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Lesson # 6 – Properties of Electric Charge and Coulomb’s Law Pre-Lesson Notes Unit: Fields Date: Concept: Properties of Electric Charge (textbook section 7.1) Definitions: Law of electric charges Law of conservation of charge Insulator Conductor Grounding Charging by induction Charging by contact Net Charge: Explanation: Lesson # 7 – Coulomb’s Law Pre Lesson Notes: Unit: Fields Date: Concept: Coulomb’s Law (textbook section 7.2) Definitions: Electric force Coulomb’s Law Coulomb’s constant Comparing Coulomb’s Law and Universal Gravitation The superposition principle Equations: In 1785 Charles Coulomb used a torsion balance to measure the force that one charged sphere exerts on another charged sphere to find how the force between two electrically charged objects depends on the magnitudes of the two charged objects and on their separation. Coulomb could not measure the absolute magnitude of the electric charge on the metal spheres. However, he could divide the charge in half by touching a charged metal sphere with a third identical uncharged metal sphere. Part A: Coulomb’s Law Two point-like objects with a net charge exert a force upon one another along a line from the centre of one object to the centre of the other. The magnitude of the electric force that object 1, with electric charge q1, exerts on object 2, with electric charge q2, when they are separated by a center-to-center distance r is given by the expression below. Note that this is also the equal magnitude electric force that object 2, with electric charge q2, exerts on object 1, with electric charge q1: k q1 q 2 Fq1 on q2 Fq 2 on q1 r2 where k = 9.0 x 109 Nm2 /C2. We assume that the objects are much smaller than their separation r (i.e. point-like objects). 1. The diagrams show two charged objects and their separation. Rank the size of the force that the left object exerts on the right object from the strongest to the weakest force. Explain how you made the ranking. A B C Ranking: Why: +2q +3q +2q d d A C +q -q d d d +q +q +q +q ½d B d Ranking: Why: 3d A -q C +q +q +q +q +q B +q Ranking: Why: d A +q d B -q +q +q C +2q +q 2d Ranking: Why: +3q +q 1/3d 2. For each of the cases above, compare the force that the left object exerts on the right object to the force that the right object exerts on the left object. Explain how you know. Basic Electric Force Sketches and Calculations In all of the following examples, the charges are shown in Coulombs and each vertical dotted line, drawn for your reference, is 1 m from the next. Sketch each force, as well as the total force (bold or in a different colour) on each charge. Don’t forget that k = 9 x 109. The 1st is done for you (& a bit of the others too): e.g. #1 F =k∙1∙1/12 = k 1 +1 +1 FNet(C) = k + k/4 +1 F2=k∙1∙1/22 = k/4 FNet (B)=0 = 1.125 x 1010 N [R] e.g. #2 +3 +1 +3 FNet(C) = 3k + 9k/4 = 4.725 x 1010 N [R] e.g. #3 +2 +2 +2 -3 +1 -3 FNet(C) = 4.5 x 1010N [R] e.g. #4 FNet(A) = 6.75 x 109 N [R] e.g. #5 +4 -2 +4 FNet(B) = 0 N e.g. #6 +1 -3 -2 +4 Lesson # 8 – Pith Ball Lab Charge Splitting 1. Imagine you had two containers, one with VA = 400 mL of water in it, and one empty, VB = 0. Now if you could connect them at the bottom so the water was free to flow, how much would be in each container once they “equalize”? 2. Create an equation to show how to calculate the final volume, V’, in each container. 3. Now imagine you have three containers this time: VA = 400 mL, VB = 0, VC = 100 mL. A and B are connected together to equalize. Then B and C are connected together to equalize. Use diagrams to prove that final volumes are 200 mL, 150 mL, and 150 mL. Notice that the total must be the same before and after. 4. Gravity is the force that causes these liquids to equalize. Electrostatic repulsion works in a very similar way to equalize charges when objects are connected to allow charges to flow. Thus, if a -10 nC charged metal sphere is touched to a neutral metal sphere, the electrons which repel each other, will spread out as much as they can. Thus, each sphere will split the total, ending up with -5 nC each so that the electrons can spread out as much as possible. 5. Try these practice questions: a) qa = -6 C, qb = -2 C, qc = -5 C. If A is touched to B and then B is touched to C, what will the final charge on each be? b) qa = -6 C, qb = +2 C, qc = -6 C. If A is touched to B and then B is touched to C, what will the final charge on each be? (As a first step, you can still just add qa and qb and divide by 2 to get the distributed average charge.) c) qa = -8 C, qb = +6 C, qc = 0 C, qd = +3 C. If A is touched to B and then B is touched to C, and then C is touched to D, what will the final charge on each be? d) qa = -8 C, qb = +6 C, qc = 0 C, qd = +3 C. If D is touched to C and then C is touched to B, and then B is touched to A, what will the final charge on each be? 6. A sphere with a -4µC charge is touched to an identical but uncharged sphere. a) If they are then placed 1.0 cm away from each other, what force will each feel? b) What mass would they need to be so that all forces would cancel? 7. Consider these four charges: qa = -8 nC, qb = +6 nC, qc = 0 C, qd = +3 nC. a) A, B, and C are laid out in order, 5.0 cm apart. Sketch the situation and the forces each would feel. b) Calculate the total electrostatic force on each of A, B, and C. c) D is touched to C and then C is touched to B, and then B is touched to A. What will the final charge on each be? d) A and B are put beside each other again. What force will they feel this time? (#6 answers : 360 N 23 tonnes) Quick Lab: Measuring the Electrostatic Force Procedure: - Get a retort stand, a test tube clamp, a straw, and two strung pith balls. - Weigh each pith ball. Hold the string in such a way that you are not measuring the weight of the string, nor are you lifting the ball. Try to get two pith balls of almost exactly the same mass. Write it here: m = ____ g = _____kg - Attach the test tube clamp to the retort stand and clamp the straw into it. If the straw isn’t already split on the unfastened end, check the other end or get a pair of scissors if necessary and cut a ½ cm slit into the free end of the straw. Slide the pith ball strings into this slit so that they hang freely beside each other. Your teacher may have an apparatus set up as a model for you. - Slide the strings up or down through the straw slit to make sure the pith balls are lined up perfectly side by side. Also try to ensure that the strings converge to a sharp point at the top where they’re clamped to the straw; i.e., ensure there is no gap between the strings there are the top. - Measure the length of the string from the straw to the center of either pith ball. - Write it here: L = ______ cm = ________ m - Touch your fingers to the pith balls to discharge them (they should be side to side, touching each other). - Rub the ebonite rod with fur. Slowly bring it near the pith balls. 1.What reaction do they first have to the approaching rod? - Allow one or both of the balls to touch the rod. 2. Now describe the reaction of the pith ball(s) to the rod. 3. Explain this reaction. - Touch the pith balls to discharge them again (they should be side to side). Recharge the ebonite rod and quickly tap it on both pith balls simultaneously so they both get charged. Remove the rod. 4. Sketch what the apparatus looks like now, with the pith balls repelling each other. - Somehow carefully measure the distance between the pith balls from their centers. Write it here: - r = ____ cm = _____ m Carefully measure the angle between the strings, or use trig to calc. it. Write it here: = ______. - Touch the pith balls to discharge them again (they should be side to side). Recharge the ebonite rod, more than before and quickly tap it on both pith balls simultaneously so they both get charged. Remove the rod. Aim to get a much larger “r” than last time. Carefully measure the distance between the pith balls from their centers. - Write it here: r = ____ cm = _____ m Carefully measure the angle between the strings. Write it here: = ________. - 5. a) What 3 forces are acting on each ball? b) Draw a F.B.D. for the left pith ball. 6. Since the pith balls are not accelerating, it must be true for each one that the net Force = _______ N. 7. Use the FBD and trig to write a relationship for the horizontal forces: 8. Use the FBD and trig to write a relationship for the vertical forces: 9. Draw a triangle with the two sides being the strings and the base being the distance between the balls. Divide this triangle down the center to create two right angled triangles sitting back to back. Label all the values you measured during the lab on this diagram. * for the first pith ball trial (not so far apart) answer these questions: 10. Show a calculation for Fg. 11. Show a calculation for T (force of tension). 12. Show a calculation for Fq. 13. Calculate q in coulombs. 14. If every electron has a charge of 1.6 x 10-19 C, how many electrons are in each pith ball? * for the second pith ball trial (when they were really far apart) answer these questions: 15. Redo #10-14 for the second pith ball trial. 16. What would it mean if one pith ball was higher than the other (if they were lop-sided)? 17. State some sources of error for this experiment. Lesson #9 – Multiple Forces in Multiple Directions! Part A: Gravity vs. Electricity – Fight! Represent and Reason. Imagine two point-like charged objects of mass m1 and m2 that have electric charges q1 and q2, respectively. Complete the table below that compares their electric and gravitational interactions. Electric Gravitational What property of the objects determines whether they participate in the interaction? What is the direction of the force between the interacting objects? Write an expression for the magnitude of the force that one object exerts on the other. It is an attractive force. How does the magnitude of the force depend on properties of the objects? How does the magnitude of the force depend on the distance between the objects? It is directly proportional to the masses m1 and m2. The electric charge determines whether they will interact. Example: 1. Calculate the force of repulsion between 2 electrons which are separated by 1m. (Remember, the charge on one electron is -1.6x10-19 C). Answer: F on left electron is 2.3x10-28 N [L] 2. Compare your answer to the gravitational force felt between them at separation. (me = 9.11 x 10-31 kg) Answer: F on left electron 5.54x10-71 N [R] Part B – Complete Multiple Charges sketching handout! Lesson #10 – The Electric Field Pre-Lesson Notes Unit: Fields Date: Concept: Electric Field (textbook section 7.3) Definitions: Electric field Electric Field Lines Electric Dipoles Uniform Electric Fields Earth’s Electric Field Electric Fields in Nature Equations: One of the reasons why electrostatic forces seem so mysterious is that they have an effect at a distance, without any objects being in contact. We are familiar with one other force that also has this property. One way of explaining how gravity can have an effect on objects far away is with the idea of a field. We say that the earth has a gravitational field that extends throughout space and that the earth is the source of the field. The earth acts on distant objects “through” its gravitational field. A: In Search of the Electric Field For this investigation you will need an electroscope, an ebonite rod, an acetate strip, and a piece of aluminum foil. An electroscope can be used as an electric field detector. If there is a fairly strong electric field at the location of the metal sphere atop the electroscope, the metal leaves inside will repel each other. 1. Observe. Charge the rod and move the electroscope in the region of space around the rod. Always be careful with the glass electroscope. Don’t let the rod and sphere touch! Describe what you observe. 2. Reason. Based on your observations, where is the electric field around the rod strongest? What happens to its strength further away? What is the source of this field? The sphere of the electroscope acts as a test particle in the rod’s electric field. The properties of a test particle are such that it does not affect the field of the source (its charge and size are small). 3. Observe. Place the electroscope on a table and bring the charged rod nearby. Place the piece of paper between the rod and sphere – don’t let them touch! Observe. Next, place the aluminum foil between the rod and sphere – don’t let them touch! Observe. What effect do different materials have on an electric field? 4. Apply. How would you protect sensitive electronics from unwanted electric fields? 5. Predict and Test. What would happen if there is aluminum foil around your cell phone? How many bars do you think you will get? Try it out! B: Picturing the Electric Field 1. Represent and Reason. For each situation pictured below, represent with arrows the gravitational force or the electric force that a test particle would experience due to the sources at the points shown. In the study of electricity we will always choose all test particles (test charges) to have a positive charge. The planet Earth ● An object with a large negative charge An object with a large positive charge ● ● Earth ● ● ● ● ● ● + ● - ● ● A force field (a field representing forces) is a way of representing the force vectors due to a source at every point in space. This would require many, many vectors so instead we draw field lines. To do this we use a set of rules for electric fields: (1) Electric field lines follow the “path” of the vectors. The vector is always tangent to the field line. (2) E-field lines start on positively charged objects and end on negatively charged objects. (3) The magnitude of the field at a point is represented by the density or concentration of the lines near that point. (4) A corollary to this idea is that the number of lines leaving or terminating on a charged object is proportional to the magnitude of its electric charge. 2. Represent and Reason. The table below gives four examples of point-like charged objects. Study the given examples and draw E-field vectors and E-field lines for each source. (a) a charge of +q (b) a charge of +2q Electric Field Lines Electric Field Vectors ● 3. ● ● ● ● ● ● + ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● + ● ● ● ● ● (c) a charge of -q ● ● ● (d) a charge of -2q ● ● ● ● - ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● - ● ● ● ● ● ● ● ● ● + Apply. A region of space has an electric field as shown to the right. Note that we do not know the location of the source of this field. A particle with a negative charge is placed at point A and then at point B. Draw the force vectors at positions A and B. Explain how to decide the direction of the force on the particle at that position in space. Explain in terms of the field line density how you determined the lengths of the vectors. A B Lesson # 11 – Electric Fields A: Problem Solving To help deepen our understanding of fields, please read the following 2 lessons online. You can use a computer in the classroom or your tablet. http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Intensity http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines In the diagram below, beads 1 and 2 carry charges 1.0 nC (nanocoulombs) and 2.0 nC, respectively. P is just a point in space, not a charge. The electric force exerted on bead 2 by bead 1 is 12 N. 1 2 P First let’s think about the effect of bead 2 on bead 1. 1. Reason. Find the electric force exerted by bead 2 on bead 1. Yes, you have enough information: use a basic law of physics! 2. Calculate. Find the electric field due to bead 2 at the location occupied by bead 1. 3. Reason. Is your D#2 answer greater than, less than, or equal to the electric field due to bead 1 at the location occupied by bead 2? Explain. 4. Reason. Given that beads 1 and 2 feel different fields, it’s reasonable to expect that they also feel different forces. But they don’t! To reconcile this apparent conflict, explain in intuitive terms how beads 1 and 2 can end up experiencing the same force even though they feel different fields. 5. Reason. The charge on bead 1 is now tripled. How does that affect: (a) the electric field due to bead 2 at the location occupied by bead 1? Explain. (b) the electric field due to bead 1 at the location occupied by bead 2? Explain. (c) the force felt by bead 2? Explain. (d) the force felt by bead 1? Explain. Definition of the Electric Field. The electric field describes the electrical effects throughout space of charged source objects on other charged particles. The strength and direction of the electric field only depend on the charges of the source and the position in space. To measure the electric field we need to use a charged particle. The value of the electric field at a point in space (for example, 10N/C) means that if we place a particle at that position, it will experience 10 N of force for each coloumb of its charge. E FE q Use the equation to find the magnitude of the electric field and use a sign convention to show the directions, just like you do with forces. Electric Field Sketches and Calculations 1. How does the magnitude of the total field at the middle x spot compare to the total field at the right x spot? (All distances are equal.) -3 x +1 x 2. Try again with a new set of charges: x -3 x +3 +2 3. What is the ratio of the total fields at the two marked spots below? x -2 x +1 +1 4. (a) Calculate the electric field midway between the +1 and the -3 charge. (All charges are in nC.) +4 +1 30 cm -3 x 30 cm 30 cm (b) Calc. FQ on a 0.5 C charge placed at the x. (c) Calc. FQ on a 20 mC charge placed at the x. Answers: (1) 6:1 (2) 13:19 (3) 59:37 (4) 500 N/C 250 N 10 N Lesson #12: Non-Collinear Fields If charges are not in a perfectly straight line, we still need to add up forces on them, or the fields around them, but now we’ll need to use trigonometry and vector components. E.g. #1) A - 3.0 nC charge is 5.0 cm above the midpoint between two +4.0 nC charges which are 10 cm apart, as shown: a) Calculate the net force on the negative charge. b) How would you calculate the ratio of the net force on each charge? c) What is the total field at the midpoint between the two positive charges? d) Imagine this diagram now as a side view, with the bottom charges sitting on the ground. What new charge would they need in order to levitate the -3.0 nC charge if it weighs 0.25 g? -3 +4 +4 E.g. #2: Three +4.0 nC charges are arranged at the corners of a square. a) What is the total field at the missing corner? b) What is the net force on the bottom right charge? +4 +4 +4 Lesson #13: Fields and Potential Energy Pre-Lesson Notes Unit: Fields Date: Concept: Potential Difference, Electric Potential, Electric Potential Energy due to Point Charges (section 7.4, 7.5) Definitions: Work and Electric Potential Difference Electric Potential Energy Electric Potential Electric Potential Due to a point charge Electric Potential energy of two point charges Equations: Electric Potential: Electric Potential Energy: Electric Potential Due to a point charge: Electric Potential energy of two point charges: Imagine forcing two repelling positive charges toward each other. This would require work. By doing work to the charges, you give them potential energy. This situation is very similar to gravitational potential energy. Consider that the force on an object near the earth can be given as Fg = m⋅g Work is F x d, so if we lift the charge up (in line with gravity) m⋅g x h (where h is the distance or height that we lift it.) Doing work results in giving something energy; that’s why we then get the equation: Eg = m⋅g⋅h. So now let’s imagine that we do work against the electrostatic force. This time, F x d would be FQ and the distance we now call r. So the equation becomes: (Remember that energy isn’t a vector so never put a “hat” on the symbol E for energy; whereas the symbol for field 𝜀 should have a vector “hat”: 𝜀⃑ ) Questions: 1. Calculate the electric potential energy stored in an electron and a proton which are: (a) 0.5 m apart; and (b) 1 m apart. 2. Calculate the work done to move the electron from 0.5 m away to 1 m away. 3. The proton and an electron are now released from rest 1 m apart. Estimate the speed of the electron when they are 0.5 m apart. (Assume the proton doesn’t move.) Answers: -4.6 x 10-28 -2.3 x 10-28 -2.3 x 10-28 22.5 m/s Electrostatic Work and the Conservation of Energy Although energy (E) is not a vector, it can be (+) or (–), and this has some significance. Consider the energy in the following two scenarios, and how much work is required to change from the 1st to the 2nd (all charges are in mC) : 1st Eq1 = +2 +4 2m 2nd You push +2 1m +4 You push Eq2 = Work = ∆E = E2 – E1 (done to go from 1st scenario to 2nd) W = Eq2 – Eq1 = =+ J (the work needed is (+) so you DO work to make it happen; i.e., you need to INPUT energy.) Now let’s consider instead what happens when they start close together and are released: 1st (now remove the barriers) +2 1m +4 2nd +2 2m +4 Work = ∆E = E2 – E1 (done to go from 1st scenario to 2nd) W = Eq2 – Eq1 = =J (the work needed is (-) so you GET work from them; i.e., an energy OUTPUT is possible!) These charges will accelerate away from each other. Using the “Law of the Conservation of Energy” we can calculate their speeds at any place. Example: What is the speed of each when they are at 2.0 m apart? Lesson #13: Using the Conservation of Energy for Field Calculations Imagine two charged particles, such as two electrons, which find themselves close to each other. They will obviously repel each other, and if they are free to move, they will accelerate away from each other. Their speed (v) is increasing, so the kinetic energy (Ek) is . Their separation (r) is increasing, so the electrostatic energy (Eq) is . This kinetic energy must come from somewhere. Indeed, the change in Ek comes from Eq. Recall that Etotal1 (initial) = Etotal 2 (final) Notice how the equation for electrostatic energy has 2 q’s in it, but the equation for kinetic energy only has one m in it. So the two electrons (at first being at rest) will have a total energy of: Total energy = 1st electron’s kinetic E + 2nd electron’s kinetic E + electrostatic energy between them (you see, Eq = k ∙ q1 ∙ q2 , so this one equation contains and takes care of both q1 and q2.) r But after accelerating for a little while, the total energy will be: (and notice here how both m’s are present, using two expressions; and both q’s are present too, but using only one expression. This way, both types of energy are represented for both particles.) Thus, ET1 = ET2 becomes : Since they’re both electrons, q1 = q2 and m1 = m2 and they will have the same speeds too. So you can re-arrange the last equation to find that: Examples: 1. Using the mass and charge of an electron (found in your textbook or on the physics classroom wall,) calculate the speed two electrons will have if they are released from r = 1.00 mm apart, when they are: a. 2.00 mm apart b. 10 cm apart c. 1.00 m apart d. Their maximum possible speed (at r2 = infinity.) Yes, it’s true that they can accelerate forever and yet reach a finite final speed! More on this later… 2. Try a similar solution backwards: starting from ET1 = ET2 find how close two electrons would need to initially be so that they will eventually reach speeds of one thousandth the speed of light. Answers: 1b) 501 m/s 1d) 503.6 m/s 2. r = 2.8 nm 8 O A “Repulsive” Question Imagine that two oxygen atoms which gained two valence electrons are repeatedly fired straight at each other in a particle accelerator, always with the same speed as each other. a) If they are fired from 1.0 m away, and they get to a minimum separation of 1.0 cm before they are stopped by their mutual repulsion, what initial velocity must they have had? (Answer: 1.85 m/s) b) If this time they are fired from really really far away (i.e., EQ is approx. zero), what speed would they need so that they will end up 1.0 mm away before stopping? (Answer: 5.9 m/s) c) This time, they are somehow pushed very close together and released. What initial separation would be necessary so that they will end up going a final maximum speed of 3.0 x 108 m/s? (Answer: 3.8 x 10-19 m) 16 Lesson #14: Milli-kan…Can you? Pre-Lesson Notes Unit: Fields Date: Concept: The Millikan Oil Drop Experiment Definitions: Fundamental physical constants Elementary charge Charge of a proton Explanation: Millikan’s experiment Equations: Electric Potential due to a point charge: Electric Potential: In 1909, Robert Millikan devised the first method for direct measurement of the electric charge of a single electron. Using the apparatus described below, he discovered that the electric charges on the drops were all whole-number multiples of a lowest value, the elementary electric charge itself, and thus that electric charge exists in basic natural units. Typical Millikan questions: (try them now!) 1. Calculate the charge on a 0.050 g oil drop which is suspended between two plates 6.0 cm apart, using a potential difference of 24 V. (Answer: 1.2 µC) 2. What voltage would accelerate a 0.00020 g oil drop which has 6000 excess electrons, upwards at 2.2 m/s2 between two plates 1.0 cm apart? (25 MV) Millikan Oil Drop Example An oil droplet with a mass 5.0 x 10-16 kg accelerates at 2.20 m/s2 upwards between two charged metal plates. The top plate is negative and the positive plate is 3.00 cm below it. A 9.0 V battery is connected to them. a) Calculate the charge on the oil droplet. b) What excess or deficit of electrons would the droplet have? Lesson # 15 – Magnetic Fields Pre-Lesson Notes Unit: Fields Date: Concept: Magnets and Electromagnets (section 8.1) Definitions: Auroras Permanent Magnets Magnetic Field Lines Earth’s magnetic field Principle of Electromagnetism Fields can describe the force of any interaction that has an effect throughout space. Another force you have no doubt played with is magnetism. We need to develop some new rules and ideas to describe magnetic effects. For example, with magnetism we won’t use test particles we will use test-compasses! A: Searching for a Field You need a bar magnet and a compass. A compass is a special device whose north end points in the direction of the magnetic field at that position in space. 1. Observe. Check that your compass works – it should jiggle at the slightest touch. Place the compass at each point in the region around the bar magnet as shown below. Draw an arrow showing the magnetic field vector ( B ) at that position in space. 2. Observe and Reason. How does the interaction of the magnet and compass depend on the distance from the bar magnet and its location around the bar magnet? How can you tell? Modify any of your field vectors if necessary. 3. Observe. Use the applet: http://falstad.com/vector3dm/. Set the field selection to “solenoid”. Set the display to “Field Vectors” and then “Field Lines”. It may be easier to see if you choose “Show Y Slice”. We can’t examine the field inside the bar magnet with a compass, but we can study it using a solenoid. A solenoid is a coil of wire whose field is very similar to that of a bar magnet. Sketch the magnetic field lines around and inside the bar magnet. Add arrows to the field lines to match your field vectors above. 4. Explain. Point out two regions where the field strength is particularly strong and weak. How can you tell? B: The Birds and the Bees of Magnetism You need an alligator wire, a compass and a 9V battery. WARNING: Never clip both ends of the wire to the battery – it will overheat! 1. Observe. Clip only one end of the wire to the battery. Move the middle of the wire above the compass. Describe your observations. Is there any evidence for a magnetic interaction between the compass and the wire? Reminder: The conventional current flows from the positive terminal to the negative terminal of the battery. We can imagine positive charges flowing through the wire in a direction opposite to that of the actual electron flow. We understand that it is only electrons that flow in most electrical circuits, but we are stuck with using the conventional current for historical reasons. 2. Observe. Place the wire over the top of the compass such that it is parallel to the needle. Briefly touch the second alligator clip to the other battery terminal – DO NOT CLIP IT ON! Hold the wire in place only long enough for the needle to stabilize. Record your observations. Illustrate the direction of the conventional current flow and the compass needle. Is there any evidence for the existence of a new magnetic field? 3. Observe. Now hold the compass just above the wire. Repeat your observations. Magnetic fields (B) are created by moving electric charges. Charges at rest (static charges) do not create magnetic fields. To illustrate the motion of the charges and the magnetic fields, we need a new set of symbols. For a vector pointing out of a page draw: For a vector pointing into a page draw: 4. Observe and Explain. Use the applet: http://falstad.com/vector3dm/. Set the field selection to “current line”. Set the display to “Field Vectors”. Select “Show Z Slice”. Sketch a sample of the field vectors around the wire. Is the current flowing into or out of the page? Does this representation agree with your compass observations? Explain. Field Vectors 5. Observe. Now set the display to “field lines”. Sketch the field lines and the current in the box to the right. Which way should the arrows on the field lines point? Now draw them! Field Lines The Right Hand Rule for Current Flow relates the motion of positive charges (conventional current) and the magnetic field they create. When the thumb of your right hand points in the direction of the conventional current, your fingers curl in the direction of the magnetic field lines. 6. Predict Imagine the wire is oriented across our page as shown below. It is difficult to draw the field lines from this point of view. We can simply show the direction of the field vectors as they pass through the page surface. Explain whether the field lines will point into or out of the page in the region around the wire. 7. Test. Switch back to field vectors and choose “Show X Slice”. You may need to rotate the image to understand what you are seeing. Explain any differences with your prediction. Draw field vectors illustrating your observations using and . Electromagnetism: Right-Hand Rules Summarized Note: All right hand rules are based on the assumption that current travels from positive to negative. For electron flow from negative to positive, simply use the left hand instead! RHR for a Straight Conductor Your thumb points in the direction of the current and your curled fingers describe the direction of the produced magnetic field. RHR for a Current Carrying Loop (or Helix) Your fingers curl around in the direction of the looping current and your thumb then points to the North side of the magnetic field produced. RHR for Electromagnetic Force Your thumb points in the direction of the current and your fingers point in the direction of the magnetic field. Then, the force on the conductor is directed outward from the palm of your hand. RHR for Moving Charge through a Uniform Field Your thumb points in the direction of the velocity of positive charge, your fingers point in the direction of the magnetic field. Then the force on the conductor is directed outward from the palm of your hand. Lenz's Law The magnetic field of an induced current always opposes the change in magnetic field that is causing the induced current Lesson #16 – Magnetic Forces on Charges Pre-Lesson Notes Unit: Fields Date: Concept: Magnetic Force on Moving Charges and on current-carrying conductor (section 8.2, 8.3) Definitions: Charge in a magnetic field Right hand rule for a moving charge in a magnetic field Equations: Magnetic force (two equations) A: Charges at Rest This is a class demonstration. We want to study the interaction between a magnetic field and a charged object that is stationary. An electromagnet will create the magnetic field and rubbing a balloon will create a static electric charge. 1. Observe. For each situation, record your observations in the chart below. Charged balloon, magnet turned Charged balloon, magnet current Charged balloon, magnet turned on reversed off 2. Reason. Is there any evidence from your observations for the existence of a magnetic interaction between a charged object at rest and a magnetic field? Justify your answer using your observations. B: Charges on the Move http://webphysics.davidson.edu/physlet_resources/bu_semester2/c12_force.html This is a class demonstration. We want to study the interaction between a magnetic field and moving charges. An old TV will produce moving charges. A permanent magnet will provide the magnetic field. 1. Observe. The bar magnet is held perpendicular to the beam with the north pole beside the TV. To a good approximation, the magnetic field points straight out from the end of the bar magnet. Record your observations. Complete the diagram showing the velocity of the current, the magnetic field and the force acting on the charges. 2. Observe. The bar magnet is held perpendicular to the beam with the north pole above the TV. Record your observations and complete the diagram. 3. Observe. The bar magnet is held parallel to the TV with the north pole pointing towards you. Record your observations and complete the diagram. Why do the field lines look so different in this example? 4. Reason. Two conditions must be met in order for magnetic fields to exert a force on a charged object. From your observations in parts A and B, what are these two conditions? The force that results from the movement of a charged object in a magnetic field is given by: Fm q v B sin , where q and v are the charge and speed of the object, B is the strength of the magnetic field the object is moving through and θ is the angle between the magnetic field lines and the velocity. The direction of the force is determined by the Right Hand Rule for the Magnetic Force on Positive Charges: The thumb points in the direction of the velocity of the object with a positive charge, the fingers in the direction of the magnetic field lines and the palm of your hand points in the direction of the force. If the object has a negative charge, just reverse you answer from the right-hand rule. 5. Reason. Use the equation for the magnetic force to verify your conditions from the previous question. Explain. 6. Predict. Use the right hand rule to predict how the beam will deflect when the bar magnet is held perpendicular to the beam with the south pole on the left side of t he tube. Draw a picture to illustrate your prediction. We will try this out as a class. Move on for now. C: The Motion of Charges in a Magnetic Field Consider an electron traveling with an initial velocity v moving through a magnetic field that is oriented into the page as shown to the right. 1. Reason. What is the direction of the force acting on the electron? Explain and draw the force on the diagram. prediction v ● B 2. Reason. A short moment later, where might the electron be located? Draw it and its new velocity vector. At this later moment, what will the direction of the force be? Draw this on the diagram. Explain. 3. Predict. Assume the electron never leaves the magnetic field. What will the complete path of the electron look like? How will the directions of the velocity and force vectors compare? What type of motion results? Explain. 4. Reason. How much work does the magnetic field do on the electron? How does the electron’s speed and kinetic energy change? Explain. D: Bubbles, Tiny Bubbles The image to the right shows the paths of particles travelling through a bubble chamber, a chamber of superheated fluid. A charged particle travelling through the fluid causes a bubble to form. This leaves a bubble trail behind the particle. A strong magnetic field passes through the bubble chamber and is oriented out of the page. The trails of five particles are labeled. A E C D B 1. Apply. Indicate the charge of the particle that produced each trail. Explain how you can tell. 2. Apply. Rank the momentum of the particles that produced each trail. Explain. 3. Apply. Why do you think trails C and D spiral inwards? Lesson #17: Electromagnetic Disturbances We have studied the creation of both electric and magnetic fields separately, but now it is time to build up the complete picture of what happens when current flows through a wire. A: The Physics of a Wire Consider a wire connected between the negative and positive terminals of a power supply. A steady current travels through the wire. The diagram to the right illustrates this situation and shows two sets of field lines. 1. Represent. Choose a colour to represent the conventional current. Draw an arrow that represents the flow of conventional current. 2. Represent and Explain. Choose a colour to represent the electric field. Explain how you decide which lines are the electric field lines. Label and colour code these lines. Add arrows showing the direction of the field lines. - + 3. Represent and Explain. Choose a colour to represent the magnetic field. Explain how you decide (as if you haven’t done question #2) which lines are the magnetic field lines. B: A Closer Look Now imagine that we zoom in to a small region of space located above the middle of the wire. The diagram below illustrates this along with the electric and magnetic field lines. Note that one field points in or out of the page and is represented with the circles. 1. Represent. Complete the diagram by: (a) labelling and colour coding the electric and magnetic field lines and (b) showing the direction of all the field lines and current 2. Reason. How do the directions of the electric and magnetic field lines compare in this region of space? wire 3. Explain. A positive test charge is placed at rest in this region of space. In this investigation, we will ignore the effects of the magnetic field since they are relatively small. What will happen to the positive charge which is initially is at rest? Explain and draw vectors representing any forces. Below is a third diagram showing the same region of space as above. But for this example, we have switched the polarity of the power supply, reversing the positive and negative terminals shown in the first diagram. 4. Represent and Explain. Complete the diagram as you did in question #4. Describe the how the fields have changed due to the reversed terminals. 5. Explain. What will happen to the positive test charge which is initially at rest? wire C: Shaking Things Up Now imagine that we continuously vary the polarity of each terminal of the wire from positive to negative. 6. Reason. Describe the motion of the charges in the wire. 7. Reason. Describe the motion of the positive test charge as the polarity varies. 8. Observe. When the polarity of the wire changes, the fields don’t change everywhere immediately. They begin to change closest to the wire and then spread outwards through space. Use the simulation: http://falstad.com/emwave1. Describe what this changing pattern looks like. 9. Observe. Freeze the simulation. Static electric or magnetic fields drop off in a steady way at positions father from the sources. Is this happening in this situation? Explain. An alternating current produces a changing pattern of electric and magnetic fields that spreads outwards from the wire through space. In 1864, James Clerk Maxwell studied these phenomena in detail and mathematically described the wave-like motion of these oscillating fields. From his calculations he determined the outwards speed of the rippling fields to be given by: v2 4k o Where k is Coulomb’s constant (k = 8.99 x 109 Nm2/C2) and o is a similar constant for magnetism (o = 1.257 x 10-6 Tm/A). 10. Predict. Determine the velocity of this wave including units (1A = 1C/s) (1T = 1 kg/Cs). What are the implications of this result? Rules for Electromagnetic Waves. What we have discovered is an electromagnetic wave which works according to these ideas: 1) An EM wave is created by the acceleration of electrical charges (our varying current in this case). 2) Energy from this wave may be absorbed by causing other charges to accelerate (our distant charge). 3) The frequency of the wave’s oscillation and the properties of the object containing the distant charge determine how much that charge will respond to the wave and how much energy it will absorb. (a kind of resonance) Lesson #18: Understanding EM Waves A genuine, real-life EM wave is very tricky to visualize – it has much more detail than a simple water wave. To help us picture what takes place when an EM wave travels through space we will use the line model of an EM wave. A: The Line Model of an Electromagnetic Wave To the right is a screen capture from the simulation we used in our previous investigation. It represents a two-dimensional slice in space and the field values along that slice. To help us talk about this image we will R define the vertical direction as the ydirection and the x- and z-directions as the horizontal plane. The wire is positioned along the y-axis with its middle at the origin. This image represents the field values in the xy-plane. G R G R G y x z + x E 1. Represent. Imagine we draw a line along the x-axis starting at the surface of the wire. Remember that the simulation does not show the field strength using the arrow lengths. Sketch a graph of the electric field vs. position along the +x-axis (the white line). Do the same for the magnetic field. Upwards is positive for the E-field and outwards is positive for the B-field. + x B - We can put these two graphs together and illustrate the field vectors to give us the line model of an electromagnetic wave. This is a picture of the electric and magnetic field vectors along one line through space at one moment in time. One extraordinary feature of an EM wave is the fact that the amplitude of the peaks decreases very, very little allowing EM waves to travel extremely far through space- even across the entire universe! 2. Reason. Examine the line model of an EM wave shown to the right. Rank the magnitude of the electric field at points A, B, C and D in space. Explain. A E B C D B 3. Reason. In what direction is the EM-wave propagating (travelling)? Draw an arrow on the diagram above. 4. Reason. How do the directions of the electric and magnetic fields compare with the direction of propagation? 5. Reason. An electron is located at point C in the diagram above. What will initially happen to it? B: Wave, Meet Conductor. Conductor, Meet Wave. Imagine that the EM wave depicted above in question #2 is a radio wave. It now passes by a long, thin conducting wire that is oriented along the y-axis. wire 1. Reason. As the wave propagates (travels) past the wire, would the electric field due to the radio wave cause the charges in the wire to move? If so, would the charges move in a direction along the wire? Explain. 2. Reason. As the wave propagates past the wire, would the magnetic field due to the wave cause the charges in the wire to move in a direction along the length of the wire? Explain. 3. Reason. Imagine a light bulb is connected to two halves of the same wire in the middle. The bulb and wire are again placed in the path of the previous radio wave and is oriented parallel to the yaxis. Would the bulb ever glow? Explain. wire 4. Reason. Would it glow if the wire was oriented parallel to the z-axis? Explain. 5. Reason. Would it glow if the wire was oriented parallel to the x-axis? Explain. An antenna is a conductor designed to capture the energy from an electromagnetic wave. The fields of the wave cause charges in the antenna to accelerate transferring electromagnetic energy from the wave into the electrical energy of the moving charges. An electric current is created or induced and this current can be passed into a circuit providing a radio station signal, a Bluetooth signal, a wireless router signal, a TV signal, a cellphone signal, … so many possible signals! 6. Predict. In order to best detect the oncoming radio wave (that is, to maximize the induced current in the circuit), how should the antenna be oriented relative to the wave? Explain. Lesson #19: Polarization When a light wave is received by our eyes, we cannot tell which direction the electric field vectors point in – but some creatures can! Certain materials react differently to electric or magnetic field vectors in different directions. A: Through the Polarizing Glass You will need a pair of polarizing filters. Be careful since some of them are glass! 1. Observe. Look at the room lights through one of the polarizing filters. Describe how the filter affects what you see. Does rotating the filter have any effect? 2. Observe. Hold a second polarizing filter in front of the first, and look at the room lights again. Describe how the filter affects the light you see. How does rotating one of the filters with respect to the other affect what you see? A light beam is linearly polarized when the electric field in all parts of the beam oscillates along a single axis. The direction along which the electric field oscillates is called the direction of polarization of a light beam. An unpolarized beam is made up of many rays of light, each of which have the electric field oscillating in different, random directions. The light transmitted by a polarizing filter (or polarizer) depends upon the relative orientation of the polarizer and the electric field in the light wave. Every polarizer has a direction of polarization, which is often marked by a line on it. The electric field of the transmitted wave is equal to the component of the electric field of the incident wave that is parallel to the direction of polarization of the polarizer. Representations of all these can be used to draw a polarization diagram. 3. Interpret. Describe what is happening according to the polarization diagram to the right. y E y x z E x z polarized beam filter with horizontal direction of polarization unpolarized beam Reason. Do the room lights produce polarized light? Explain how you can tell from your observations. Draw a polarization diagram illustrating how the room lights can be completely blocked by the filters. x 5. Observe. Describe other sources of polarized light from around the classroom or amongst your belongings. 6. Reason. Describe how you should orient the polarizers with respect to one another so that the light transmitted through the polarizers would have (1) maximum intensity or (2) minimum intensity. 7. Represent. Draw a polarization diagram for the situation of minimum intensity you described above. x When two polarizers are oriented with respect to one another such that the light is at a minimum intensity, the polarizers are said to be crossed. 8. Reason. Suppose that you had a polarizer with its direction of polarization marked. How could you use this polarizer to determine the direction of polarization of another unmarked polarizer? Explain your reasoning. 9. Reason. A beam of polarized light is incident on a polarizer, as shown in the side view diagram. The direction of E-field of the light makes an angle with respect to the polarizer’s direction of polarization (see front view diagram). The amplitude of the electric Polarizer Incident light E vector Side View Front view field of the incident light is Eo. Write an expression for the amplitude of the transmitted component of the electric field. Represent this in a polarization diagram. x 10. Predict. A beam of unpolarized light travels through a pair of crossed filters – no light is transmitted from the last filter. What will happen when a third filter whose direction of polarization makes a 45o angle with both filters is placed between two crossed filters? Explain and draw a polarization diagram. We will test this as a class.