PHYS 251 Physics II Instructor: Dr. Johnny B. Holmes, Professor of Physics Office: CW 103 Phone:...

PHYS 251Physics II

Instructor:Dr. Johnny B. Holmes, Professor of Physics

Office: CW 103 Phone: 321-3448Course 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.

http://facstaff.cbu.edu/~jholmes/P251/intro.html

http://www.cbu.edu/

Review of Physics IIn 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 – overviewcontinued

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 ForceThe 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 10-27kgNote: 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 ChargesNote 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 Conceptcontinued

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 Conceptcontinued

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 - DefinitionThe 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 / r12

2 .

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:

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 = 2r = 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 Coulr = 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.


Thus, when we add the fields due to several charges, we get:

Ex = {(kqi / ri2 ) cos(i)} and

Ey = {(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.


r1 = [(xf-x1)2 + (yf-y1)2]1/2 θ1= tan-1[(yf-y1)/(xf-x1)]

+ q1 at (x1,y1)

- q2 at (x2,y2)

* field point at (xf,yf)E1

E2

r1

r2

x

y

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, x and used the limiting process:

v = LIMITas t goes to 0 [xt] = 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 ChargeEx = (k qi / ri

2 ) cos(i) and

Ey = (k qi / ri2 ) sin(i) where the ri is

the distance from qi, 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, qi) = Q = total charge.* field pointri

qi

θi line of charge

Line of uniformly distributed charge

If the charge is uniformly distributed on the line, then Q/L = qi/xi = constant = .

Thus, Q = L and qi = xi . Therefore,

Ex = (k xi / ri2 ) cos(i) and

Ey = (k xi / ri2 ) sin(i) .

We can now replace the sum with an integral:

Ex = xlxr (k dx / r2 ) cos() and

Ey = xlxr (k dx / r2 ) sin() .



Ey = xlxr (k 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).



Ey = xlxr (k dx / r2 ) sin() .A diagram should help with the geometry:r = (x2 + a2) cos() = -x/r sin()= a/r .

* (field point)

ri a

i

qi xi < 0



Ey = xlxr (k dx / r2 ) sin() with

r = (x2 + a2) , cos() = -x/r , sin()= a/r becomes:

Ex = xlxr (k x) dx / (x2 + a2)3/2 and

Ey = xlxr (k a dx / (x2 + a2)3/2 .It is probably not obvious what these integrals reduce

to, but they are integrable.


Ex = xlxr (k x) dx / (x2 + a2)3/2 and

Ey = xlxr (k 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 / a) * [cos(L) - cos(R)] .

Extra Credit

For up to three points extra credit on your regular collected homework score, starting from Ey = xlxr (k a dx / (x2 + a2)3/2 , get the result: Ey = (k /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 / 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 R

xL<0 xR

Line of uniformly distributed charge: limiting case

Ey = (k / 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 2k /a .

*

a

L R

xL<0 xR

Line of charge: extended

Ey = (k / 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 / 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


We again start by breaking the continuous ring up into finite bits of charge: qi = xi and then converting to an integral.

Ei

* field point

r a

qi R


By symmetry, we can see that the Eix ‘s on opposite sides will cancel in pairs, leaving the horizontal component of the Electric field zero. Eiy

* Eix

r r a

qi R R qi


However, the Eiy ‘s will all add. In addition, we see that all the Eiy ‘s will be the same and equal to (k qi / r2) * sin().

Eiy

* Eix

r a

qi R

Another Case: a Ring of ChargeThus Ey = Eix = [(k qi / r2) * sin()] .

Since sin() = a/r and r = [R2 + a2] , both do not depend on which qi we are considering, soEy = [k sin() / r2]* [ qi] = k*Q*sin() / r2. Note that Q = 2R.

Eiy

* Eix

r a qi R

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 chargeWhen 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!

PHYS 251 Physics II Instructor: Dr. Johnny B. Holmes, Professor of Physics Office: CW 103 Phone:...

Documents

Transcript of PHYS 251 Physics II Instructor: Dr. Johnny B. Holmes, Professor of Physics Office: CW 103 Phone:...