Transcript Charge and Electric Field

Electrostatics
Chapter 23
Week-1-2
What’s Happening
• Clicker use will start on Friday
(maybe). We will use them today
informally.
• There will NOT be a quiz this
week.
• There WILL be a quiz a week
from Friday.
• WebAssigns are now active.
Get to work!!
Probable First Observation Electricity
Idiot!
If lightening had actually traveled down the kite
string, old Ben Franklin would have been toast!
Probably never happened, but good story!
A Quick Experiment
Experimental Procedure
The sequence of Experiments
1. Identify the two rods
2. Treat each rod
3. Bring one rod near to the other
4. PREDICT WHAT WILL HAPPEN
5. VOTE ON POSSIBILITIES
6. Observe what happens
7. Did you learn anything? What?
Pivot
Allowable Predictions
(Use your clicker if you have one.)
A. Rods will attract each other
B. Rods will repel each other
C. Nothing will happen
D. Something not listed above will happen
Experiment #1
motion
Rubber rod
Pivot
Rubber rod
A.
B.
C.
D.
Rods will attract each other
Rods will repel each other
Nothing will happen
Something not listed above will happen
Experiment #2
Rubber rubbed with
skin of dead rabbit
Pivot
Rubber rubbed with
skin of dead rabbit
A.
B.
C.
D.
Rods will attract each other
Rods will repel each other
Nothing will happen
Something not listed above will happen
The charges on the two rods are ..
A. Since we treated both rods in the same way, they
should be of the same type
B. ……. different types
C. I have no idea what you are asking for.
D. Leave me alone … I’m napping!
If you rubbed the rods longer
and/or harder, do you think the
effect that you see would be
Stronger
B. Weaker
C. The same
A.
If the two rods are brought closer together, the force
acting between them will get …
A. Stronger
B. Weaker
C. The same
Definition of sorts
We DEFINE the “stuff” that we put on the rods
by the rubbing process as CHARGE.
We will try to understand what charge is and
how it behaves.
We add to the properties of materials:
Mass
Charge
Experiment #3
Glass rubbed with wool
Pivot
Glass rubbed with wool
A.
B.
C.
D.
Rods will attract each other
Rods will repel each other
Nothing will happen
Something not listed above will happen
Experiment #4
Rubber rubbed with
skin of dead rabbit
Pivot
Glass rubbed with wool
1.
2.
3.
4.
Rods will attract each other
Rods will repel each other
Nothing will happen
Something not listed above will happen
What’s Going On?
 All of these effects involve rubbing two surfaces
together.
 Or pulling two surfaces apart.
 Something has “happened “to each of these objects.
 These objects have a new PROPERTY

Other properties are mass, color
 We call this NEW PROPERTY .……….
………CHARGE.
 There seems to be two types of charge.
We call these two types of charge
Positive
Negative
An object without either a (+) or (-) charge is referred
to as being
NEUTRAL.
Example - Tape
Separation
An Example
Effect of Charge
We have also observed that there
must be TWO kinds of charge.
 Call these two types


positive (+)
negative(-)
 We “define” the charge that winds up
on the rubber rod when rubbed by
the dead cat to be NEGATIVE.
 The charge on the glass rod or the dead cat
is consequently defined as POSITIVE.
Old Ben screwed up more than
once!!
++++++++++------------+++---++---+-++-
From whence this charge???
Easily Removed
+
Materials
 Two kinds of materials:

Insulators


Conductors



Electrons and Protons are tightly bound to their
positions. Hard to move them around.
Electrons are easily removed and moved around.
Electrons are said to be MOBILE charges.
There are other kinds of materials that we will
not discuss: semiconductors, semi-metals
Experiment #5
Rubber rubbed with
skin of dead rabbit
Pivot
Metal Rod
A.
B.
C.
D.
Rods will attract each other
Rods will repel each other
Nothing will happen
Something not listed above will happen
What about a charged rod and a
piece of wood??
A.
B.
C.
D.
Rods will attract each other
Rods will repel each other
Nothing will happen
Something not listed above will happen
Ways to charge an object
 Rubbing or bond breaking (same thing)
 Transfer



Direct transfer
Polarization
Induction
Neutral Object POLARIZATION
Positive charge attracts negative charges.
Rod becomes “polarized.
Negative end is closer to positive charge
Distance effect causes attraction.
Induction
Polarize
Ground
Remove Ground
Positive !
Balloon Physics
Same as before: Polarization
What happens when two
surfaces touch or rub?
Bonding!
The Triboelectric Series
When two of the following materials are rubbed together under
ordinary circumstances, the top listed material becomes
positively charged and the lower listed material becomes
negatively charged.
No! No!
No! No!
MORE POSITIVE
rabbit's fur
glass
mica
nylon
wool
cat's fur
silk
paper
cotton
wood
acrylic
cellophane tape
polystyrene
polyethylene
rubber balloon
saran wrap
MORE NEGATIVE
So far we have found?
 There are TWO types of charge.
 Positive
 Negative
 Like Charges Attract
 Un-Like charges repel
 The force between charges increases as they are
brought closer together.
 This charge separation results from chemical bonds
which are severed.
Forces Between Charges
Q1
Q2
Attract
+
+
+
-
X
-
+
X
-
-
Repel
X
X
Coulomb’s Law –
Force between charges
 The force between two charges is proportional to the
product of the two charges and inversely proportional
to the square of the distance between them.
 The force acts along the line connecting the two
charges.
q1q2
1 q1q2
q1q2
F 2 
k 2
2
r
40 r
r
k
1
40
Remember
Coulomb’s Law
The Unit of Charge is called
THE COULOMB
1 q1q2
F
runit
2
40 r
1
 k  9 x109 Nm2 / C 2
40
Smallest Charge: e ( a positive number)
1.6 x 10-19 Coul.
electron charge = -e
Proton charge = +e
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?
The Electric Field
Fields
 Imagine an object is placed at a particular point in





space.
When placed there, the object experiences a force F.
We may not know WHY there is a force on the object,
although we usually will.
Suppose further that if we double some property of
the object (mass, charge, …) then the force is found
to double as well.
Then the object is said to be in a force field.
The strength of the field (field strength) is defined as
the ratio of the force to the property that we are
dealing with.
Example – Gravitational Field.
 Property is MASS (m).
 Force is mg.
 Field strength is defined
as Force/Property
F  mg
Gravitatio nal Field Strength 
Force
Gravitatio nal Force mg


g
Property
mass
m
The Gravitational Field That We
Live In.
M
m
mg
Mg
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.
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
Q
F
E   k 2 runit
r
q
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
(r 2  x 2 )3 / 2
L/2

0
dx
(r 2  x 2 )3 / 2
(standard integral)
Completing the Math
Doing the integratio n :
kLm
Ex 
 L2 
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
thk=dr
dq=rdV=r x surface area x thickness
=r x 4r2 x dr
Summary
qQ
runit
2
r
F
Q
E
 k 2 runit
q
r
General
Fj
Qj
E  E j  
  k 2 r j ,unit
q
rj
Fk
E  k
rdV (r )
r
2
 k
sdA(r )
r
2
 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!)
 Using some
advanced mathematics we can derive
from this equation, the fact that:
F  qE