PHYS 251 Physics II Instructor Dr Johnny B

PHYS 251 Physics II Instructor: Dr. Johnny B. Holmes, Professor of Physics Office: AH 004 Phone: 321 -3448 Course web page: http: //facstaff. cbu. edu/~jholmes/P 251/intro. html If you have not received a link to the course web page or have misplaced the link, send me an e-mail (to jholmes@cbu. edu) and I’ll send you the link.

PHYS 251: Physics II Catalog description: A second course in physics covering electric forces, electric fields, voltage, capacitance, current, resistance, magnetic forces, magnetic fields, induction, oscillations, and waves. Prerequisite: PHYS 150 (Physics I). Corequisite: PHYS 251 L (Physics II Lab).

Will this course be easier or harder than Physics I? • This course will be easier than Physics I since we will use the same basic ideas such as Newton’s Laws and Conservation of Energy and the same way of thinking. • This course will be harder than Physics I because we will not be able to see and put our hands on most of what we are working with – instead we will use electrical meters to measure things like voltage and current.

Grading: (explained on syllabus) • 5 tests, each counts as one grade • 1 set of 9 regular collected homework problems which counts as one grade [Honors students will have 14 problems. ] • 2 sets of computer homework programs (17 total) where each set counts as one grade • final exam, which counts as 3 grades Total: 11 grades, the final grade will be based on the average of these 11 grades. 100 – A – 93 – B – 82 – C – 70 – D – 65 – F – There is a lot of graded homework. See last page of syllabus

Absence policy If you miss 3 or fewer classes, your lowest single score will be dropped (not counting homework scores). If the final is lowest, it will count only 2 instead of 3 times. Thus, if you have 3 or fewer class absences, the total will be based on the remaining 10 grades. Note: If you can not come to class due to a CBU approved reason (such as the pandemic or athletic games), you will be counted as absent unless you submit a review of the powerpoint slides assigned for the day you miss – see the syllabus for the daily slides. Unlike PHYS 150, I will not offer the opportunity for you to improve one test grade by fixing the problems you got wrong.

Regular Homework There are 9 [or 14 for honors] regular collected homework problems for the semester. These must be done and written up using the 7 step paradigm described in the syllabus. This paradigm is not good for the problems with obvious solutions, but is good for attacking problems that do not have obvious solutions. It is also a good way of communicating your thinking. This paradigm will be modeled in the examples done in class. The 7 steps are given in the next slide.

7 -step problem solving paradigm 1. 2. 3. 4. I want to and I can: motivation. What do you know: draw a diagram! What are you looking for: define symbols. Brainstorm: how is what you are looking for related to what you know: what laws (e. g. Newton’s 2 nd Law) and/or definitions apply. 5. Plan the solution: write the starting equations (e. g. ΣF=ma and/or vx≡dx/dt). Notation: “≡” means “defined to be”. 6. Execute the solution (do the math) - be sure to include units with all numerical values. 7. Check your answer – is it reasonable?

Computer Homework The 17 computer homework programs (each program consists of a problem set) are designed to give you graded practice. They emphasize getting the answer right the first time. If you get an answer wrong, the computer will tell you right away, and often tell you how to get it right. It is your task to actually get them correct. A random number generator will change the numbers so you will have to learn how to do them and not just remember the right answer.

Tests The 5 tests and final exam emphasize familiarity, recognition and speed. The material on the tests should be somewhat familiar. You should be able to recognize the type of problem, the basic principles involved, and determine which techniques to apply. Partial credit for wrong or missing answers may be given if you show your work on the test paper.

Study sheet This course emphasizes basic principles and problem solving, not memorization. To reduce the perceived need to memorize, you are permitted to bring to the tests one 8. 5” x 11” sheet of paper with information on one side. You may bring two study sheets to the final (writing on one side only, or one sheet with writing on both sides). Calculators are recommended, but calculators can not be shared and no internet connected devices are permitted during tests. You should keep your calculator, hands, and eyes on the desk and not in your lap.

OFFICE HOURS I would enjoy seeing you outside of class – whether to talk about homework, upcoming tests, or just talk about physics – whatever you are interested in. So please feel that you are welcome to stop by my office anytime you see me in my office! You can also e-mail me or call me at my office (321 -3448) with questions or suggestions. My office is right outside the classroom in AH 004. My availability is posted right by my office door and on the course web page. No formal appointment is necessary. By the way, I won’t think that you are “dumb” for doing this! On the contrary, it shows me you are interested in learning!

Review of Science I will not ask you what you believe. I will ask you about the generally accepted theories and have you work within those theories. To be a scientific theory, theory must first explain the repeatable experiments. This is useful in organizing the information. So even an incomplete theory may be useful. But this organization is not enough; a good scientific theory must predict new things to look for. If theory fails in those predictions, it must be either modified or replaced. This means that scientific theories must be falsifiable.

Review of Science If a theory is not falsifiable, it is not considered a scientific theory. Such untestable theories are more religious and/or philosophical. That does not mean that such ideas are not useful. Science often starts with very creative ideas that can not yet be tested. A current example is the idea of parallel universes. However, until theory makes testable predictions, it is not considered a scientific theory. Such untested ideas are still useful since they suggest new things to look for and/or help in designing better instruments with which to look. They are also fun to play with!

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 angular acceleration, of torque (rotational force), of moment of inertia (rotational inertia), of rotational kinetic energy, and of angular momentum. Note: Computer Homework Vol. 0 #5 and Vol. 0 #6 are math review programs on exponentials and logs. Computer Homework Vol. 1 #2 is a Physics I review program on adding vectors. Computer Homework Vol. 2 #1 is a Physics I review program on energy and power.

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, transformers, the circuit element of inductor, and basic AC circuits. In part 5 we look at oscillations and waves, and in particular, at oscillating electric and magnetic fields and electromagnetic waves. This will be the basis for our understanding of light and cell phone signals.

Part 1: Electricity Need for a New Force In dealing with electricity, we start with the realization that 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. The other forces, such as friction and contact force, are really due to collisions and electrical forces and so are not considered basic forces. Newton’s Law of gravity said that every mass attracts every other mass according to the relation: Fgravity = G M 1 m 2 / r 122 (attractive) where G describes the strength of the force in the appropriate units, M 1 and m 2 are the two masses involved, and r 12 is the distance between M 1 and m 2. The force depends on the square of the distance since space is 3 -D and the force spreads out and so weakens as the area over which it spreads through increases (area of sphere is 4 r 2).

A new fundamental quantity: Electric Charge It took a lot longer with electricity than with gravity, since electricity is not as obvious as gravity except for lightning, 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. How can we account for this? 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: electric charge.

Electric Charge – cont. How can we account for both attractive and repulsive forces? In order to account for both attractive and repulsive forces and describe electricity fully, the idea is to have two different kinds of charge, which we call positive and negative. [Note: we could have called the two kinds of charges something besides positive and negative; perhaps male and female, red and blue, etc. ] 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 which determines the direction of this electric force. We also found that the magnitude of the force decreases with distance between the charges just like gravity, so we have Coulomb’s Law: Felectricity = k q 1 q 2 / r 122 where k, like G in gravity, describes the strength of the force in terms of the units used; q 1 and q 2 are the charges; and r 12 is the distance between the two charges.

Electric Force Charge is a fundamental quantity, like length, mass and time. The unit of charge in the MKS system is called the Coulomb. (recall in the MKS system we have length in Meters, mass in Kilograms, time in Seconds, and now we add charge in Coulombs) 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*m 2 / Coul 2. (Compare this to G which is 6. 67 x 10 -11 Nt*m 2 / kg 2 !)

Electric Force The huge 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 it’s 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! [Note that using positive and negative for the two kinds of charge works great with balancing out the forces. ]

Fundamental Charges When we break matter up, we find there are just a few fundamental particles: electrons, protons and neutrons. (We’ll consider whether these are really fundamental or not and whethere 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 -31 kg proton: qp = +1. 6 x 10 -19 Coul; mp = 1. 67 x 10 -27 kg neutron: qn = 0; mn = 1. 67 x 10 -27 kg 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. The charges of the proton and electron are exactly opposite so when paired together their charges cancel.

Fundamental Charges Note that the electron and proton both have the same amount of charge, with the electron’s charge being negative and the proton’s being positive. [This assignment of positive to the proton charge is arbitrary – it was made by Benjamin Franklin. ] This amount of basic 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: qelectron = -e; qproton = +e.

Electric Forces Recall that the + or – sign on the charge is important in determining the direction of the electric force vector but not its magnitude. The + or – sign will be important later when we consider electric energy (a scalar). 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. We consider this next.

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 electric force on a second nearby charge should also wiggle. Does this second charge feel the wiggle in the electric 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. We’ll look at these wiggles (or waves) in the last part of this course.

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 of the object: (recall that “≡” means “defined to be”) or, Fon 2 = q 2 * Efrom 1 Fg = m*g Efrom 1 ≡ Fon 2 due to 1 / q 2 g ≡ Fg/m. Note that since F is a vector and q is a scalar, E must be a vector (that is, the electric field has a direction). The direction of E will be in the same direction as F if we have a positive q and opposite the direction of F if we have a negative q.

Electric Field: UNITS From the definition of Electric Field: Efrom 1 ≡ Fon 2 due to 1 / q 2 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, g = Fg/m, the unit is Nt/kg = m/s 2, so we often call the gravitational field, g, the acceleration due to gravity on that planet instead of the gravitational field of the planet. However, the gravitational field acts on any mass in its field always, but a mass will have its acceleration equal to g only if there are no other forces on the mass.

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 q 1 q 2 / r 122 (Coulomb’s Law) and Efrom 1 ≡ Fon 2 / q 2 (def. of Electric field) which gives us the formula for a point charge: Efrom 1 = k q 1 / r 1 f 2. where r 1 f is the distance from the charge that sets up the field to the field point. This is like Fgravity = Weight = mg where Fgravity = GMm/r 2 and so g = GM/r 2. We should really start thinking of “g” as being the gravitational field around the planet instead of the acceleration due to gravity.

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 -10 m (this is one of the excited states of hydrogen). How fast, v=? , will the electron be going in its orbit? Diagram: Steps 2 and 3 qproton = +e = 1. 6 x 10 -19 Coul qelectron = -e = -1. 6 x 10 -19 Coul v=? r = 1 x 10 -10 m, melectron = 9. 1 x 10 -31 kg of the problem solving paradigm r +

Example, cont. We first recognize this [step 4 of the problem solving paradigm] as a circular motion problem and a Newton’s Second Law v - F problem where the electric r force causes the circular motion: + Step 5 of the problem solving paradigm: [steps 2&3: diagram] S F = ma where Fcenter = Felec = k e e / r 2 directed towards the center, m is the mass of the electron since the electron is the particle that is moving, and acirc = w 2 r = v 2/r.

Example, cont. Step 6 of the problem solving paradigm: SF = ma becomes ke 2/r 2 = m(v 2/r), or v = [ke 2/mr]1/2 = [{9 x 109 Nt*m 2/C 2 * (1. 6 x 10 -19 C)2}/{9. 1 x 10 -31 kg* 1 x 10 -10 m}]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. Step 7 of the problem solving paradigm: Is this speed reasonable? At this point, you probably don’t have any idea if this is reasonable or not since you probably don’t have any experience dealing with the insides of atoms.

Electric Field Example What is the strength of the electric field, E=? , due to the proton at the position where the electron is in the previous problem (r = 1 x 10 -10 m)? [diagram: steps 2&3] E (magnitude) = ? - r 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 [step 4], so we can use [step 5]: E = kq/r 2. [step 6]: E (mag) = (9 x 109 Nt-m 2/C 2) * (1/6 x 10 -19 C) / (1 x 10 -10 m)2 = 1. 44 x 1011 Nt/Coul. E(dir) points away from the positive charge. Note that the field of the proton points away, but the force on the electron is towards the proton since the electron has a negative charge. [step 7]: Recognize how big the field is in this case.

Electric Fields due to several point charges Since electric fields are vectors, we need to add the fields of individual point charges as vectors to get the total electric field. This is how we worked with individual forces to get the total (or net) force in Physics I. 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/r 2, and is the angle the direction of 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 = Si {(kqi / ri 2 ) cos( i)} and Ey = Si {(kqi / ri 2 ) 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. A diagram showing the r’s and θ’s is on the next slide.

Electric Field due to two point charges r 1 = [(xf -x 1)2 + (yf -y 1)2]1/2 θ 1= tan-1[(yf -y 1)/(xf -x 1)] r 2 = [(xf -x 2)2 + (yf -y 2)2]1/2 θ 2= tan-1[(y 2 -yf)/(x 2 -xf)] note that E 1 is directed away from +q 1 and E 2 is directed towards –q 2 q 1 at (x 1, y 1)+ Note that in determining the θ for the positive charge y 1 1 we use (yf – y 1) and (xf – x 1), but since the field points towards the negative charge, we reverse this for θ 2 and use (y 2 – yf) and (x 2 – xf). x r 1 r 2 - q 2 at (x 2, y 2) E 2 2 field point at (xf, yf) E 1 1

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! There is a worked out example in Part 1 Slide Set extra. Computer Homework, Vol 1 #2, gives you a review of adding vectors. 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.

Continuous Distribution of Charges Is it possible to deal with a continuous distribution, say a line of charge, or a plate of charge? We’ll consider first the simpler case of a line of charge, then we’ll consider the more complicated plate of charge. The idea of a continuous distribution of charges is not consistent with the idea of individual electrons and protons. However, since the charges of these particles are so small, it would take a huge number to give a reasonable total charge. We then use an approximation that the charges are distributed continuously instead of in discrete packages of electrons or protons.

Continuous Distribution of Charges 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. Here, however, we do not divide (differentiate) the small pieces but add (integrate) the small pieces.

Continuous Distribution of Charges We saw in Physics I that we can use the calculus to provide analytical solutions to problems, but we can now also NOT use the limiting process but instead keep the small bits and use a computer to deal with each small bit – a numerical solution. For our line of charge, we could keep the little bits of charge, the Dqi , and add all of the little electric fields due to all of the little charge bits using a computer. We actually do this in our first lab – use the computer in a numerical way to find electric fields. The analytical way has the advantage of giving us nice formulas, so we’ll look at how this might work in some common situations. However, the analytical way may be too hard to solve in many real cases, but we can always get at least approximate answers to specific problems using the numerical way.

Line of Charge Ei-x = Si (k Dqi / ri 2 ) cos( i) and Ei-y = Si (k Dqi / ri 2 ) 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 line, ri, (line from the Dqi to the field point) makes with the horizontal. Here, Si (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 = dq/dx ≡ = constant. Thus, Q = L and Dqi = Dxi. This allows the sum to be over a distance instead of over a charge. Therefore, Ei-x = S (k Dxi / ri 2 ) cos( i) and Ei-y = S (k Dxi / ri 2 ) sin( i). We can now replace the sum with an integral: Ex = x-left x-right (k dx / r 2 ) cos( ) and Ey = x-left x-right (k dx / r 2 ) sin( ). Note that both r and depend on x!

Line of uniformly distributed charge Writing this in more standard form with the dx at the end: Ex = x-left x-right (k / r 2) cos( ) dx and Ey = x-left x-right (k / r 2) sin( ) dx. The tricky part is relating r and to x , that is, finding r as a function of x: r(x), and finding as a function of x: (x), since the integral is over dx (assuming the charge is distributed over a wire that runs horizontally).

Line of uniformly distributed charge Ex = x-left x-right (k / r 2 ) cos( ) dx and Ey = x-left x-right (k / r 2 ) sin( ) dx. A diagram should help with the geometry: to make the geometry simpler, we choose the x=0 position to be directly below the field point. r = (x 2 + a 2) cos( ) = -x/r sin( )= a/r. (field point) ri a Note here that cos( ) is positive but x is negative, so cos( ) = -x/r. i Dqi xi < 0 Note also that we really didn’t have to find (x), only cos( ) and sin( ) in terms of x. line of charge

Line of uniformly distributed charge r = (x 2 + a 2) cos( ) = -x/r sin( )= a/r. If x is positive (to the right of center), we see that the x component of the electric field is negative (points to the left) so cos( ) = -x/r. In terms of the angle within the o - , and using the facts that triangle, φ, we note that φ = 180 cos( ± 180 o) = -cos( ) and cos(- ) = cos( ), we have cos( ) = cos(180 o- φ) = -cos(180 o- φ - 180 o) = -cos(- φ) = -cos(φ) = -x/r = cos( ). (field point) a ri φi i xi > 0 Dqi line of charge

Line of uniformly distributed charge Ex = x-left x-right (k / r 2) cos( ) dx and Ey = x-left x-right (k / r 2) sin( ) dx with r = (x 2 + a 2) , cos( ) = -x/r , sin( )= a/r becomes: Ex = x-left x-right [k (-x) / (x 2 + a 2)3/2 ]dx and Ey = x-left x-right [k a / (x 2 + a 2)3/2 ] dx . It is probably not obvious what these integrals reduce to, but they can be integrated.

Line of uniformly distributed charge - special case using symmetry Ex = x-left x-right [k (-x) / (x 2 + a 2)3/2 ] dx and Ey = x-left x-right [k a / (x 2 + a 2)3/2 ]dx. 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 Ey = (k / a) * [cos( Left) - cos( Right)] if we change variables from x to (see next slide).

Extra Credit For up to three points extra credit on your regular collected homework score, starting from Ey = xl xr [k a / (x 2 + a 2)3/2 ] dx , get the result: Ey = (k /a) * [cos( Left) - cos( Right)]. 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 of a long wire Ey = (k / a) * [cos( Left) - cos( Right)]. What does long mean? Whenever a « -x. Left , Left 0 o, and whenever a « x. Right , Right 180 o. Left x. L<0 a Right x. R

Line of uniformly distributed charge: limiting case of a long wire Ey = (k / a) * [cos( Left) - cos( Right)]. As Left 0 o and Right 180 o, then [cos( Left) - cos( Right)] [cos(0 o) – cos(180 o)] = 2, and so Ey goes to the value 2 k /a and by symmetry Ex = 0. L x. L<0 a R x. R

Line of charge: extended Ey = (k / a) * [cos( Left) - cos( Right)]. 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( Left) - cos( Right)]. Yes! In the case shown, both Left and Right are greater than 90 o, but otherwise you proceed as usual. Note that Ex will no longer be zero by symmetry. field point a Right Left 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 = 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 DEi-x’s on opposite sides will cancel in pairs, leaving the horizontal component of the Electric field zero Ex = 0. DEi-y DEi-x r r a Dqi R R Dqi

Another Case: a Ring of Charge However, the DEi-y’s will add. In addition, we see that all the DEi-y’s will be the same and equal to (k Dqi / r 2) * sin( ). DEiy DEix r a Dqi R

Another Case: a Ring of Charge Thus Ei-y = Si DEi-y = Si [(k Dqi / r 2) * sin( )]. Since sin( ) = a/r and r = [R 2 + a 2] , both do not depend on which Dqi we are considering, so Ey = [k sin( ) / r 2]* [Si Dqi] = k*Q*sin( ) / r 2. Note that Q = 2 R. DEi-y DEi-x Note that Ey = k*Q*sin( ) / r 2 r a can be rearranged to become Ey = k*( 2 R)*(a/r) / r 2 Ey = 2 k Ra / (a 2+R 2)3/2. Dqi 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 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! In the next slide set, we’ll see a new way to attack such problems if they have a symmetry that we can use.