PHYS 251 Physics II

1. PHYS 251Physics 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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)

7. 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.

8. 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 positiveand negative. Gravity with only attractive forces needed only one kind of mass. Electricity, with attractive and repulsive forces, needs two kinds of charge.

9. 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 / r122where k, like G in gravity, describes the strength of the force in terms of the units used.

10. 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 !)

11. 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?

12. 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!

13. 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.

14. 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.

15. 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

16. 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!

17. 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?

18. 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.

19. Electric Field - Definition The field strength should depend on the charge or charges that set it up. The forcedepends 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 1Fg = m*g or, Efrom 1 = Fon 2 / q2g = Fg/m. Note that since F is a vector and q is a scalar, E must be a vector.

20. 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.

21. 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)andF on 2 = Efrom 1 * q2(def. of Electric field)which gives us the formula for a point charge: E from1 = k q1 / r122 .

22. 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!

23. 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.

24. 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

25. 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 = mawhere Fcenter = Felec = k e e / r2 directed towards the center, m is the mass of the electronsince the electron is the particle that is moving,and acirc = w2r = v2/r.

26. 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.

27. 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.

28. 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 fromthe positive charge.

29. 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(q) where E = kq/r2, and q is the angle the direction the field makes with the horizontal.

30. Electric Fields due to several point charges Thus, when we add the fields due to several charges, we get: Ex = S {(kqi / ri2) cos(qi)} and Ey = S {(kqi / ri2) sin(qi)}where theri is the distance from the ith charge to the point where we are calculating the field, and qi is the angle the field at that point makes with the horizontal.

31. Electric Fields due to several point charges 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) y x r1 r2 E2 E1

32. 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.

33. 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.

34. Line of Charge Ex = S (k Dqi / ri2) cos(qi) and Ey = S (k Dqi / ri2) sin(qi)where theri is the distance from Dqi, the ith charge, to the point where we are calculating the field, and qi is the angle the field at that point makes with the horizontal. Here, S (Dqi) = Q = total charge. * field point ri θi line of charge Dqi

35. 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(qi) and Ey = S (k l Dxi / ri2) sin(qi). We can now replace the sum with an integral: Ex = xlxr (k l dx / r2 ) cos(q) and Ey = xlxr (k l dx / r2 ) sin(q).

36. Line of uniformly distributed charge Ex = xlxr (k l dx / r2 ) cos(q) and Ey = xlxr (k l dx / r2 ) sin(q). The tricky part is relating r and q to x since the integral is over dx (assuming the charge is distributed over a wire that runs horizontally).

37. Line of uniformly distributed charge Ex = xlxr (k l dx / r2 ) cos(q) and Ey = xlxr (k l dx / r2 ) sin(q). A diagram should help with the geometry: r = (x2 + a2) cos(q) = -x/r sin(q)= a/r . * (field point) ria qi Dqixi< 0

38. Line of uniformly distributed charge Ex = xlxr (k l dx / r2 ) cos(q)and Ey = xlxr (k l dx / r2 ) sin(q)with r = (x2 + a2) , cos(q) = -x/r , sin(q)= a/r becomes: Ex = xlxr (k l (-x) dx / (x2 + a2)3/2and 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.

39. Line of uniformly distributed charge Ex = xlxr (k l (-x) dx / (x2 + a2)3/2and 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 q: Ey = (k l / a) * [cos(qL) - cos(qR)] .

40. 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(qL) - cos(qR)]. Hint: relate x to q, then use the method of substitution (instead of relating r and q to x, relate r and x to q).

41. Line of uniformly distributed charge:limiting case Ey = (k l / a) * [cos(qL) - cos(qR)] . What does very large mean? Whenever a « -xL , qL approaches 0o, and whenever a « xR , qR approaches 180o . * a qLqR xL<0xR

42. Line of uniformly distributed charge: limiting case Ey = (k l / a) * [cos(qL) - cos(qR)] . As qL goes to 0o and qR goes to 180o, then [cos(qL) - cos(qR)] goes to [cos(0o) – cos(180o)] = 2, and so Ey goes to the value 2kl /a . * a qLqR xL<0xR

43. Line of charge: extended Ey = (k l / a) * [cos(qL) - cos(qR)] . Does this formula work when the field point is not directly above the wire (see figure below) ? * field point wire

44. Line of charge: extended Ey = (k l / a) * [cos(qL) - cos(qR)] . Yes! In the case shown, both qL and qR are greater than 90o, but otherwise you proceed as usual. * field point aqL qR wire

45. 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

46. 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 ra DqiqR

47. 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 rr a DqiqR R qDqi

48. 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(q). DEiy * DEix ra DqiqR

49. Another Case: a Ring of Charge Thus Ey = SDEix = S [(k Dqi / r2) * sin(q)] .Since sin(q) = a/r and r = [R2 + a2] , both do not depend on which Dqi we are considering, soEy= [k sin(q) / r2]*[SDqi] = k*Q*sin(q) / r2. Note that Q = l 2pR. DEiy * DEix ra DqiqR

50. 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 q change for each different position.