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Syllabus and teaching strategy General Physics (PHY 2140) Introduction Lecturer: Dr. Alan A. Sebastian, Physics Building Phone: 313-577-2720 (to leave a message with a secretary) e-mail: [email protected], Web: http://www.physics.wayne.edu/~alan Office Hours: TBD, Physics Building, , or by appointment. Grading: Reading Quizzes Quiz section performance/Homework Best Hour Exam Second Best Hour Exam Final ¾ Syllabus and teaching strategy ¾ Electricity and Magnetism • Properties of electric charges • Insulators and conductors • Coulomb’s law bonus 10% 25% 25% 40% Reading Quizzes: It is important for you to come to class prepared! Homework and QUIZ Sessions: The quiz sessions meet once a week; quizzes will count towards your grade. Hour Exams and Final Exam: There will be THREE (3) Hour Exams and one Final Exam. Online Content: Lectures will be made available to you as a supplemental reference. Lecture 1. Chapter 15 7/31/2007 1 7/31/2007 2 15.1 Properties of Electric Charges Discovery Introduction Knowledge of electricity dates back to Greek antiquity (700 BC). Began with the realization that amber (fossil) when rubbed with wool, attracts small objects. This phenomenon is not restricted to amber/wool but may occur whenever two nonnon-conducting substances are rubbed together. Observation of “Static Electricity” Electricity” A comb passed though hair attracts small pieces of paper. An inflated balloon rubbed with wool. “Electrically charged” charged” Rub shoes against carpet/car seat to charge your body. Remove this charge by touching another person/a piece of metal. Two kinds of charges Named by Benjamin Franklin (1706(1706-1790) as positive and negative. negative. Like charges repel one another and unlike charges attract one another. another. 7/31/2007 3 7/31/2007 4 1 15.1 Properties of Electric Charges Nature of Electrical Charge 15.1 Properties of Electric Charges Quantization Origin of charge is at the atomic level. Robert Millikan found, in 1909, that charged objects may only have have an integer multiple of a fundamental unit of charge. Nucleus : “robust” robust”, positive. positive. Electrons : mobile, negative. negative. Usual state of the atom is neutral. neutral. Charge has natural tendency to be transferred between unlike materials. Electric charge is however always conserved in the process. Units Charge is not created. created. Usually, negative charge is transferred from one object to the other. 7/31/2007 5 15.2 Insulators and Conductors –Material classification 7/31/2007 7/31/2007 6 Identify substances or materials that can be classified as Conductors ? Insulators? Glass, Rubber are good insulators. Copper, aluminum, and silver are good conductors. Semiconductors are a third class of materials with electrical properties somewhere between those of insulators and conductors. In SI, electrical charge is measured in coulomb ( C). The value of |e| |e| = 1.602 19 x 10-19 C. Mini-quiz: Materials/substances may be classified according to their capacity capacity to carry or conduct electric charge Conductors are material in which electric charges move freely. Insulator are materials in which electrical charge do not move freely. Charge is quantized. quantized. An object may have a charge ±e, or ± 2e, or ± 3e, etc. but not ±1.5e. Proton has a charge +1e. +1e. Electron has a charge –1e. 1e. Some particles such a neutron have no (zero) charge. A neutral atom has as many positive and negative charges. Silicon and germanium are semiconductors used widely in the fabrication of electronic devices. 7 7/31/2007 8 2 15.2 Insulators and Conductors – Charging by Conduction. 15.2 Insulators and Conductors – Earth/Ground. Consider negatively charge rubber rod brought into contact with a neutral conducting but insulated sphere. Some electrons located on the rubber move to the sphere. Remove the rubber rod. Excess electrons left on the sphere. It is negatively charged. charged. This process is referred as charging by conduction. conduction. 7/31/2007 9 15.2 Insulators and Conductors – Charging by Induction. 7/31/2007 10 15.2 Insulators and Conductors – Charging by Induction. Consider a negatively charged rubber rod brought near a neutral conducting sphere insulated from the ground. Repulsive force between electrons causes redistribution of charges on the sphere. Electrons move away from the rod leaving an excess of positive charges near the rod. Connect a wire between sphere and Earth on the far side of the sphere. Repulsion between electrons cause electrons to move from sphere to Earth. Disconnect the wire. The sphere now has a positive net charge. This process is referred as charging by induction. induction. Charging by induction requires no contact with the object inducing the charge. 7/31/2007 When a conductor is connected to Earth with a conducting wire or pipe, it is said to be grounded. grounded. Earth provides a quasi infinite reservoir of electrons: can accept or supply an unlimited number of electrons. Consider a negatively charged rubber rod brought near a neutral conducting sphere insulated from the ground. Repulsive force between electrons causes redistribution of charges on the sphere. Electrons move away from the rod leaving an excess of positive charges near the rod. Connect a wire between sphere and Earth on the far side of the sphere. Repulsion between electrons cause electrons to move from sphere to Earth. Disconnect the wire. The sphere now has a positive net charge. This process is referred as charging by induction. induction. Charging by induction requires no contact with the object inducing the charge. 11 7/31/2007 Q: How does this mechanism work if we use a positively charged glass rod instead? 12 3 15.2 Insulators and Conductors – Polarization. Mini-quiz A positively charged object hanging from a string is brought near near a non conducting object (ball). The ball is seen to be attracted to the the object. Polarization is realignment of charge within individual molecules. Produces induced charge on the surface of insulators. how e.g. rubber or glass can be used to supply electrons. 1.Explain 1.Explain why it is not possible to determine whether the object is is negatively charged or neutral. 2.What 2.What additional experiment is needed to reveal the electrical charge charge state of the object? ? 7/31/2007 13 Explain why it is not possible to determine whether the object is negatively charged or neutral. Bring a known neutral ball near the object and observe whether there is an attraction. ? + Attraction between a charged object and a neutral object subject to polarization. ++ -++ - - 7/31/2007 14 What additional experiment is needed to reveal the electrical charge state of the object? Attraction between objects of unlike charges. 7/31/2007 Two Experiments: Two possibilities: + + 15 7/31/2007 Bring a known negatively charge object near the first one. If there is an attraction, the object is neutral, and the attraction is achieved by polarization. -+++ -- -++ ++ 0 16 4 15.3 Coulomb’s Law - Observation Charles Coulomb discovered in 1785 the fundamental law of electrical force between two stationary charged particles. An electric force has the following properties: Inversely proportional to the square of the separation, separation, r, between the particles, and is along a line joining them. Proportional to the product of the magnitudes of the charges |q1| and |q2| on the two particles. Attractive if the charges are of opposite sign and repulsive if the charges have the same sign. sign. 15.3 Coulomb’s Law – Mathematical Formulation q q F = ke 1 2 2 r ke known as the Coulomb constant. Value of ke depends on the choice of units. SI units q1 Force: the Newton (N) Charge: the coulomb ( C). Current: the ampere (A =1 C/s). Distance: the meter (m). Experimentally measurement: ke = 8.9875×109 Nm2/C2. Reasonable approximate value: ke = 8.99×109 Nm2/C2. q2 r 7/31/2007 17 7/31/2007 18 Example Charge and Mass of the Electron, Proton and Neutron. 7/31/2007 Particle Charge ( C) Mass (kg) Electron -1.60 ×10-19 9.11 ×10-31 Proton +1.60 ×10-19 1.67 ×10-27 Neutron 0 1.67 ×10-27 1e = -1.60 ×10-19 c Takes 1/e=6.6 ×1018 protons to create a total charge of 1C Number of free electrons in 1 cm3 copper ~ 1023 Charge obtained in typical electrostatic experiments with rubber or glass 10-6 C = 1 μc A very small fraction of the total available charge 19 7/31/2007 20 5 15.3 Coulomb’s Law – Remarks Mini-Quiz Name the first action at a distance force you have encountered in physics so far. The electrostatic force is often called Coulomb force. It is a force (thus, a vector): vector): a magnitude a direction. r F 21 + q1 + q2 r F 21 + q1 F 21 F 21 q2 Second example of action at a distance. 7/31/2007 21 7/31/2007 22 Question: The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3x10-11 m. Find the magnitude of the electric force that each particle exerts on the other. Example: Electrical Force Question: The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3x10-11 m. Find the magnitude of the electric force that each particle exerts on the other. Observations: We are interested in finding the magnitude of the force between two particles of known charge, and a given distance of each other. The magnitude is given by Coulomb’ Coulomb’s law. F = ke q1 q2 r2 q1 =-1.60x10-19 C q2 =1.60x10-19 C r = 5.3x10-11 m 7/31/2007 23 7/31/2007 24 6 Question: The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3x10-11 m. Find the magnitude of the electric force that each particle exerts on the other. Observations: We are interested in finding the magnitude of the force between two particles of known charge, and a given distance of each other. The magnitude is given by Coulomb’ Coulomb’s law. q1 =-1.60x10-19 C q2 =1.60x10-19 C r = 5.3x10-11 m Solution: Fe = ke e 2 r2 = 8.99 × 10 9 Nm 2 C2 (1.6 ×10 ( 5.3 ×10 −19 −11 ) m) C Superposition Principle From observations: one finds that whenever multiple charges are present, the net force on a given charge is the vector sum of all forces exerted by other charges. Electric force obeys a superposition principle. principle. 2 2 = 8.2 × 10 −8 N Attractive force with a magnitude of 8.2x10-8 N. 7/31/2007 25 7/31/2007 Consider three point charges at the corners of a triangle, as shown shown below. Find the resultant force on q3. Example: Using the Superposition Principle y Consider three point charges at the corners of a triangle, as shown shown below. Find the resultant force on q3 if q1 = 6.00 x 10-9 C q2 = -2.00 x 10-9 C q3 = 5.00 x 10-9 C y q2 3.00 m q1 + 7/31/2007 q2 4.00 m F32 + q3 F31 - 3.00 m q1 + F31 - 26 37.0o 4.00 m F32 + q3 37.0o x Observations: The superposition principle tells us that the net force on q3 is the vector sum of the forces F32 and F31. The magnitude of the forces F32 and F31 can calculated using Coulomb’ Coulomb’s law. x 27 7/31/2007 28 7 Lightning Review Consider three point charges at the corners of a triangle, as shown shown below. Find the resultant force on q3. y q2 3.00 m F32 = ke q3 q2 F31 = ke q3 q1 r 37.0o + q3 5.00 m q1 + Solution: F31 F32 4.00 m 2 r2 x ( 5.00 ×10 C )( 2.00 ×10 C ) = 5.62 ×10 −9 = 8.99 ×109 = 8.99 ×109 Nm 2 C2 Nm 2 C2 1. Properties of electric charge 9 two types: positive and negative 9 always conserved and quantized ( 4.00m ) ( 2. Insulators and conductors 9 charges move freely in conductors; opposite is true for insulators 9 conductors can be charged by conduction and induction; insulators can be polarized −9 )( 5.00 ×10−9 C 6.00 ×10−9 C ( 5.00m ) −9 2 2 ) = 1.08 ×10 −8 N N Fx = − F32 + F31 cos 37.0o = 3.01× 10−9 N Review Problem: OperatingOperating-room personnel must wear special conducting shoes while working around oxygen. Why? What might happen if personnel wore ordinary rubber shoes (sneakers)? Fy = F31 sin 37.0o = 6.50 × 10−9 N F = Fx2 + Fy2 = 7.16 × 10−9 N θ = 65.2o 7/31/2007 29 Recall that units can be manipulated: q1 q2 r 2 [ F ] = [ ke ] [ Newton] = [ ke ] Electric forces act through space even in the absence of physical contact. Suggests the notion of electrical field (first introduced by Michael Faraday (1791(1791-1867). An electric field is said to exist in a region of space surrounding a charged object. If another charged object enters a region where an electrical field is present, it will be subject to an electrical force. [ q1 ][ q2 ] 2 [r ] [Coulomb][Coulomb] 2 [ meter ] [ ke ] = N ⋅ m 7/31/2007 30 15.4 Electric Field - Discovery Example: Fun with units F = ke 7/31/2007 2 C2 31 7/31/2007 32 8 15.4 Electric Field – Quantitative Definition 15.4 Electric Field – Quantitative Definition (2) Direction defined as the direction of the electrical force exerted on a small positive charge placed at that location. A field : generally changes with position (location) A vector quantity : magnitude and direction. Magnitude at a given location Expressed as a function of the force imparted by the field on a given test charge. E = E F + E qo - - - - - - + - + - - - 7/31/2007 33 15.4 Electric Field – Electric Field of a Charge “q” Given One finds 7/31/2007 F = ke E = ke + + + + + + + + + + + - + + + + + + + + + + + + 7/31/2007 34 • If q>0, field at a given point is radially outward from q. q qo r r2 + q q qo E • If q<0, field at a given point is radially inward from q. r r2 q 35 7/31/2007 E qo 36 9 Problem-Solving Strategy Example: An electron moving horizontally passes between two horizontal planes, the upper plane charged negatively, and the lower positively. A uniform, upwardupward-directed electric field exists in this region. This field exerts a force on the electron. Describe the motion of the electron in this region. Electric Forces and Fields Units: For calculations that use the Coulomb constant, ke, charges must be in coulombs, and distances in meters. Conversion are required if quantities are provided in other units. units. Applying Coulomb’ Coulomb’s law to point charges. It is important to use the superposition principle properly. Determine the individual forces first. Determine the vector sum. Determine the magnitude and/or the direction as needed. - - 37 - - - - - + + + + - vo + + + - - - - - + + + + + + + + + + + + + + + + + + + Ex = 0 + + + + + + + + + + + + + + + + + 38 vo No electric field No force No acceleration Constant horizontal velocity + + + - - - - - - - - - - - - - - - - - - vo + + + + + + + + + + + + + + + + + + Observations: Vertically: Fx = 0 ax = 0 vx = vo x = vot 7/31/2007 + - - - - - - - - - - - - - - - - - 7/31/2007 - - - - - - - - - - - - - - - - - Observations: Horizontally: - - + 7/31/2007 - - Constant electric field Constant force Constant acceleration Vertical velocity increase linearly with time. E y = Eo Fy = qo Eo a y = qo Eo / mo v y = qo Eo t / mo y= 39 7/31/2007 1 qo Eot 2 / mo 2 40 10 - - - - - - - - - - - - - - - - - - - - - - Example: Electric Field Due to Two Point Charges Question: Charge q1=7.00 μC is at the origin, and charge q2=-10.00 μC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. - + + + + + y + + + + + + + + + + + + + + + + + E1 Conclusions: The charge will follow a parabolic path downward. Motion similar to motion under gravitational field only except the the downward acceleration is now larger. 0.400 m P q1 7/31/2007 41 E E2 0.300 m x q2 7/31/2007 42 Question: Charge q1=7.00 μC is at the origin, and charge q2=-10.00 μC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Question: Charge q1=7.00 μC is at the origin, and charge q2=-10.00 μC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Observations: First find the field at point P due to charge q1 and q2. Field E1 at P due to q1 is vertically upward. Field E2 at due to q2 is directed towards q2. The net field at point P is the vector sum of E1 and E2. The magnitude is obtained with Solution: E1 = ke E2 = ke q1 r12 q2 r22 ( 7.00 ×10 C ) = 3.93 ×10 N / C −6 = 8.99 ×109 Nm2 C2 ( 0.400m ) 5 2 (10.00 ×10 C ) = 3.60 ×10 N / C −6 = 8.99 ×109 Nm2 C2 ( 0.500m ) 5 2 Ex = 53 E2 = 2.16 ×105 N / C E y = E1 − E2 sin θ = E1 − 54 E2 = 1.05 ×105 N / C E = ke 7/31/2007 q r E = Ex2 + E y2 = 2.4 ×105 N / C φ = arctan( E y / Ex ) = 25.9o 2 43 7/31/2007 44 11 15.5 Electric Field Lines 15.5 Electric Field Lines (2) A convenient way to visualize field patterns is to draw lines in the direction of the electric field. Such lines are called field lines. lines. Remarks: 1. 2. Electric field lines of single positive (a) and (b) negative charges. a) Electric field vector, E, is tangent to the electric field lines at each point in space. The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the the electric field in a given region. b) + q - q E is large when the field lines are close together and small when far apart. 7/31/2007 45 15.5 Electric Field Lines (3) 2. 3. 7/31/2007 46 15.5 Electric Field Lines (4) Electric field lines of a dipole. dipole. Rules for drawing electric field lines for any charge distribution. 1. 7/31/2007 Lines must begin on positive charges (or at infinity) and must terminate on negative charges or in the case of excess charge at infinity. The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. No two field lines can cross each other. 47 + 7/31/2007 - 48 12 Application: Measurement of the atmospheric electric field The electric field near the surface of the Earth is about 100 N/C downward. Under a thundercloud, the electric field can be as large as 20000 N/C. How can such a (large) field be measured? A A 7/31/2007 49 15.6 Conductors in Electrostatic Equilibrium 50 15.6 Conductors in Electrostatic Equilibrium Good conductors (e.g. copper, gold) contain charges (electron) that are not bound to a particular atom, and are free to move within the material. When no net motion of these electrons occur the conductor is said to be in electroelectro-static equilibrium. equilibrium. 7/31/2007 7/31/2007 Properties of an isolated conductor (insulated from the ground). 1. Electric field is zero everywhere within the conductor. 2. Any excess charge field on an isolated conductor resides entirely on its surface. 3. The electric field just outside a charged conductor is perpendicular to the conductor’ conductor’s surface. 4. On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest – at sharp points. 51 7/31/2007 52 13 1. Electric field is zero everywhere within the conductor. 2. This property is a direct result of the 1/r2 repulsion between like charges. If an excess of charge is placed within the volume, the repulsive force pushes them as far apart as they can go. They thus migrate to the surface. If this was not true, true, the field inside would be finite. Free charge there would move under the influence of the field. A current would be induced. The conductor would not be in an electrostatic state. 7/31/2007 3. 53 7/31/2007 4. The electric field just outside a charged conductor is perpendicular to the conductor’ conductor’s surface. Any excess charge field on an isolated conductor resides entirely entirely on its surface. 54 On an irregular shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest smallest – at sharp points. Consider, for instance, a conductor fairly flat at one end and relatively relatively pointed at the other. Excess of charge move to the surface. Forces between charges on the flat surface, tend to be parallel to the surface. Those charges move apart until repulsion from other charges creates creates an equilibrium. At the sharp ends, the forces are predominantly directed away from from the surface. There is less of tendency for charges located at sharp edges to move away from one another. Produces large fields (and force) near sharp edges. If not true, the field would have components parallel to the surface of the conductor. This field component would cause free charges of the conductor to move. A current would be created. There would no longer be a electroelectro-static equilibrium. - 7/31/2007 55 7/31/2007 - 56 14 Remarks Faraday’s ice-pail experiment Property 4 is the basis for the use of lightning rods near houses and buildings. (Very important application) Most of any charge on the house will pass through the sharp point of the lightning rod. First developed by B. Franklin. +++++ + + + + + + + - - + +++++ - - - + + + + + + + + + + + + + - + + + - - + + + - + + + + + + + Demonstrates that the charge resides on the surface of a conductor. 7/31/2007 57 7/31/2007 58 Mini-quiz Question: Suppose a point charge +Q is in empty space. Wearing rubber gloves, gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge? Question: Suppose a point charge +Q is in empty space. Wearing rubber gloves, gloves, we sneak up and surround the charge with a spherical conducting shell. What effect does this have on the field lines of the charge? Answer: Negative charge will build up on the inside of the shell. Positive charge will build up on the outside of the shell. There will be no field lines inside the conductor but the field lines will remain outside the shell. ? + + - + - + q - + - + 59 7/31/2007 + - - + 7/31/2007 + - - + + q + + - + + 60 15 Question: Is it safe to stay inside an automobile during a lightning storm? storm? Why? Answer: Yes. It is. The metal body of the car carries the excess charges on its external surface. Occupants touching the inner surface are in no danger. Mini-Quiz Question: Is it safe to stay inside an automobile during a lightning storm? Why? SAFE 7/31/2007 61 7/31/2007 62 16