we
will look at later.
The Sphere
dq
r
thk=dr
dq=rdV=r x surface area x thickness
=r x 4r2 x dr
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 )
r2
 k
sdA(r )
r2
 k
mds (r )
r2
(Note: I left off the unit vectors in the last
equation set, but be aware that they should
be there.)
To be remembered …
 If the ELECTRIC FIELD at a point is E, then
 E=F/q (This is the definition!)
advanced
 Using some
mathematics we can
derive from this equation, the fact that:
F  qE