Transcript Lecture #2 01/13/05

Announcements
Physics Department Seminar:
TITLE: "The fluid dynamics of climatic variations."
SPEAKER: Professor Walter A. Robinson,
Department of Atmospheric Sciences
University of Illinois at Urbana-Champaign
TIME: Thursday Jan. 13, 2005 at 4 PM
Essays on Quizzes/Surveys:
I do read and comment on them when appropriate.
Reserve Books
Several books are on reserve in the library:
There will be required, supplemental readings from::
Biomedical Applications of Introductory Physics
Optional, suggested, readings from::
Electricity and Magnetism, Purcell
Div, Grad, Curl and all that
For review:
First-year Calculus, Hille and Salas
Other optional books may be placed for your enjoyment.
Matter and Charges
•All matter is made of positive and negative charges (or neutral)
•An object’s total charge is very close to zero
•When an object becomes charged, a tiny fraction of its charged
particles (usually electrons) are lost or gained
•These particles (usually electrons) can flow through objects
•Some materials are better at allowing the flow of electrons than
others
Conductor
A material that allows
electrons or other
charged particles to
flow freely
Insulator
A material that resists
the flow of electrons
and other charged
particles
Elementary Charge
•Charges seem to come only in integer multiples of a fundamental
charge unit called e
•We will treat e as a positive number (some sources treat it as
negative)
e = 1.602  10-19 C
Particle
Proton
Neutron
Electron
Oxygen nuc.
Calcium ion
Chlorine ion
q
e
0
-e
8e
2e
-e
know these
Some ways to charge objects
•By rubbing dissimilar
objects
•By chemical
processes
•By proximity between a
charge and a conductor –
charging by induction
•By physical contact
between a charge and
a conductor
+
-
+
Quiz
Threeballs
pithballs
are suspended
thin threads.
Three
are suspended
fromfrom
thin threads.
Various
Variousare
Objects
are thenagainst
rubbedother
against
other
Objects
then rubbed
objects
objectsagainst
(nylonsilk,
against
silk,
glasspolyester,
against etc.)
(nylon
glass
against
polyester,
and each
of the balls
is charged
and
each ofetc.)
the balls
is charged
by touching
by touching
with objects.
one of these
objects.
It is
them
with onethem
of these
It is found
that
found1 that
1 and
2 repel
eachballs
other2 and
balls
and pithballs
2 repel each
other
and that
that3pithballs
2 and
3 repel each other.
and
repel each
other.
Fromthis
thiswe
wecan
canconclude
concludethat:
that:
From
B) All
the1 balls
same sign
A)
Balls
and 3carry
carrycharges
chargesofofthe
opposite
sign.
Charges
equalcarry
signcharges
repel, soof1the
andsame
2 have
the
B)
All theofballs
sign
sameball
sign,
as dono
2 and
3, and so they all have the
C) One
carries
charge.
same sign. to determine without more
D) Impossible
information.
Coulomb’s Law
•Like charges repel, unlike charges attract
•Force is directly along a line joining the two charges
q1
q2
r
ke q1q2
Fe 
2
r
ke = 8.988109 Nm2/C2
•An inverse square law, just like gravity
•Can be attractive or repulsive – unlike gravity
•Constant is enormous compared to gravity
•Obeys the superposition principle just like gravity
Coulomb’s Law: Applied
A Helium nucleus (charge +2e) is separated from one of its
electrons (charge –e) by about 3.00 10-11m. What is the force the
nucleus exerts on the electron? Is it attractive or repulsive?
ke q1q2
9 Nm2/C2
k
=
8.98810
Fe 
e
2
r
r = just
3.00
10-11m the force
We
calculated
the electron
q1 =on
3.204
10-19Cfrom the
nucleus. How does this
q2 =compare
-1.602with
10-19the
C force
on the nucleus from the electron?
A)
on the
nucleus is twice as big
Fe=The- force
0.513
N
Attractive Force
B) The force on the nucleus is half as big
C) The forces are equal in magnitude
Newton’s Laws and Kinematics
Newton’s laws and all the kinematics you learned in 113 are still
true!
Fnet  ma
F12   F21
A body in motion tends to stay in motion,
therefore changing velocity, i.e. acceleration,
requires a force!
t'
t'
dx
dv
v  ;a 
 v   adt  vt0 ; x   vdt  x0
dt
dt
t0
t0
If a does not depend on
time, then
1 2
v  at  v 0 ; x  at  v 0 t  x 0
2
Coulomb’s Law vs. Gravity
A Helium nucleus (charge +2e) is separated from one of its
electrons (charge –e) by about 3.00 10-11m, and we just
calculated the electrostatic forces involved.
ke q1q2
9 Nm2/C2
k
=
8.98810
Fe 
e
2
r
Suppose we could adjust the distance between the
nucleus (considered as a point particle) and one
electron. Can we find a point at which the
electric and gravitational forces are equal?
A) Yes, move the particles apart.
B) Yes, move the particles together.
C) No, they will never be equal.
Coulomb’s Law – more versions
ke q1q2
Fe 
rˆ
2
r
•Force is a vector
(remember from last semester)
q1q2
-12 C2/ (N●m2)
ˆ
Fe 
r

=
8.85410
2
0
4 0 r
•Permittivity of free space
ke 
1
4 0
•Force can be positive or negative.
No absolute value signs unless
you just want the magnitude!
Conductors redistribute charge
•Conductor
Q
Add charges,
Q
•Conductors allow free flow of
charge
•Like charges repel
•So the charges will redistribute
themselves over the sphere
Spherical Shells
•Just like with gravity, a charge
outside feels the charge from the
sphere as if it were concentrated
in the middle
Q
q
•Consider the forces from
opposite charge elements and the
vector decomposition
Coulomb’s Law
My body contains about 31028 electrons, all repelling each other.
How come I don’t explode?
A) Electrons attract each other, not repel each other
B) The gravitational force is so strong it holds me together,
overcoming electric forces
C) There are also positive charges that cancel out the negative
charges
D) Electrical forces are too weak to consider
Electric Fields
•Electric Field is the ability to extert a force at a distance on a
charge
•It is defined as force on a test charge divided by the charge
•Denoted by the letter E
•Units N/C
+ +
– ––
+ +
+
F
Small test
charge q
E  F /q
E-Field: Why?
•When we have a charge distribution, and we want to know
what effect they would have external charges, we can either
•Do many sums (or integrations) every time a charge comes
in to find the force on that charge
•Or calculate the field from the charge distribution, and
multiply the field by the external charge to obtain the force
•Simplification!
Electric Field from a Point Charge
keQq
F  2 rˆ
r
E  F /q
ke Q
E  2 rˆ
r
Point
charge
Q
Small test
charge q
Note: the field is a vector!
Quiz
Two test charges are brought separately into the vicinity of a charge +Q.
First, test charge +q is brought to point A a distance r from charge +Q. Next, the +q
charge is removed and a test charge +2q is brought to point B a distance 2r from
charge +Q. Compared with the electric field of the charge at A, the electric field of
the charge at B is:
• T
+Q
+q
+Q
A
+2q
B
A) Greater
B) Smaller
C) The same.