Transcript Part 1 Set 1 - FacStaff Home Page for CBU

PHYS 251
Physics II
Instructor:
Dr. Johnny B. Holmes, Professor of Physics
Office: CW 103 Phone: 321-3448
Course web page:
http://facstaff.cbu.edu/~jholmes/P251/intro.html
navigation to find the course web page:
start with www.cbu.edu ;
choose Academics;
choose School of Sciences;
choose either Faculty or Physics Dept.;
choose my name (Dr. Holmes);
choose my home page;
choose PHYS 251.
Review of Physics I
In Physics I, we began with the fundamentals
(things we couldn’t define in terms of other things):
distance, time, and mass;
We found that space was three dimensional, so we
developed the idea of vectors and found out how
to work with them (add them in rectangular form).
We then considered relations between space and time
– motion with velocity and acceleration.
We next looked at how to cause or predict motion
using forces and worked with Newton’s Laws of
Motion. We also introduced Newton’s Law of
Gravity as one of the fundamental forces in nature.
Review of Physics I
(continued)
To make some problems simpler, we defined the
concepts of work and energy, and then introduced
the Law of Conservation of Energy.
In order to work with collisions (and explosions), we
developed the idea of momentum.
We then expanded all of these ideas to include
rotations (spinning), and developed the ideas of
angular speed and angular acceleration, of torque
(rotational force), of moment of inertia (rotational
inertia), of rotational kinetic energy, and of
angular momentum.
Physics II - overview
In Physics II, we are going to expand these concepts to
include another basic force: electromagnetism.
In part 1, we will start this study by considering electric
forces and define the concept of electric field.
In part 2, we then extend the electric force to electric
energy and define the concept of voltage which we
then use to work with basic DC circuits including the
circuit elements of resistors and capacitors.
Physics II – overview
continued
In part 3 we consider the magnetic force and introduce
the idea of magnetic fields which we then use to
look at electric motors, cyclotrons, and mass
spectrometers.
In part 4 we look at electromagnetism: the interaction
of electric and magnetic fields which leads to the
basic design of an electric generator, the circuit
element of inductor, and basic AC circuits.
In part 5 we look at oscillating electric and magnetic
fields and electromagnetic waves.
Electricity: need for a New Force
In dealing with electricity, we start with the
realization we are dealing with a new type of
force.
Before we consider this new force, let’s review what
we covered in Physics I where we considered one
of the other basic forces in nature: gravity.
Newton’s Law of gravity said that every mass
attracts every other mass according to the relation:
Fgravity = G M1 m2 / r122 (attractive)
A new fundamental quantity:
Electric Charge
It took a lot longer with electricity than with gravity, but
we finally realized that there is an Electric Force that
is basic and works in a way similar to gravity.
But unlike gravity where the force was ONLY
ATTRACTIVE, we find that the electric force is
sometimes attractive but also sometimes
REPULSIVE.
Also, the force wasn’t between the masses of two
objects. Instead, we found that there was another
property associated with matter that we name: charge.
Electric Charge – cont.
In order to account for both attractive and
repulsive forces and describe electricity
fully, we needed to have two different kinds
of charge, which we call positive and
negative.
Gravity with only attractive forces needed
only one kind of mass. Electricity, with
attractive and repulsive forces, needs two
kinds of charge.
Electric Force: Coulomb’s Law
After experimenting with different charges, we
found that like charges repel, and unlike
charges attract.
We also found that the force decreases with
distance between the charges just like
gravity, so we have Coulomb’s Law:
Felectricity = k q1 q2 / r122 where k, like G in
gravity, describes the strength of the force in
terms of the units used.
Electric Force
Charge is a fundamental quantity, like
length, mass and time. The unit of charge
in the MKS system is called the Coulomb.
When charges are in Coulombs, forces in
Newtons, and distances in meters, the
Coulomb constant, k, has the value:
k = 9.0 x 109 Nt*m2 / Coul2 . (Compare this to
G which is 6.67 x 10-11 Nt*m2 / kg2 !)
Electric Force
The big value of k compared to G indicates
that electricity is VERY STRONG
compared to gravity. Of course, we know
that getting hit by lightning is a BIG DEAL!
But how can electricity be so strong, and
yet normally we don’t realize its there in
the way we do gravity?
Electric Force
The answer comes from the fact that, while
gravity is only attractive, electricity can be
attractive AND repulsive.
Since positive and negative charges tend to attract,
they will tend to come together and cancel one
another out. If a third charge is in the area of the
two that have come together, it will be attracted to
one, but repulsed from the other. If the first two
charges are equal, the attraction and repulsion on
the third will balance out, just as if the charges
weren’t there!
Fundamental Charges
When we break matter up, we find there are
just a few fundamental particles: electron,
proton and neutron. (We’ll consider whether these
are really fundamental or not and whether there are other
fundamental particles in the last part of the next physics
course, PHYS 252.)
electron: qe = -1.6 x 10-19 Coul; me = 9.1 x 10-31kg
proton: qp = +1.6 x 10-19 Coul; mp = 1.67 x 10-27kg
neutron: qn = 0;
mn = 1.67 x 10-27kg
Note: The mass of the proton and the neutron appear to be the same, but
in fact the neutron is slightly more massive; this will be important in
nuclear physics, but not for us now.
Fundamental Charges
Note that the electron and proton both have
the same charge, with the electron being
negative and the proton being positive. This
amount of charge is often called the
electronic charge, e. This electronic
charge is generally considered a positive
value (just like g in gravity). We add the
negative sign when we need to:
qe = -e; qp = +e.
Electric Forces
Unlike gravity, where we usually have one big
mass (such as the earth) in order to have a
gravitational force worth considering, in
electricity we often have lots of charges
distributed around that are deserving of our
attention!
This leads to a concept that can aid us in
considering many charges: the concept of
Electric Field
Concept of “Field”
How does the electric force (or the
gravitational force, for that matter), cause a
force across a distance of space?
In the case of gravity, are there “little devils”
that lasso you and pull you down when you
jump? Do professional athletes “pay off the
devils” so that they can jump higher?
Answer: We can develop a better theory than this!
Electric Field Concept
continued
One way to explain this “action at a distance”
is this: each charge sets up a “field” in
space, and this “field” then acts on any
other charges that go through the space.
One supporting piece of evidence for this idea
is: if you wiggle a charge, the force on a
second charge should also wiggle. Does
this second charge feel the wiggle in the
force instantaneously, or does it take a little
time?
Electric Field Concept
continued
What we find is that it does take a little time for
the information about the “wiggle” to get to
the other charge. (It travels at the speed of light,
so it is fast, but not instantaneous!)
This is the basic idea behind radio
communication: we wiggle charges at the
radio station, and your radio picks up the
“wiggles” and decodes them to give you the
information.
Electric Field - Definition
The field strength should depend on the charge or
charges that set it up.
The force depends on the field set up by those
charges and the amount of charge of the particle
at that point in space (in the field) just like the
force of gravity depends on the gravitational field
of the planet and the mass:
Fon 2 = q2 * Efrom 1
Fg = m*g
or,
Efrom 1 = Fon 2 / q2
g = Fg/m .
Note that since F is a vector and q is a scalar, E must
be a vector.
Electric Field: UNITS
From the definition of Electric Field:
Efrom 1 = Fon 2 / q2
we see that the units of Electric Field are
Nt/Coul.
Note: The unit of field does not (yet) have its own name like
some other things do, such as the unit for energy is a Nt*m
= Joule, or the unit for power is Joule/sec = Watt. For
gravitational fields, the unit is Nt/kg = m/s2, so we often
called the gravitational field, g, the acceleration due to
gravity on that planet instead of the gravitational field.
Electric Field for a point charge
If we have just one point charge setting up the
field, and a second point charge comes into the
field, we know that:
Fon 2 = k q1 q2 / r122
(Coulomb’s Law)
and
F on 2 = Efrom 1 * q2
(def. of Electric field)
which gives us the formula for a point charge:
E from1 = k q1 / r122 .
Finding Electric Fields
We can calculate the electric field in space due to
any number of charges in space by simply adding
together the many individual Electric fields due to
the point charges!
Computer Homework, Vol 3 #1, gives you graded
practice with working with fields due to one or
several charges. Remember that electric fields
have directions and so they are vectors and thus
must be added as vectors!
Finding Electric Fields
In the first laboratory experiment,
Simulation of Electric Fields, we use a
computer to perform the many vector
additions required to look at the total
electric field due to several charges in
several geometries.
With the calculus, we can (and will)
determine the electric fields due to certain
continuous distributions of charges, such as
charges on a wire or a plate.
Force Example
Suppose that we have an electron orbiting a
proton such that the radius of the electron in
its circular orbit is 1 x 10-10m (this is one of
the excited states of hydrogen). How fast will
the electron be going in its orbit?
qproton = +e = 1.6 x 10-19 Coul
qelectron = -e = -1.6 x 10-19 Coul
r = 1 x 10-10 m,
melectron = 9.1 x 10-31kg
Example, cont.
We first recognize this as a circular motion
problem and a Newton’s Second Law
problem where the electric force causes the
circular motion:
S F = ma where Fcenter = Felec = k e e / r2
directed towards the center, m is the mass
of the electron since the electron is the
particle that is moving, and acirc = w2r =
v2/r.
Example, cont.
ke2/r2 = m(v2/r), or v = [ke2/mr]1/2 =
[{9x109 Nt*m2/C2 * (1.6x10-19C)2}/{9.1x10-31kg* 1x10-10m}]1/2
= 1.59 x 106 m/s = 3.56 million mph.
Note that we took the + and - signs for the
charges into account when we determined that
the electric force was attractive and directed
towards the center. The magnitude has to be
considered as positive.
Electric Field Example
What is the strength of the electric field due to
the proton at the position where the electron
is in the previous problem (r = 1 x 10-10m)?
E (magnitude) = ?
E (direction) = ?
Q = +e (due to proton) = 1.6 x 10-19 Coul
r = 1 x 10-10 m.
Electric Field Example
Here we recognize this as an example of the
electric field due to a point charge, so we
can use:
E = kq/r2 .
E (mag) =
(9 x 109 Nt-m2/C2) * (1/6 x 10-19C) / (1 x 10-10m)2
= 1.44 x 1011 Nt/Coul.
E(dir) points away from the positive charge.
Electric Fields due to
several point charges
Since electric fields are vectors, we need to add
the fields of individual point charges as vectors.
And since we add vectors by adding the
rectangular components, we must first express
the individual point charge fields in rectangular
form:
Ex = E cos() and Ey = E sin()
where E = kq/r2, and  is the angle the direction
the field makes with the horizontal.
Electric Fields due to several
point charges
Thus, when we add the fields due to several
charges, we get:
Ex = S {(kqi / ri2 ) cos(i)}
and
Ey = S {(kqi / ri2 ) sin(i)}
where the ri is the distance from the ith
charge to the point where we are calculating
the field, and i is the angle the field at that
point makes with the horizontal.
Electric Fields due to several
point charges
r1 = [(xf-x1)2 + (yf-y1)2]1/2
θ1= tan-1[(yf-y1)/(xf-x1)]
y
+ q1 at (x1,y1)
x
r1
r2
- q2 at (x2,y2)
* field point at (xf,yf)
E2
E1
Continuous Distribution of
Charges
Is it possible to deal with a continuous
distribution, say a line of charge, or a plate
of charge?
Let’s consider first the simpler case of a line
of charge, then we’ll consider the more
complicated plate of charge.
Line of Charge
How do we deal with a continuous situation?
How did we deal with the continuous motion
in PHYS 150? We started with a small unit,
Dx and used the limiting process:
v = LIMITas Dt goes to 0 [Dx / Dt] = dx/dt .
We can do the same thing here with charge:
we break the continuous charge into bits,
and treat each bit as a point charge.
Line of Charge
Ex = S (k Dqi / ri2 ) cos(i)
and
Ey = S (k Dqi / ri2 ) sin(i) where the ri is the
distance from Dqi, the ith charge, to the point
where we are calculating the field, and i is
the angle the field at that point makes with
the horizontal.
Here, S (Dqi) = Q = total charge.
* field point
ri
θi
Dqi
line of charge
Line of uniformly distributed
charge
If the charge is uniformly distributed on the
line, then Q/L = Dqi/Dxi = constant = l.
Thus, Q = lL and Dqi = l Dxi . Therefore,
Ex = S (k l Dxi / ri2 ) cos(i)
and
Ey = S (k l Dxi / ri2 ) sin(i) .
We can now replace the sum with an integral:
Ex = xlxr (k l dx / r2 ) cos()
Ey = xlxr (k l dx / r2 ) sin() .
and
Line of uniformly distributed
charge
Ex = xlxr (k l dx / r2 ) cos()
and
Ey = xlxr (k l dx / r2 ) sin() .
The tricky part is relating r and  to x since
the integral is over dx (assuming the charge is
distributed over a wire that runs horizontally).
Line of uniformly distributed
charge
Ex = xlxr (k l dx / r2 ) cos()
and
Ey = xlxr (k l dx / r2 ) sin() .
A diagram should help with the geometry:
r = (x2 + a2)
cos() = -x/r sin()= a/r .
* (field point)
ri
a
i
Dqi xi < 0
Line of uniformly distributed
charge
Ex = xlxr (k l dx / r2 ) cos()
and
Ey = xlxr (k l dx / r2 ) sin()
with
r = (x2 + a2) , cos() = -x/r , sin()= a/r
becomes:
Ex = xlxr (k l (-x) dx / (x2 + a2)3/2
and
Ey = xlxr (k l a dx / (x2 + a2)3/2 .
It is probably not obvious what these integrals reduce
to, but they are integrable.
Line of uniformly distributed
charge
Ex = xlxr (k l (-x) dx / (x2 + a2)3/2
and
Ey = xlxr (k l a dx / (x2 + a2)3/2 .
In the case of Ex, if the field point is directly
above the middle of the line of charge, the
symmetry makes the value of Ex = zero!
In the case of Ey, we can find an answer if we
change variables from x to :
Ey = (k l / a) * [cos(L) - cos(R)] .
Extra Credit
For up to three points extra credit on your
regular collected homework score, starting
from Ey = xlxr (k l a dx / (x2 + a2)3/2 , get
the result: Ey = (kl /a) * [cos(L) - cos(R)].
Hint: relate x to , then use the method of
substitution (instead of relating r and  to x,
relate r and x to ).
Line of uniformly distributed charge:
limiting case
Ey = (k l / a) * [cos(L) - cos(R)] .
What does very large mean?
Whenever a « -xL , L approaches 0o, and
whenever a « xR , R approaches 180o .
*
a
L
xL<0
R
xR
Line of uniformly distributed charge:
limiting case
Ey = (k l / a) * [cos(L) - cos(R)] .
As L goes to 0o and R goes to 180o, then
[cos(L) - cos(R)] goes to [cos(0o) – cos(180o)] = 2,
and so Ey goes to the value 2kl /a .
*
a
L
xL<0
R
xR
Line of charge: extended
Ey = (k l / a) * [cos(L) - cos(R)] .
Does this formula work when the field point is
not directly above the wire (see figure
below) ?
* field point
wire
Line of charge: extended
Ey = (k l / a) * [cos(L) - cos(R)] .
Yes! In the case shown, both L and R are greater
than 90o, but otherwise you proceed as usual.
* field point
a
L
R
wire
Another Case: a Ring of Charge
What is the electric field due to a ring of
charge, with the field point on axis at a
distance a above the center of the ring?
* field point
a
R
Another Case: a Ring of Charge
We again start by breaking the continuous
ring up into finite bits of charge: Dqi = l Dxi
and then converting to an integral.
DEi
* field point
r
a
Dqi  R
Another Case: a Ring of Charge
By symmetry, we can see that the DEix ‘s on
opposite sides will cancel in pairs, leaving
the horizontal component of the Electric
field zero.
DEiy
* DEix
r
r
a
Dqi  R R  Dqi
Another Case: a Ring of Charge
However, the DEiy ‘s will all add. In addition,
we see that all the DEiy ‘s will be the same
and equal to (k Dqi / r2) * sin().
DEiy
* DEix
r
a
Dqi  R
Another Case: a Ring of Charge
Thus Ey = S DEix = S [(k Dqi / r2) * sin()] .
Since sin() = a/r and r = [R2 + a2] , both do not
depend on which Dqi we are considering, so
Ey = [k sin() / r2]* [S Dqi] = k*Q*sin() / r2.
Note that Q = l 2pR.
DEiy
DEix
*
r
Dqi
R
a
Ring - extended discussion
Note that the problem becomes much harder
if the field point is off-axis:
1. The Ex is no longer zero due to the lack of
symmetry.
2. The Ey integral is no longer trivial because
both r and  change for each different
position.
Plate of charge
When we had a line of charge, we had a
problem that involved an integral. Unless
the geometry was pretty simple, the integral
was hard to do.
For a plate of charge, we need to consider
small areas of charge, instead of small lines
of charge. This makes the resulting integral
a double integral (over both x and y), and
the geometry is now three dimensional
instead of sometimes just two!