General Physics PDF for First Level Science Students

General Physics. Electricity and Magnetism, Oscillations, Mechanical Waves and Wave Optics. For First Level Science Students. A Course Abstracted From Physics for Scientists and Engineers by R. A. Serway and J. W. Jewett, Jr. Dear Student...

General Physics. Electricity and Magnetism, Oscillations, Mechanical Waves and Wave Optics. For First Level Science Students. A Course Abstracted From Physics for Scientists and Engineers by R. A. Serway and J. W. Jewett, Jr. Dear Student Physics is the most fundamental science. It helps you to complete your knowledge in any other science. To study Physics is a continuous process of teaching by your instructor (important) and of learning by yourself (more important). For your maximum benefit from the Physics course, we advise you the following: 1- Attend every lecture, do not miss any. Take your own written notes for what is explained in the lecture. 2- Make sure to understand every point in the lecture and the solved examples in the topic. Do not hesitate to ask politely for more explanation of any difficult point you may not fully understand. 3- Summarize the concepts, rules and laws of each lecture then try to solve the problems assigned to the topic of the lecture, before going to its section. Correct what you have solved in the corresponding tutorial (section for solving problems). Try to be active in the section by preparing the problems of the chapter. 4- Do not miss tutorials. 5- Try also to study the solved examples in the assigned text book (you can download a softcopy of it from the internet as a pdf file). The marks distribution for Physics 101 and 102 is as follows: 60 % for the final theoretical exam, 10 % for the midterm theoretical exam, 10 % for practical coursework, and 20 % for final lab practical exam. Try hard to get every mark, you and your instructors will be very proud when you get an A+ mark in Physics……. that can be done. Good Luck. Your Physics Team. References: Raymond A. Serway, John W. Jewett - Physics for Scientists and Engineers with Modern Physics-Cengage Learning (2018), 9th edition  Contents. Chapter 1: Electric fields……………..……………………. 1 Chapter 2: Gauss’s law…………………………………….. 8 Chapter 3: Electric potential …….…..…………………… 15 Chapter 4: Capacitance and dielectrics …..……………. 24 Chapter 5: Magnetic fields …………….............…….….. 33 Chapter 6: Oscillatory motion...…………….…………... 40 Chapter 7: Wave motion ………….……...………………. 49 Chapter 8: Superposition of waves …...…...……………. 62  Chapter 1. Electric fields 1.1 Properties of charges:  Charges are either positive or negative.  A neutral object will become positively charged when some electrons are removed from it. A neutral object will become negatively charged when some electrons are added from it.  Charges of the same sign repel one another and charges with opposite signs attract one another.  Electric charge is always conserved in an isolated system. One object gains some amount of negative charge while the other gains an equal amount of positive charge.  The electric charge q is said to be quantized. That is, electric charge exists as integral multiple of electronic charge e, Q = ne. 1.2 Coulomb’s law: Figure 1  The force of attraction or repulsion between any two point charges q1 and q2 placed in space or air and separated by a distance r is: q1 q2 F = ke r2 1 where ke is Coulomb’s constant = 8.99x109 N. m2 / C2 and ke = where εo is a 4π εo constant called the permittivity of free space and equals 8.85x10−12 C2/ N. m2. Serway & Jewett 8th edition 1   We must note that the force between the two charges is a mutual force according  to Newton’s third law. This means that if q1 is acting on q2 by a force F12 then q2  must act on q1 by an equal and opposite force F21 such that:   F12 = − F21 When more than two charges are present, the resultant force on any one of them equals the vector sum of the forces exerted by the other individual charges. Example 1.1: Consider three point charges located at the corners of a right triangle as shown in Figure 2, where q1 = q3 = 5.00 µC, q2 = − 2.00 µC, and a = 0.100 m. Find the resultant force exerted on q3. Solution: q2 q3 F23 = ke a2 −6 −6 9 (2.00 x10 )(5.00 x10 ) (8.99 x10 = ) (0.100)2 8.99 N q1 q3 (5.00 x10−6 )(5.00 x10 −6 ) =F13 = ke ( 8.99 x10 ) 9 = 11.24 N ( ) 2 2a (2)(0.100)2  The x and y components of F13 are:  =F13 x (11.24) = cos 45 7.95 N  =F13 y (11.24) = sin 45 7.95 N  F3 x = 7.95 + ( −8.99) = − 1.04 N  F3= y 7.95 + = 0 7.95 N  F3 = ( − 1.04 iˆ + 7.95 ˆj N ) Example 1.2: Three point charges lie along the x axis as shown in Figure. The positive charge q1 = 15.0 µC is at x = 2.00 m, the positive charge q2 = 6.00 µC is at the origin, and the net force acting on q3 is zero. What is the x coordinate of q3? Figure 3 Serway & Jewett 8th edition 2 Solution: q1 q3 q2 q3 F13 = ke 2 F23 = ke ( 2.0 − x ) x2 q1 q3 q2 q3 ke 2 = ke ( 2.0 − x ) x2 x2 q x= q2 6.0 = 2 = = 0.63 ( 2.0 − x ) 2 q1 ( 2.0 − x ) q1 15.0 x = 0.775 m 1.3 The electric field: Figure 4 shows a small positive test charge qo placed near a second object carrying a much greater positive charge Q. We define the electric field due to the source charge at the location of the test charge as: Figure 4  The strength (or magnitude) of the electric field is the magnitude of the force per unit positive test charge placed in the field. Thus we can write: E = F N/C qo The force on a charged particle q placed in an electric field of strength E is given by: F = qE The force by which Q acts on qo, separated by distance r , in Figure 4 is given by: Q qo F = ke r2 So, the strength of the field due to Q at distance r from Q is given by: Q E = F = ke 2 qo r Figure 5 Serway & Jewett 8th edition 3 The vector form of the electric field is expressed as:   E = F = ke Q2 rˆ qo r where rˆ is a unit vector directed from Q toward point P as shown in Figure 5. If the source charge Q is positive, Figure 5 shows the situation with the test charge removed: the source charge sets up an electric field at P, directed away from Q. If Q is negative, the electric field at P is directed toward the source charge. 1.4 Patterns of the electric field lines: Figure 6 Important notes:   The electric field vector E is tangent to the electric field line at each point. The line has a direction, indicated by an arrowhead, that is the same as that of the electric field vector. The direction of the line is that of the force on a positive test charge placed in the field.  The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region. Therefore, the field lines are close together where the electric field is strong and far apart where the field is weak.  The lines must begin on a positive charge and terminate on a negative charge.  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 or touch. Serway & Jewett 8th edition 4 1,5 Motion of a Charged Particle in a Uniform Electric Field  When a particle of charge q and mass m is placed in an electric field E , the electric   force exerted on the charge is F = q E. If that is the only force exerted on the particle, it must be the net force, and it causes the particle to accelerate according to the particle under a net force model. Therefore,    = F q= E ma and the acceleration of the particle is:   qE a = m  If E is uniform (that is, constant in magnitude and direction), the electric force on the particle is constant and we can apply the particle under constant acceleration model to the motion of the particle. If the particle has a positive charge, its acceleration is in the direction of the electric field. If the particle has a negative charge, its acceleration is in the direction opposite the electric field. Example 1.3:  A uniform electric field E is directed along the x axis between parallel plates of charge separated by a distance d as shown in Figure 7. A positive point charge q of mass m is released from rest at a point A next to the positive plate and accelerates to a point B next to the negative plate. (A) Find the speed of the particle at B by modeling it as a particle under constant acceleration. (B) Find the speed of the particle at B by modeling it as a non- isolated system. Solution: 2 (A) v= f v i2 + 2 a d v f2= 0 + 2 a d Figure 7 2q E d =vf 2a d = m (B) W = ∆K 1 2F d 2(q E ) d Fd = m v f2 =vf = 2 m m Serway & Jewett 8th edition 5  Problems (1) Rank the magnitudes of the electric field at points A, B, and C shown in Figure (greatest magnitude first). (2) Three charged particles are arranged on corners of a square as shown in Figure, with charge − Q on both the particle at the upper left corner and the particle at the lower right corner and with charge + 2Q on the particle at the lower left corner. (i) What is the direction of the electric field at the upper right corner, which is a point in empty space? (a) It is upward and to the right. (b) It is straight to the right. (c) It is straight downward. (d) It is downward and to the left. (e) It is perpendicular to the plane of the picture and outward. (ii) Suppose the + 2Q charge at the lower left corner is removed. Then does the magnitude of the field at the upper right corner (a) become larger, (b) become smaller, (c) stay the same, or (d) change unpredictably? (3) Three point charges are arranged as shown in Figure. Find the magnitude and the direction of the electric force on the particle at the origin. (4) Two small beads having positive charges q1 = 3 q and q2 = q are fixed at the opposite ends of a horizontal insulating rod of length d = 1.50 m. The bead with charge q1 is at the origin. As shown in Figure, a third small, charged bead is free to slide on the rod. At what position x is the third bead in equilibrium? Serway & Jewett 8th edition 6 (5) A small object of mass 3.80 g and charge − 18.0 µC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field? (6) Three charged particles are located at the corners of an equilateral triangle as shown in Figure. Calculate the total electric force on the 7.00-µC charge. (7) Three point charges are located on a circular arc as shown in Figure, (a) What is the total electric field at P, the center of the arc? (b) Find the electric force that would be exerted on a − 5.00-nC point charge placed at P. (8) A proton (m = 1.67x10-27 kg, q = 1.6x10-19 C) accelerates from rest in a uniform electric field of 640 N/C. At one later moment, its speed is 1.20 Mm/s. (a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval? (9) Three identical point charges, each of mass m = 0.100 kg, hang from three strings as shown in Figure. If the lengths of the left and right strings are each L = 30.0 cm and the angle θ is 45.0o, determine the value of q. Answers (1) A then B then C. (2)(i) (d) (ii) (a) (3) 1.38x10 N, 77.5 below the negative x-axis. -5 o (4) 0.951 m (5) 2.07x10 N/C, vertically downward. 3 (6) 0.872 N at 330°. (7) (a) 1.80x10 N/C to the right (b) 8.98x10 N to the left. 4 -5 (8) (a) 6.13x1010 m/s2 (b) 1.96x10-5 s (c) 11.7 m (d) 1.20x10-15 J (9) 1.98x10-6 C. Serway & Jewett 8th edition 7  Chapter 2. Gauss’s Law. 2.1 The Electric flux: Consider an electric field that is uniform in both magnitude and direction as shown in Figure 1. The field lines penetrate a rectangular surface of area A, whose plane is oriented perpendicular to the field. The number of lines per unit area (in other words, the line density) is proportional to the magnitude of the electric field. Therefore, the total number of lines penetrating the surface is proportional to the product EA. This product of the magnitude of the electric field E and surface area A perpendicular to the field is called Figure 1. the electric flux ΦE. ΦE =E A Electric flux is proportional to the number of electric field lines penetrating some surface and has the units of N m2/ C. If the surface under consideration is not perpendicular to the field, the flux through it must be less than that given by above equation. Consider the shown Figure 2, where the normal to the surface of area A is at an angle θ. We take the component of E in the direction perpendicular to A to get the electric flux. Φ = E A cos θ  E The vector A representing the area is a vector Figure 2. perpendicular to the area. The electric flux, which is a scalar quantity, can thus be written as:   ΦE =E.A Serway & Jewett 8th edition 8  2.2 Gauss’s law: Consider a positive point charge q located at the center of a sphere of radius r as shown in Figure 3. We know that the magnitude of the electric field everywhere on the surface of 1 q the sphere is E =. The field lines are directed 4π εo r 2 radially outward and hence are perpendicular to the surface at every point on the surface. That is, at each surface point, E is parallel to normal of the surface surrounding the charge. Therefore, the flux through the surface is:  1 q  q =ΦE  =  4π εo r  2  4 π r 2 εo ( ) The net flux through the spherical surface is proportional to the charge inside the surface. The flux is independent of the Figure 3. 2 radius r because the area of the spherical surface is proportional to r , whereas the electric field is proportional to 1/r 2. Now consider several closed surfaces surrounding a charge q as shown in Figure 4. Surface S1 is spherical, but surfaces S2 and S3 are not. The flux q that passes through S1 has the value. As discussed in the εo preceding section, flux is proportional to the number of electric field lines passing through a surface. The construction shown in Figure 4 shows that the number of lines through S1 is equal to the number of lines through the non-spherical surfaces S2 and S3. Therefore, Gauss’s law states that: Figure 4. The net flux through any closed surface surrounding a point q charge q is given by and is independent of the shape of εo that surface. Now consider a point charge located outside a closed surface of arbitrary shape as shown in Figure 5. As can be seen from this construction, any electric field line entering the surface leaves the surface at another point. The number of electric field lines entering the surface equals the number leaving the surface. Therefore, the net electric flux through a closed surface that surrounds no charge is zero. Figure 5. Serway & Jewett 8th edition 9  Example 2.1: An insulating solid sphere of radius ‘a’ has a uniform volume charge density ρ (C/m3) and carries a total positive charge Q. (A) Calculate the magnitude of the electric field at a point outside the sphere. (B) Find the magnitude of the electric field at a point inside the sphere. Solution: (A) let’s choose a spherical gaussian surface of radius (r > a), concentric with the sphere, as shown in Figure 6. The amount of charge enclosed by the gaussian surface is the total charge on the sphere Q. To calculate the field on the gaussian surface we apply Gauss’s law: Q Q E A= or E (4 π r 2 ) = εo εo Q Q =E = 2 ke 2 (r > a) 4π εo r r (B) Now let’s choose a spherical gaussian surface of radius Figure 6. (r < a), concentric with the sphere (Figure 7). First we calculate the density ρ of charges: Q ρ= 4 3 π a3 The amount of charge enclosed by the gaussian surface is given by: Q Q 3 =qin ρ= Vsurface 4 π a3 (= 4 3π r ) 3 a3 r 3 To calculate the field on the gaussian surface we apply Gauss’s law: qin qin Q E A= εo or ( 2 E 4 π r= ) = εo εo a 3 r3 Q Q =E = r ke r (r < a) 4 π ε o a3 a3 The electric field can thus be plotted versus r as shown in Figure 8. The electric field inside the sphere (r < a) varies linearly with r. The field outside the sphere (r > a) is the same as that of a point charge Q located at (r = 0). Figure 8. Serway & Jewett 8th edition 10 Example 2.2: Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length λ (C/m) (Figure 9). Solution: The total charge inside the Gaussian surface = λ To calculate the field on the gaussian surface we apply Gauss’s law: qin λ  E= A = εo εo The area of the Gaussian surface cut perpendicularly by the Figure 9. electric field is the side area of the cylinder: A = (2π r )  λ E (2π r )  = εo λ λ =E = 2 ke (2π ε o ) r r Example 2.3: Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ (C/m2) (Figure 10). Solution: The total charge inside the Gaussian surface = σA To calculate the field on the gaussian surface we apply Gauss’s law: qin σ A E= A = εo εo σ This gives: E= Figure 10. εo Serway & Jewett 8th edition 11 2.3 Conductors in electrostatic equilibrium: A good electrical conductor contains charges (electrons) that are not bound to any atom and therefore are free to move about within the material. When there is no net motion of charge within a conductor, the conductor is in electrostatic equilibrium. A conductor in electrostatic equilibrium has the following properties: 1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow. 2. If the conductor is isolated and carries a charge, the charge resides on its surface. 3. The electric field at a point just outside a charged conductor is perpendicular to the surface of the conductor and has a magnitude σ / ε o , where σ is the surface charge density at that point. 4. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. To explain the first property, consider a conducting slab  placed in an external field E as shown in Figure 11. The electric field inside the conductor must be zero, assuming electrostatic equilibrium exists. If the field were not zero, free electrons in the conductor would experience   an electric force F = q E and would accelerate due to this force. This motion of electrons, however, would mean that the conductor is not in electrostatic equilibrium. Therefore, the existence of electrostatic equilibrium is Figure 11. consistent only with a zero field in the conductor. Let’s investigate how this zero-field is accomplished. Before the external field is applied, free electrons are uniformly distributed throughout the conductor. When the external field is applied, the free electrons accelerate to the left, causing a plane of negative charge to accumulate on the left surface. The movement of electrons to the left results in a plane of positive charge on the right surface. These planes of charge create an additional electric field inside the conductor that opposes the external field. As the electrons move, the surface charge densities on the left and right surfaces increase until the magnitude of the internal field equals that of the external field, resulting in a net field of zero inside the conductor. The time it takes a good conductor to reach equilibrium is on the order of 10−16 s, which for most purposes can be considered instantaneous. Serway & Jewett 8th edition 12  Problems (1) Charges of 3.00 nC, − 2.00 nC, − 7.00 nC, and 1.00 nC are contained inside a rectangular box with length 1.00 m, width 2.00 m, and height 2.50 m. Outside the box are charges of 1.00 nC and 4.00 nC. What is the electric flux through the surface of the box? (a) 0 (b) − 5.64 x 102 N. m2/C (c) − 1.47 x 103 N. m2/C (d) 1.47 x 103 N. m2/C (e) 5.64 x 102 N. m2/C (2) A uniform electric field of 1.00 N/C is set up by a uniform distribution of charge in the xy- plane. What is the electric field inside a metal ball placed 0.500 m above the xy-plane? (a) 1.00 N/C (b) − 1.00 N/C (c) 0 (d) 0.250 N/C (e) varies depending on the position inside the ball. (3) Rank the electric fluxes through each gaussian surface shown in Figure from largest to smallest. Display any cases of equality in your ranking. (4) A vertical electric field of magnitude 2.00 x 104 N/C exists above the Earth’s surface on a day when a thunderstorm is brewing. A car with a rectangular size of 6.00 m by 3.00 m is traveling along a dry gravel roadway sloping downward at 10.0o. Determine the electric flux through the bottom of the car. (5) Four closed surfaces, S1 through S4, together with the charges − 2Q , Q , and − Q are sketched in Figure. (The colored lines are the intersections of the surfaces with the page.) Find the electric flux through each surface. (6) A charge of 170 µC is at the center of a cube of edge 80.0 cm. No other charges are nearby. (a) Find the flux through each face of the cube. (b) Find the flux through the whole surface of the cube. (c) What If? Would your answers to either part (a) or part (b) change if the charge were not at the center? Explain. Serway & Jewett 8th edition 13 (7) The charge per unit length on a long, straight filament is − 90.0 µC/m. Find the electric field (a) 10.0 cm, (b) 20.0 cm, and (c) 100 cm from the filament, where distances are measured perpendicular to the length of the filament. (8) A solid sphere of radius 40.0 cm has a total positive charge of 26.0 µC uniformly distributed throughout its volume. Calculate the magnitude of the electric field (a) 0 cm, (b) 10.0 cm, (c) 40.0 cm, and (d) 60.0 cm from the center of the sphere. (9) In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C directed downward. At 600 m above the ground, the electric field is 100 N/C downward. What is the average volume charge density in the layer of air between these two elevations? Is it positive or negative? (10) A particle with charge Q = 5.00 µC is located at the center of a cube of edge L = 0.100 m. In addition, six other identical charged particles having q = − 1.00 µC are positioned symmetrically around Q as shown in Figure. Determine the electric flux through one face of the cube. (11) A long, straight metal rod has a radius of 5.00 cm and a charge per unit length of 30.0 nC/m. Find the electric field (a) 3.00 cm, (b) 10.0 cm, and (c) 100 cm from the axis of the rod, where distances are measured perpendicular to the rod’s axis. Answers (1) b (2) c (3) c then (a = b) then d (4) 3.55x105 N. m2 /C (5) − Q / ε o for S1, 0 for S2, − 2 Q / ε o for S3, 0 for S4. (6) (a) 3.20x106 N. m2 /C (b) 1.92x107 N. m2 /C (c) Flux is the same. (7) (a) 1.62x10 N. m /C (b) 8.09x10 N. m /C 7 2 6 2 (c) 1.62x106 N. m2 /C (8) (a) 0 (b) 3.65 x 105 N/C (c) 1.46 x 106 N/C (d) 6.49 x 105 N/C (9) 1.77 x 10−12 C/m3; positive (10) 18.8x103 N. m2 /C (11) (a) 0 (b) 5.39 x 103 N/C outward (c) 539 N/C outward Serway & Jewett 8th edition 14  Chapter 3. Electric Potential 3.1 Electric potential and potential difference: When a mass m is moved by a gravitational field from point A to point B, through a displacement ∆s, the work done by the field is:     Wint = Fg. ∆s= m g. ∆s The potential energy of the mass will change by an amount   ∆U =− Wint =− m g. ∆s The negative sign indicates a decrease in the potential energy of the mass. Figure 1. In analogy to a mass in a gravitational field, a positive test charge qo  placed in an electric field E is moved from point A to point B, through a displacement ∆s, so the work done by the field is:     Wint = Fe. ∆s = qo E. ∆s The electric potential energy of the charge will change by an amount   ∆U =− Wint =− qo E. ∆s Again, the negative sign indicates a decrease in the potential energy of the charge. Figure 2. The potential difference ∆V = VB − VA between two points A and B in an electric field is defined as the change in potential energy of the system when a test charge qo moved between the points divided by the test charge: ∆U   ∆V = =− E. ∆ s J / C or Volt qo In all these equations we assumed that the gravitational or the electric field is uniform. The magnitude of the electric field is thus equal to the negative of the rate of change of electric potential with distance, namely, the potential gradient. ∆V E= − V /m ∆s It’s your task to show that the units of the electric field N / C and V / m are equivalent. Serway & Jewett 8th edition 15  If an external agent moves a charge from A to B without changing the kinetic energy of the charge, the agent performs work that changes the potential energy of the system: W =∆U =q ∆V The electron-Volt (eV): it is a unit of energy, which is equal to the energy gained or lost when an electron or a proton of charge e (= 1.6x10−19 C) is moved through a potential difference of 1 Volt. it is equal to = x10−19 C )(1 V ) 1.6 x10 −19 J. q ∆V (1.6= Let a charge q to be pushed from point A to point B  along the displacement vector s that makes an angle θ with the field direction (Figure 3). The potential difference between A and B is   ∆V =− E. s =− E s cos θ =− E d The change in potential energy of q is ∆ U =q ∆V =− q E d This is the same change in potential energy if q were pushed from A to C since B and C are at the same electric potential. Note that if q is positive, the negative sign means decrease in potential energy of the charge when pushed Figure 3. in direction of the field. There is no work done, neither change in potential energy if any charge is pushed between B and C. Example 3.1: A 12-V battery is connected between two parallel plates as shown in Figure 4. The separation between the plates is d = 0.30 cm, and we assume the electric field between the plates to be uniform. Find the magnitude of the electric field between the plates. ∆V 12 Solution: = E = = 4000 V / m Figure 4. d 0.30 x10 −2 Serway & Jewett 8th edition 16 Example 3.2: A proton is released from rest at point A in a uniform electric field that has a magnitude of 8.0x104 V/m (Figure 5). The proton undergoes a displacement of magnitude d  = 0.50 m to point B in the direction of E. Find the speed of the proton after completing the displacement. Solution: ∆V =− E d =− (8.0 x104 )(0.50) =− 4.0 x10 4 V For this isolated system, total energy is conserved. Figure 5. ∆K + ∆U =0 ( 1 2 mv 2 ) − 0 + e ∆V =0 − 2 e ∆V − 2 (1.6 x10−19 )( − 4.0 x10 4 ) =v = = 2.77 x106 m / s −27 m 1.67 x10 Because ∆V is negative, ∆U is also negative for the proton–field system. The negative value of ∆U means the potential energy of the system decreases as the proton moves in the direction of the electric field. As the proton accelerates in the direction of the field, the gain in its kinetic energy is equal to the loss in the electric potential energy of the system. 3.2 Electric potential and potential energy of point charges:  An isolated positive point charge q produces an electric field E directed radially outward from the charge. The electric potential at point P at distance r12 from a point charge q2 is defined as the work done to move unit positive charge from infinity to P. It is given by: q2 VP = ke r12 The sign of VP will be the same as the sign of q2. Figure 6. It is customary to choose the reference of electric potential for a point charge to be V = 0 at r = ∞. Figure 7 shows the variation of the electric potential V with distance r from charge q in case that charge q is positive or negative. Figure 7. Serway & Jewett 8th edition 17 The potential difference between two points in the field A and B (Figure 8) is given by:   VA ke q  1 − 1  VB −=  rB rA  This is the work done to move unit positive charge from A to B. Figure 8. The total electric potential at some point P due to several point charges is the sum of the potentials due to the individual charges. For a group of point charges, we can write the total electric potential at P as: q V = ke ∑ ri i i Notice that this sum is an algebraic sum of scalars rather than a vector sum. Each charge should be substituted with its proper sign. Now let a charge q1 to be brought from infinity and placed at position P in Figure 6. The potential energy of a system of two charges q1 and q2 separated by distance r12 (Figure 9) is given by: q1 q2 =U q= 1 VP ke Figure 9. r12 Again, charges in this equation must be substituted with their proper signs. If q1 and q2 are of similar charges, work must be done by an external agent to bring q1 from infinity to P, because of the repulsive force between q1 and q2. This work increases the potential energy of the system and has a positive sign. If q1 and q2 are of opposite charges, work is given out to an external agent to bring q1 from infinity to P, because of the attractive force between q1 and q2. This work decreases the potential energy of the system and has a negative sign. Serway & Jewett 8th edition 18 If the system consists of more than two charged particles, we can obtain the total potential energy of the system by calculating U for every pair of charges and summing the terms algebraically. For example, the total potential energy of the system of three charges shown in Figure 10 is: q q q q q q  = ke  1 2 + 2 3 + 3 1  U Figure 10.  r12 r23 r13  Example 3.3: Figure 11(a). Figure 11(b). As shown in Figure 11(a), a charge q1 = 2.00 µ C is located at the origin and a charge q2 = − 6.00 µ C is located at (0, 3.00) m. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. (B) Find the change in potential energy of the system of two charges plus a third charge q3 = 3.00 µ C as the latter charge moves from infinity to point P (Figure 11(b)). Solution: −6  q1 q2  9  2.00 x10 − 6.00 x10 −6  VP= ke  + = (8.99 x10 )  +  (A) r  1 r2   4.00 5.00  = − 6.29 x103 V (B) ∆U =q3 VP =(3.00 x10 −6 )( − 6.29 x103 ) =− 1.89 x10 −2 J Therefore, because the potential energy of the system has decreased, an external agent has to do positive work to remove the charge q3 from point P back to infinity. Serway & Jewett 8th edition 19 3.3 Electric potential of a spherical conductor: Every point on the surface of a charged conductor in equilibrium is at the same electric potential. Furthermore, because the electric field is zero inside the conductor, the electric potential is constant everywhere inside the conductor and equal to its value at the surface. As determined in Example 2.1, the electric field outside 2 the sphere is ke Q r and points radially outward. Because the field outside a spherically symmetric charge distribution is identical to that of a point charge, we expect the potential to also be that of a point charge, ke Q r. At the surface of the conducting sphere in Figure 12(a), the potential must be ke Q R. Because the entire sphere must be at the same potential, the potential at any point within the sphere must also be ke Q R. Figure 12(b) is a plot of the electric potential as a function of r, and Figure 12(c) shows how the electric field varies with r. Figure 12 Serway & Jewett 8th edition 20  Problems Use the following useful information: For the proton (q = +1.6x10-19 C, m = 1.67x10-27 kg) For the electron (q = −1.6x10-19 C, m = 9.11x10-31 kg) (1) proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. What is the change in the electric potential energy of the proton–field system when the proton travels to x = 2.50 m? (a) 3.40x10−16 J (b) − 3.40x10−16 J (c) 2.50x10−16 J (d) − 2.50x10−16 J (e) − 1.60x10−19 J (2) The electric potential at x = 3.00 m is 120 V, and the electric potential at x = 5.00 m is 190 V. What is the x component of the electric field in this region, assuming the field is uniform? (a) 140 N/C (b) − 140 N/C (c) 35.0 N/C (d) − 35.0 N/C (e) 75.0 N/C (3) Rank the potential energies of the four systems of charges shown in Figure from largest to smallest. Include equalities if appropriate. (4) In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x = 20.0 cm to x = 60.0 cm. (i) Does the electric potential energy of the charge–field system (a) increase, (b) remain constant, (c) decrease, (d) change unpredictably? (ii) Has the particle moved to a position where the electric potential is (a) higher than before, (b) unchanged, (c) lower than before, (d) unpredictable? (5) (a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 120 V. (b) Calculate the speed of an electron that is accelerated through the same electric potential difference. Serway & Jewett 8th edition 21 (6) A uniform electric field of magnitude 325 V/m is directed in the negative y direction in Figure. The coordinates of point A are (− 0.200, − 0.300) m, and those of point B are (0.400, 0.500) m. Calculate the electric potential difference VB − VA using the dashed-line path. (7) An electron moving parallel to the x axis has an initial speed of 3.70x106 m/s at the origin. Its speed is reduced to 1.40x105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) State the direction of the electric field. (c) Which point is at the higher potential? (8) Given two particles with 2.00-µC charges as shown in Figure and a particle with charge q = 1.28x10−18 C at the origin, (a) what is the net force exerted by the two 2.00-µC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-µC particles? (c) What is the electric potential at the origin due to the two 2.00-µC particles? (9) Two particles each with charge +2.00 µC are located on the x axis. One is at x = 1.00 m, and the other is at x = − 1.00 m. (a) Determine the electric potential on the y axis at y = 0.500 m. (b) Calculate the change in electric potential energy of the system as a third charged particle of − 3.00 µC is brought from infinitely far away to a position on the y axis at y = 0.500 m. (10) Two parallel plates having charges of equal magnitude but opposite sign, are separated by 12.0 cm. Each plate has a surface charge density of 36.0 nC/m2. A proton is released from rest at the positive plate. Determine (a) the magnitude of the electric field between the plates from the charge density, (b) the potential difference between the plates, (c) the kinetic energy of the proton when it reaches the negative plate, (d) the speed of the proton just before it strikes the negative plate, (e) the acceleration of the proton, and (f) the force on the proton. (g) From the force, find the magnitude of the electric field. (h) How does your value of the electric field compare with that found in part (a)? Serway & Jewett 8th edition 22  Answers. (1) b (2) d (3) c then a then d then b (4)(i) a (ii) c (5)(a) 1.52x105 m/s (b) 6.49x106 m/s (6) +260 V (7)(a) − 38.9 V (b) Positive x axis (c) The origin (8)(a) zero (b) zero (c) 4.50x10 V 4 (9)(a) 3.22x10 V 4 (b) − 9.66x10−2 J (10)(a) 4.07x103 N/C (b) 488 V (c) 7.81x10-17 J (d) 3.06x105 m/s (e) 3.90x1011 m/s2 (f) 6.50x10-16 N (g) 4.07x103 N/C (h) They are equal. Serway & Jewett 8th edition 23  Chapter 4. Capacitance and Dielectrics 4.1 The parallel plate capacitor: A parallel-plate capacitor (Figure 1) consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is connected to the terminals of a battery, electrons transfer between the plates and the wires so that the plates become charged. When the potential difference between the plates becomes equal to the voltage of the battery, flow of electrons in the wires stops and the capacitor is fully charged. In this case one of the plates will carry a positive charge +Q while the other a negative charge − Q. Experiments show that the quantity of charge Q on each plate is linearly proportional to the potential difference between the plates; that is, Q ∝ ∆V. Figure 1. This relationship can be written as: Q= C ∆V where C is the constant of proportionality called the capacitance of the capacitor and is defined as the ratio of the magnitude of the charge on either plate to the magnitude of the potential difference between the plates. Q C= ∆V We see that capacitance has SI units of coulombs per volt, or the farad (F). 1 F = 1C V 4.2 Calculating Capacitance of a parallel plate capacitor: Considering the parallel plate capacitor shown in Figure 1, the surface charge density on each plate is σ = Q A. The value of the electric field between the plates is: σ Q = E = εo εo A Because the field between the plates is uniform, the magnitude of the potential difference between the plates equals E d ; therefore, Qd ∆V= E d= εo A So, from definition of capacitance: Q ε A = C = o ∆V d Serway & Jewett 8th edition 24 That is, the capacitance of a parallel-plate capacitor is proportional to the area of its plates and inversely proportional to the plate separation. Example 4.1: An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. A 20.0-V potential difference is applied to these plates. Calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate. Solution: ∆V 20 (a) = = E =−3 1.11x10 4 V / m d 1.80 x10 (b) σ ε= = oE (8.85 x10 −12 )(1.11x10=4 ) 9.82 x10 −8 C / m 2 εo A (8.85 x10 −12 )(7.60 x10 −4 ) (c) = C = −3 = 3.74 x10 −12 = F 3.74 pF d 1.80 x10 (d) Q = C ∆V = (3.74 x10−12 )(20) = 7.47 x10 −11 C = 74.7 pC 4.3 Combination of Capacitors: Parallel Combination: Figure 2 shows two capacitors connected in parallel to a battery. The individual potential differences across capacitors connected in parallel are the same and are equal to the potential difference applied across the combination. That is, ∆V1 =∆V2 =∆V where ∆V is the battery terminal voltage. When being fully charged, the maximum charges on the two capacitors are Q1 and Q2. The total charge Qtot stored by the two capacitors is the sum of the charges on the individual capacitors: = Q1 + Q2 Qtot Figure 2. Serway & Jewett 8th edition 25 Suppose you wish to replace these two capacitors by one equivalent capacitor having a capacitance Ceq as the one in Figure 3. The equivalent capacitor must store charge Qtot and the voltage across the equivalent capacitor is also ∆V when connected to the same battery. Therefore, for the equivalent capacitor, Q= tot Ceq ∆V Therefore, Ceq ∆V = C1 ∆V + C2 ∆V = C1 + C2 Ceq,parallel Figure 3. This rule extends to any number of capacitors connected in parallel, that is, Ceq,parallel = C1 + C2 + C3 +....... Series Combination: Figure 4 shows two capacitors connected in series to a battery. When fully charged, they end up with the charges on the capacitors being the same: = Q1 = Q2 Q where Q is the charge on any plate of the capacitors. Figure 4 shows that the total voltage ∆Vtot across the combination is split between the two capacitors: Figure 4. ∆Vtot = ∆V1 + ∆V2 Suppose you wish to replace these two capacitors by one equivalent capacitor having a capacitance Ceq as the one in Figure 5. After it is fully charged, the equivalent capacitor must have a charge of − Q on its right plate and a charge of + Q on its left plate. Applying the definition of capacitance to the circuit in Figure 5 gives: Q ∆Vtot = Figure 5. Ceq Therefore, Q Q1 Q2 1 1 1 = + = + Ceq C1 C2 Ceq,series C1 C2 Serway & Jewett 8th edition 26 This rule extends to any number of capacitors connected in series, that is, 1 1 1 1 = + + +........ Ceq,series C1 C2 C3 Example 4.2: Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 6. All capacitances are in microfarads. Figure 6. Solution: The 1.0-µF and 3.0-µF capacitors (upper red circle in Fig. 6a) are in parallel. Their equivalent capacitance is: C1,3 = 1 + 3 = 4 µ F The 2.0-µF and 6.0-µF capacitors (lower red circle in Fig. 6a) are also in parallel. Their equivalent capacitance is: C2,6 = 2 + 6 = 8 µ F The circuit now looks like Figure 6b. The two 4.0-µF capacitors (upper green circle in Fig. 6b) are in series. Their equivalent capacitance is: 1 1 1 = + = 0.50 µ F C4,4 = 2 µ F C4,4 4 4 The two 8.0-µF capacitors (lower green circle in Fig. 6b) are in series. Their equivalent capacitance is: 1 1 1 = + = 0.25 µ F C8,8 = 4 µ F C8,8 8 8 The circuit now looks like Figure 6c. The 2.0-µF and 4.0-µF capacitors are in parallel. Their equivalent capacitance is: C2,4 = 2 + 4 = 6 µ F This final value is that of the single equivalent capacitor shown in Figure 6d. Serway & Jewett 8th edition 27 4.4 Electric energy stored in a charged capacitor: Because positive and negative charges are separated in the system of two conductors in a capacitor, electric potential energy is stored in the system. When a capacitor is connected to a battery, the flow of electrons causes one plate to be positively charged while the other plate to be negatively charged. Some of the chemical potential energy in the battery is converted to electric potential energy associated with the separation of positive and negative charges on the plates. If the plates of a charged capacitor are connected by a conductor such as a wire, charge moves between each plate and its connecting wire until the capacitor is uncharged. The recombination of the positive and negative charged produces an electric current in the wire. This makes the system to lose its electric potential energy stored which appears as heat energy in the wires. 1 The relation ∆V = Q shows that there is a direct proportionality between the C voltage of the capacitor ∆V and the amount of charge it carries Q. This relation appears graphically as shown in Figure 7. The energy stored in the capacitor is qual to the work done by the battery to charge the capacitor from q = 0 to q = Q. This is represented by the yellow shaded triangle and is given by: 1 = U Q ∆V 2 Figure 7. Q = C ∆V or ∆V = Now substituting with Q this can be written as: C 1 2 1 Q2 U= C ( ∆V ) = 2 2 C Example 4.3: A 12.0-V battery is connected to a capacitor, resulting in 54.0 µC of charge stored on the capacitor. How much energy is stored in the capacitor? Solution: 1 1 U= Q ∆V= (54.0 x10−6 )(12.0)= 3.24 x10 −4 J 2 2 Serway & Jewett 8th edition 28 4.5 Capacitors with dielectric: A dielectric is a non-conducting material such as rubber, glass, or waxed paper. We can perform the following experiment to illustrate the effect of a dielectric in a capacitor. Consider a parallel-plate capacitor that without a dielectric has a charge Qo and a capacitance Co (Figure 8a). The potential difference across the capacitor is ∆Vo = Qo Co. The potential difference is measured by a device called a voltmeter. Hence, there is no path by which charge can flow and alter the charge on the capacitor. If a dielectric is now inserted between the plates, the voltmeter indicates that the voltage between the plates decreases to a value ∆V (Figure 8b). The voltages with and without the dielectric are related by a factor κ as follows: ∆V ∆V = o κ Because ∆V < ∆Vo, we see that κ > 1. The dimensionless factor κ is called the dielectric constant of the material. Because the charge Qo on the capacitor does not change, the capacitance must change to the value Qo Qo Qo = C = = κ = κ Co ∆V ∆ Vo / κ ∆ Vo That is, the capacitance increases by the factor κ when the dielectric completely fills the region between the plates. εo A ε A =C κ= d d Where ε = κ ε o is called the permittivity of the dielectric material. Figure 8. Serway & Jewett 8th edition 29 Therefore, a dielectric provides the following advantages: An increase in capacitance. An increase in maximum operating voltage. Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C. Example 4.4: A parallel-plate capacitor is charged with a battery to a charge Qo. The battery is then removed, and a slab of material that has a dielectric constant κ is inserted between the plates. Identify the system as the capacitor and the dielectric. Find the energy stored in the system before and after the dielectric is inserted. Solution: 1 Qo2 1 Qo2 1 Qo2 Uo Uo = = U = = 2 Co 2 C 2 κ Co κ Because κ > 1, the final energy is less than the initial energy. We can account for the decrease in energy of the system by performing an experiment and noting that the dielectric, when inserted, is pulled into the device. The kinetic energy gained by the accelerating slab comes at the expense of the energy stored. Serway & Jewett 8th edition 30  Problems (1) Given a 2.50-µF capacitor, a 6.25-µF capacitor, and a 6.00-V battery, find the charge on each capacitor if you connect them (a) in series across the battery and (b) in parallel across the battery. (2) For the system of four capacitors shown in Figure, find (a) the equivalent capacitance of the system, (b) the charge on each capacitor, and (c) the potential difference across each capacitor. (3) Four capacitors are connected as shown in Figure. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, taking ∆Vab = 15.0 V. (4) (a) A 3.00-µF capacitor is connected to a 12.0-V battery. How much energy is stored in the capacitor? (b) Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored? (5) A parallel-plate capacitor in air has a plate separation of 1.50 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine (a) the charge on the plates before and after immersion, (b) the capacitance and potential difference after immersion, and (c) the change in energy of the capacitor. (6) A 2.00-nF parallel-plate capacitor is charged to an initial potential difference ∆Vi = 100 V and is then isolated. The dielectric material between the plates is mica, with a dielectric constant of 5.00. (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference across the capacitor after the mica is withdrawn? Serway & Jewett 8th edition 31 (7) The values of capacitances shown in Figure are: C1 = 8 µ F , C2 = 4 µ F , C3 = 6 µ F , C4 = 10 µ F. Calculate the potential difference across each of the four capacitors. Answers (1)(a)= = Q2.5 Q6.25 10.7 µC µC , Q6.25 = 37.5 µC (b) Q2.5 = 15.0 (2)(a) 3.33 µF = (b) Q3 = Q6 180 µC , Q =2 Q= 4 120 µC = (c) V3 V= 2 60 V , V =6 V= 4 30 V (3)(a) 5.96 µF (b) Q20 = 89.5 µC , Q6 = 63.2µC , = = Q3 Q15 26.3 µC (4)(a) 216 µJ (b) 54 µJ (5)(a) 369 pC (b) 1.2x10−10 F, 3.1 V (c) − 45.5 nJ (6)(a) 40.0 µJ (b) 500 V (7) V1 = 5.4 V , V2 = 10.7 V , V3 = 16.1 V , V4 = 13.9 V Serway & Jewett 8th edition 32  Chapter 5. Magnetic fields. 5.1 Magnetic fields and magnetic force: Figure 1 shows magnetic field lines around a permanent  magnet. The direction of the magnetic field B at any location is the direction in which a compass needle points at that location. Notice that the magnetic field lines outside the magnet point away from the north pole and toward the south pole. Figure 1.  A charged particle q moving with velocity v in a magnetic field  B also experiences a magnetic force given by:    FB = q v x B    where v x B denotes the cross (vector) product of v and    B, which gives a vector perpendicular to both v and B. The magnitude FB of the magnetic force exerted on the particle is proportional to the charge q, the speed v of the particle and the strength of the magnetic field B. The magnitude FB of the magnetic force is also proportional to sin θ, where θ is the angle the particle’s  velocity vector makes with the direction of B. Figure 2. When a charged particle moves parallel to the magnetic field vector, the magnetic force acting on the particle is zero. When the particle’s velocity vector makes any angle θ ≠ 0 with the magnetic field, the magnetic force acts in    a direction perpendicular to both v and B ; that is, FB is   perpendicular to the plane formed by v and B (Figure. 2). The magnetic force exerted on a positive charge is in the direction opposite the direction of the magnetic force exerted on a negative charge moving in the same direction (Figure. 3).  The magnitude of FB is given by: FB = q v B sinθ Figure 3. Serway & Jewett 8th edition 33 The right-hand rule shown in Figure 4 determines  the direction of FB. Point the four fingers of your right hand along the   direction of v and curl them toward B. Your extended thumb, which is at a right angle to your  fingers, points in the direction of FB if q is positive and is opposite the direction of your thumb if q is negative. Units of magnetic flux density B: Figure 4. FB We can write: B= q v sinθ N N The unit of B is = = 1 Tesla (T ) C.m s A.m 5.2 Motion of charged particle in a uniform magnetic field, Orbiting charges: Figure 5 shows a positively charged particle q projected with v in a direction speed perpendicular to a magnetic field B directed into the plane of the page.  Because the force FB is always perpendicular  to v , a charged particle moving in a magnetic field will travel along a circular path.  FB here is acting as a centripetal force because it is always directed towards the center of the circle. Figure 5.   Since v is perpendicular to B (or θ = 90°): m v2 mv qvB= and r = r qB The radius of the orbit increases with increase of: (a) Increase of the particle momentum m v. (b) Decrease of magnetic field flux density B. (c) Decrease of the particle charge. Serway & Jewett 8th edition 34  The period and the frequency of rotation are given by: 2π r 2π m 1 qB =T = =f = v qB T 2π m This shows that T and f do not depend neither on the velocity of charge nor on the radius of the circle. Note that the work of the magnetic force acting on a charged particle is always zero, because the force is always at perpendicular to the motion. The only effect of a magnetic force is to change the direction of motion, never to increase or decrease the magnitude of the velocity nor the kinetic energy of the particle. Example 5.1: An electron in an old-style television picture tube moves toward the front of the tube with a speed of 8.0x106 m/s along the x axis as in Figure. Surrounding the neck of the tube are coils of wire that create a magnetic field of flux density 0.025 T, directed at an angle of 50o to the x axis and lying in the xy plane. Calculate the magnetic force on the electron. Solution: =FB v B sinθ (1.6 x10−19 )(8.0 x106 )(0.025)(sin60 q= =  ) 2.8 x10 −14 N Example 5.2: Electrons are accelerated from rest through a potential difference of 350 V and then enter a uniform magnetic field that is perpendicular to the velocity vector of the electrons. The electrons travel along a curved path because of the magnetic force exerted on them, and the radius of the path is measured to be 7.5 cm. (A) What is the magnitude of the magnetic field? (B) What is the period of rotation of the electron. Solution: 1 (A) q ∆V = me v 2 2 2 q ∆V 2(1.6 x10−19 )(350) so =v = = −31 1.11x107 m / s me 9.11x10 me v (9.11x10 −31)(1.11x107 ) =B = −19 =−2 8.4 x10 −4 T qr (1.6 x10 )(7.5 x10 ) 2 π me 2 π (9.11x10 −31) = (B) T = −19 = −4 10−8 s 42.6 ns 4.26 x= qB (1.6 x10 )(8.4 x10 ) Serway & Jewett 8th edition 35 5.3 Applications involving charged particles moving in a magnetic field:   A charge moving with a velocity v in the presence of both an electric field E and a   magnetic field B experiences both an electric force q E and a magnetic force   q v x B. The total force (called the Lorentz force) acting on the charge is     = F qE +qv xB The velocity selector: Assume a combination of an electric field and a magnetic field oriented as shown in Figure 6. A uniform electric field is directed to the right, and a uniform magnetic field is applied in the direction perpendicular to the electric field (into the page).  If q is positive and the velocity v is upward, the magnetic force is to the left and the electric force is to the right. When the magnitudes of the two fields are chosen so that q E = q v Bin , the charged particle is in equilibrium and moves in a straight vertical line through the region of the fields. We find that E v = Bin Only those particles having this speed pass undeflected Figure 6. through the mutually perpendicular electric and magnetic fields. The Mass Spectrometer: A mass spectrometer separates ions according to their mass-to-charge ratio. A beam of ions first passes through a velocity selector and then enters a second  uniform magnetic field Bo that has the same direction  as the magnetic field in the selector Bin (Figure 7). Upon entering the second magnetic field, the ions move in a semicircle of radius r before striking a detector at P. mv Since r = , we can express the ratio m/q as q Bo m r Bo r Bo Bin = = Figure 7. q v E Therefore, we can determine m/q by measuring the radius of curvature and knowing the field magnitudes Bin, Bo, and E. Serway & Jewett 8th edition 36  5.4 Magnetic force on a current-carrying conductor: Figure 8 shows a segment of wire of length L and cross-sectional area A having charge density (number of charge carriers per unit volume) n. The number of charge carriers contained in this length is n L A. All these charges experience a magnetic force due  to B in given by:    ( ) F B = q v d x B in n L A  where v d is the drift velocity of charges. Substituting for n = I ( q vd A) Figure 8.    F B = I L x B in  where L is a vector that points in the direction of the current I and has a magnitude equal to the length L of the segment.  The magnitude of F B is given by: FB = I L Bin sinθ   where θ is the angle between L and B in. Example 5.3: A wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0o, (b) 90.0o, and (c) 120o. Solution: FB = I L Bin sinθ =FB (5.00)(2.80)(0.390) = sin 60 4.73 N =FB (5.00)(2.80)(0.390) = sin120 5.46 N =FB (5.00)(2.80)(0.390) = sin120 4.73 N Serway & Jewett 8th edition 37  Problems. Use the following useful information: For the proton (q = +1.6x10-19 C, m = 1.67x10-27 kg) For the electron (q = −1.6x10-19 C, m = 9.11x10-31 kg) (1) Determine the initial direction of the deflection of charged particles as they enter the magnetic fields shown in Figure below. (2) Find the direction of the magnetic field acting on a positively charged particle moving in the various situations shown in Figure below if the direction of the magnetic force acting on it is as indicated.  (3) A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00x107 m/s and experiences an acceleration of 2.00x1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field. (4) An electron moves in a circular path perpendicular to a uniform magnetic field with flux density of 2.00 mT. If the speed of the electron is 1.50x107 m/s, determine (a) the radius of the circular path and (b) the time interval required to complete one revolution. (5) In a mass spectrometer the magnitude of the electric field between the plates of the velocity selector is 2.50x103 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.0350 T. Calculate the radius of the path for a singly charged ion having a mass m = 2.18x10−26 kg. Serway & Jewett 8th edition 38 (6) A velocity selector consists of electric and magnetic fields described by the   expressions E = E k and B = (15.0 k ) mT Find the value of E such that a 750- eV electron moving in the negative x direction is undeflected. (7) A conductor carrying a current I = 15.0 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A magnetic force per unit length of 0.120 N/m acts on the conductor in the negative y direction. Determine: (a) the magnitude and (b) the direction of the magnetic field in the region through which the current passes. (8) A horizontal power line of length 58.0 m carries a current of 2.20 kA northward as shown in Figure. The Earth’s magnetic field at this location has a magnitude of 5.00x10−5 T. The field at this location is directed toward the north at an angle 65.0o below the power line. Find (a) the magnitude and (b) the direction of the magnetic force on the power line. (9) A wire carries a steady current of 2.40 A. A straight section of the wire is 0.750 m  long and lies along the x axis within a uniform magnetic field, B = (1.60 k ) T. If the current is in the positive x direction, what is the magnetic force on the section of wire? Answers. (1)(a) upwards (b) Out of page. (c) No deflection (d) Into the page. (2)(a) Into the page. (b) To the right. (c) Downwards. (3) ( −20.9 j ) mT (4)(a) 4.27 cm (b) 1.79x10−8 s (5) 0.278 m (6) 244 kV/m (7)(a) 8.00 mT (b) In the positive z direction. (8)(a) 5.78 N (b) into the page. (9) ( −2.88 j ) N Serway & Jewett 8th edition 39  Chapter 6. Oscillatory Motion. Periodic motion is motion of an object that regularly returns to a given position after a fixed time interval. The aim of this chapter is to study a special kind of periodic motion occurs in mechanical systems called the simple harmonic motion. 6.1 Motion of an object attached to a spring: As a model for simple harmonic motion, consider a block of mass m attached to the end of a spring, with the block free to move on a frictionless, horizontal surface (Figure 1). When the spring is neither stretched nor compressed, the block is at rest at the position called the equilibrium position of the system, which we identify as x = 0 (Figure 1b). Figure 1. We can understand the oscillating motion of the block in Figure 1 by recalling that when the block is displaced to a position x, the spring exerts on the block a force that is proportional to the position and given by Hooke’s law: Fs = − k x We call Fs a restoring force because it is always directed toward the equilibrium position and therefore opposite the displacement of the block from equilibrium. The resultant force Fs produces an acceleration a such that Fs = m a , so ma = − k x k and a= − x m Serway & Jewett 8th edition 40 At this point we can define the simple harmonic motion as: An object moves with simple harmonic motion whenever its acceleration is proportional to its position and is oppositely directed to the displacement from equilibrium. At the extreme positions x = +A and x = − A, the accelerations of the object are − k A m and k A m respectively. At the equilibrium position x = 0 the speed of the object is maximum and uniform and the acceleration equals zero. 6.2 Analysis Model, Particle in S.H.M.: (a) Motion in which timing starts at the extreme position, xi = A at t = 0 (Figure 2a): Figure 2  The variation of displacement with time can be expressed as: x = A cos ω t  The velocity is given as the rate of change of displacement with time: v = dx = − A ω sin ω t dt  The acceleration is given as the rate of change of velocity with time or as the second derivative of the displacement: 2 a = d 2x = − A ω 2 cos ω t dt a = − ω2 x F = m a = − mω2 x This motion is considered a simple harmonic if we take for simplicity ω2 = k m. Serway & Jewett 8th edition 41 (b) Motion in which timing starts at the equilibrium position, xi = 0 at t = 0 (Figure 3a): Figure 3  The variation of displacement with time can be expressed as: x = A sin ω t  The velocity is given as the rate of change of displacement with time: v = dx = A ω cos ω t dt  The acceleration is given as the rate of change of velocity with time or as the second derivative of the displacement: 2 a = d 2x = − A ω 2 sin ω t dt a = − ω2 x F = m a = − mω2 x Again, this motion is considered a simple harmonic if we take ω 2 = k m and 2π ω = k m , where ω = = 2 π f is called the angular frequency. T The graphs of position, velocity, and acceleration versus time for both cases are shown in Figures 2b and 3b. Serway & Jewett 8th edition 42 Example 6.1: A 200-g block connected to a light spring for which the force constant is 5.00 N/m is free to oscillate on a frictionless, horizontal surface. The block is displaced 5.00 cm from equilibrium and released from rest. (A) Find the period of its motion. (B) Determine the maximum speed of the block. (C) What is the maximum acceleration of the block? (D) Express the position, velocity, and acceleration as functions of time in SI units. Solution: 5.00k (A) ω = = 5.00 rad / s = 0.200m 2π 2π = T = = 1.26 s ω 5.00 (B) v= max ω= A (5.00)(0.0500) = 0.25 m / s 2 (C) amax ω= = A (5.00)2 (0.0500) = 1.25 m / s 2 (D) x = (0.0500) cos(5.00 t ) v = − (0.25) sin(5.00 t ) a = − (1.25) cos(5.00 t ) 6.3 Energy of the simple harmonic oscillator: Let us examine the mechanical energy of the block–spring system illustrated in Figure 1. Because the surface is frictionless, we expect the total mechanical energy of the system to be constant. We assume a massless spring, so the kinetic energy of the system corresponds only to that of the block. We can express the kinetic energy of the block as 1 1 =K = m v2 m ω 2 A2 sin2 ω t 2 2 The elastic potential energy stored in the spring for any elongation x is given by 1 2k x 2. 2 Substituting for k = m ω the elastic potential energy is 1 U = m ω 2 A2 cos2 ω t 2 The total mechanical energy of the simple harmonic oscillator is expressed as 1 m ω 2 A2 sin2 ω t + cos2 ω t E = K +U = 2 ( ) 1 1 or =E = m ω 2 A2 k A2 2 2 That is, the total mechanical energy of a simple harmonic oscillator is a constant of the motion and is proportional to the square of the amplitude. Serway & Jewett 8th edition 43 The total mechanical energy is equal to the maximum potential energy stored in the spring when x = ± A because v = 0 at these points and there is no kinetic energy. At the equilibrium position, where U = 0 because x = 0, the total energy, all in the form of kinetic energy, is again 1 2k A2. Plots of the kinetic and potential energies versus time appear in Figure 4a. At all times, the sum of the kinetic and potential energies is a constant equal to 1 2k A2. Figure 4 The variations of K and U with the position x of the block are plotted in Figure 4b. Energy is continuously being transformed between potential energy stored in the spring and kinetic energy of the block. Finally, we can obtain the velocity of the block at an arbitrary position by expressing the total energy of the system at some arbitrary position x as 1 1 1 E = K +U = m v 2 + k x 2 = k A2 2 2 2 k v = ± m ( ) A2 − x 2 = ±ω (A 2 − x2 ) When you check this equation to see whether it agrees with known cases, you find that it verifies that the speed is a maximum at x = 0 and is zero at the turning points x = ± A. Serway & Jewett 8th edition 44 Figure 5 illustrates the position, velocity, acceleration, kinetic energy, and potential energy of the block–spring system for one full period of the motion. Most of the ideas discussed so far are incorporated in this important figure. Serway & Jewett 8th edition 45 Example 6.2: A 0.500-kg cart connected to a light spring for which the force constant is 20.0 N/m oscillates on a frictionless, horizontal air track. (A) Calculate the maximum speed of the cart if the amplitude of the motion is 3.00 cm. (B) What is the velocity of the cart when the position is 2.00 cm? (C) Compute the kinetic and potential energies of the system when the position of the cart is 2.00 cm. Solution: 1 1 (A) =E = k A2 2 mv max 2 2 k 20.0 = v max = A ( 0.0300 = ) 0.190 m / s m 0.500 k 20.0 (B) v = ± m ( A2 − x 2 = ± ) 0.500 ( 0.03002 − 0.02002 = ± 0.141 m / s ) The positive and negative signs indicate that the cart could be moving to either the right or the left at this instant. 1 1 (C) = K m=v2 (0.500)(0.141) = 2 5.00 × 10 −3 J 2 2 1 1 = U x2 k= (20.0)(0.0200) = 2 4.00 × 10 −3 J 2 2 Example 6.3: The cart in example 6.2 could have been set into motion by releasing the cart from rest at x = 3.00 cm. What if the cart were released from the same position, but with an initial velocity of v = − 0.100 m/s? What are the new amplitude and maximum speed of the cart? Solution: 1 1 1 1 E = m v 2 + k x 2 = (0.500)( −0.100)2 + (20.0)(0.0300)2 = 1.15 × 10 −2 J 2 2 2 2 1 2 1 But E = k A , then 1.15 × 10 −2 = (20.0) A2 and A = 0.0339 m 2 2 k 20.0 = v max = A ( 0.0339 = ) 0.214 m / s m 0.500 The amplitude and maximum velocity are larger than the previous values because the cart was given an initial velocity at t = 0. Serway & Jewett 8th edition 46  Problems (1) The position of an object moving with simple harmonic motion is given by x = 4 cos(6 π t ) , where x is in meters and t is in seconds. What is the period of the oscillating system? 1 1 (a) 4 s (b) 6 s (c) 3 s (d) 6π s (2) A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 6.0 cm has an energy of 12 J. If the block is replaced by one whose mass is twice the mass of the original block and the amplitude of the motion is again 6.0 cm, what is the energy of the system? (a) 12 J (b) 24 J (c) 6 J (d) 48 J (3) A 0.60-kg block attached to a spring with force constant 130 N/m is free to move on a frictionless, horizontal surface as in Figure 1 of this chapter. The block is released from rest when the spring is stretched 0.13 m. At the instant the block is released, find the force on the block and its acceleration. (4) A vertical spring stretches 3.9 cm when a 10-g object is hung from it. The object is replaced with a block of mass 25 g that oscillates up and down in simple harmonic motion. Calculate the period of motion. (5) A 1.00-kg object is attached to a horizontal spring. The spring is initially stretched by 0.100 m, and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500 s later. What is the maximum speed of the object? (6) A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.00 cm, and the frequency is 1.50 Hz. (a) Find an expression for the position of the particle as a function of time. Determine (b) the maximum speed of the particle and (c) the earliest time (t > 0) at which the particle has this speed. (d) the maximum positive acceleration of the particle (e) the earliest time (t > 0) at which the particle has this acceleration. (f) the total distance traveled by the particle between t = 0 and t = 1.00 s. (7) A 1000-kg automobile is driven into a brick wall. The car’s bumper behaves like a spring with a force constant 5.00x106 N/m and compresses 3.16 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall? Serway & Jewett 8th edition 47 (8) A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm. (9) A 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system, (b) the speed of the object when its position is 1.00 cm, and (c) the kinetic energy and the potential energy when its position is 3.00 cm. Answers (1) c (2) a (3) 16.9 N to the left, 28.2 m/s2 to the left (4) 0.63 s (5) 0.628 m/s (6)(a) v = (2.00 cm ) sin(3.00 π t ) (b) 0.188 m/s (c) 0.333 s (d) 1.78 m/s2 (e) 0.500 s (f) 12.0 cm (7) 2.23 m/s (8)(a) 0.400 m/s (b) 1.60 m/s2 (c) 0.32 m/s (d) − 0.96 m/s2 (e) 0.232 s (9)(a) 0.028 J (b) 1.02 m/s (c) K = 0.0122 , U = 0.0158 J Serway & Jewett 8th edition 48  Chapter 7. Wave Motion. 7.1 Introduction, what is a wave? A wave is defined as a periodic disturbance traveling through a medium or vacuum. In a general classification, waves are divided into two categories; namely; mechanical waves and electromagnetic waves. Mechanical waves are those which need a medium to propagate by vibration of the particles of the medium. Electromagnetic waves are those which can propagate in vacuum by vibration of combined electric and magnetic fields. The scope of this chapter is confined to the study of mechanical waves. These waves are divided to two sub-categories, transverse and longitudinal waves. A transverse wave is a traveling wave or pulse that causes the elements of the disturbed medium to move perpendicular to the direction of wave propagation. This is illustrated in Figure 1, where particle P on the string vibrates up and down in a simple harmonic motion as the wave propagates horizontally along the string. Figure 1 Figure 2 A longitudinal wave is a traveling wave or pulse that causes the elements of the disturbed medium to move parallel to the direction of wave propagation. When the left end of the long, stretched spring, shown in Figure 2, is pushed to the right and then pulled to the left, a pulse is moving down the spring. This movement creates a sudden compression of a region of the coils. The direction of the displacement of the coils is parallel to the direction of propagation of the compressed region. Serway & Jewett 8th edition 49 7.2 Travelling waves: The sinusoidal wave is the simplest example of a periodic continuous wave. The brown curve in Figure 3 represents a snapshot of a traveling sinusoidal wave at t = 0, and the blue curve represents a snapshot of the wave at some later time t. A wave travelling with speed v will move a distance v t during this time. Imagine two types of motion that can occur. First, the entire waveform in Figure 3 moves to the right so that the brown curve moves toward the right and Figure 3 eventually reaches the position of the blue curve. This movement is the motion of the wave. If we focus on one particle of the medium, such as the particle at x = 0, we see that each particle moves up and down along the y axis in simple harmonic motion. This movement is the motion of the particles of the medium. Figure 4 Figure 5 Figure 4 shows a snapshot of a wave moving through a medium. Figure 5 shows a graph of the position of one particle of the medium as a function of time. The displacement (y): This is the distance of any particle on the wave from its undisturbed (equilibrium) position. It has the unit of meter (m). The amplitude (A): This is the maximum displacement of any point on the wave from its undisturbed (equilibrium) position. It has the unit of meter (m). The wavelength (λ): This is the minimum distance between any two adjacent points in the wave (e.g.) crest to crest or trough to trough. It has the unit of meter (m). The period (T): This is the time taken by a particle on the wave to make one complete vibration. It is also the time taken by the wave to move a complete wavelength. It has the unit of second (s). Serway & Jewett 8th edition 50 The frequency (f): This is the number of crests (or troughs, or any other point on the wave) that pass a given point in a unit time interval. The frequency of the wave is the same as the frequency of the simple harmonic oscillation of one particle of the medium. The frequency of a sinusoidal wave is related to the period by the expression f = 1 T The unit of the frequency is Hertz (Hz) or s−1. It is clear from Figures 4 and 5 that the displacement y of the particles in the wave is a function of both position x of the particle from the source of the wave (placed at x = 0) and the time t. For a wave moving to the right with speed v, this function can be expressed as:  2π  y ( x , t ) = A sin  ( x − v t )  λ  If the wave were traveling to the left, then  2π  y ( x , t ) A sin  = ( x + v t )  λ  The wave speed, wavelength, period and frequency are related by the expression: λ = v = λf T Substituting this expression for v gives  x t  y ( x , t ) = A sin 2 π  −    λ T  To simply this equation we define two more quantities, 2π The angular wave number: k = rad/m λ 2π The angular frequency: ω = rad/s T Then y ( x , t ) = A sin (k x − ω t ) If the vertical position y of a particle of the medium is not zero at x = 0 and t = 0, we generally express the wave function in the form ( x , t ) A sin ( k x − ω t + φ ) y= where φ is the phase constant and can be determined from the initial conditions, as will be shown in the next example. It represents by which angle is one particle shifted relative to the other. The path difference represents by which distance is one particle shifted relative to the other. It is related to the phase constant by φ φ path difference = = λ 2π k Serway & Jewett 8th edition 51  Example 7.1: A sinusoidal wave traveling in the positive x direction has an amplitude of 15.0 cm, a wavelength of 40.0 cm, and a frequency of 8.00 Hz. The vertical position of a particle of the medium at t = 0 and x = 0 is also 15.0 cm as shown in Figure 6. (A) Find the wave number k, period T, angular frequency ω, and speed v of the wave. (B) Determine the phase constant φ and write a Figure 6 general expression for the wave function. Solution: