Physics 2 - Lecture 3: Electromagnetism 1 - Pham Tan Thi

Nội dung text: Physics 2 - Lecture 3: Electromagnetism 1 - Pham Tan Thi

1. Vietnam National University Ho Chi Minh City Ho Chi Minh City University of Technology General Physics A1 Week 11-15: Electromagnetism
2. Maxwell Equations and Light
3. Atomic Structure q Benjamin unified the concepts of lightning and static electricity (but not quite right) q In particular his choice of signs will come back to haunt us when we get to circuits q Let’s briefly review atomic structure
4. Conductors and Insulators q If outermost electrons are free to move within the material, we call the material a conductor q If they’re not free to move (but can still be rubbed off), we call the material an insulator q Third possibility is that a material can be a semiconductor q Now what happened with the glass rod and silk, plastic rod and wool
5. Extracting Charges The van de Graaff machine: q Has a belt which transfers charge to the metal dome q Does it take electrons up to the dome or transfer electrons away from the dome?
6. How does it work
7. Charging a Human q Body consists largely of salt water and thus is a reasonably good conductor q When touching the metal, some of Cl- ions on skin surface transfer extra electron to metal q Leaving body with an excess of Na+ ions, and thus a net positive charge q By grounding objects, we are trying to prevent the build up of any significant charge on them q Excess charge shared with the earth
8. Electric Dipoles q We can induce a dipole on an insulator by bringing a charge close
9. QUIZ q Metal spheres 1 and 2 are touching. Both are initially neutral. a. The charged rod is brought near. b. The charged rod is then removed. c. The spheres are separated. q Afterward, the charges on the sphere are: A. Q1 is + and Q2 is +. B. Q1 is + and Q2 is –. C. Q1 is – and Q2 is +. D. Q1 is – and Q2 is –. E. Q1 is 0 and Q2 is 0.
10. QUIZ q Metal spheres 1 and 2 are touching. Both are initially neutral. a. The charged rod is brought near. b. The spheres are separated. c. The charged rod is then removed. q Afterward, the charges on the sphere are: A. Q1 is + and Q2 is +. B. Q1 is + and Q2 is –. C. Q1 is – and Q2 is +. D. Q1 is – and Q2 is –. E. Q1 is 0 and Q2 is 0.
11. Coulomb’s Law q There is a force between 2 charges along the line joining them (a central force). q The size of the force drops as 1/r2
12. Coulomb Force
13. Force on q due to a continuous charge distribution q Consider the interaction of an infinitesimal volume element with the point charge q q Assume charge density ρ(r’); the charge in the volume element is q So the force on q due to the volume element is q The net fore is
14. Fields: Proposal by Michael Faraday q Faraday was struck by the way iron filings lined up around a bar magnet. q Perhaps the magnet is altering space around itself, and this alteration is responsible for the long range force. q Faraday’s idea (he called it “lines of force”) came to be called a field. q It’s very intuitive (like most of Faraday’s thoughts) but the idea of a field was placed on a firm mathematical footing later by James Clerk Maxwell.
15. Electric Field q Let’s consider the idea of an electric field. q I put a positive test charge q0 at some point in space and then use the force on this test charge to measure the electric field due to another charge Q. q The electric field is present at the position of the test charge q0, whether or not q0 is there q The electric field is everywhere q The force on the test charge is just a convenient way of measuring the field
16. Electric Field due to a Point Charge q The source of the electric field = a point with charge Q. q q0 = a small positive test charge, located at r with respect to Q. q The force on q0 (Coulomb force) is The electric field at r is, by definition, Now forget about the test charge. The electric field s present even if there is no test charge. The test charge is just a way to measure E(r).
17. Electric Field from a Point Charge q Let’s put a test charge q0 at some point in space, and measure the force: q We can determine the electric field from the point charge q q The electric field that we measure has nothing to do with test charge q0
18. Unit vector, r-hat
19. QUIZ q At which point is the electric field stronger? A. Point A B. Point B C. Not enough information to tell.
20. QUIZ q What is the direction of the electric field at the black dot? A. A B. B C. C D. D E. None of these
21. Electric Field from 3 Equal + Point Charges q Enet = E1 + E2 + E3 q Start by simplifying: q No component along z direction so Ez = 0 q (E3)y and (E1)y will cancel each other out so there is only a component along the x-axis (Enet)x = (E1)x + (E2)x + (E3)x = 2(E1)x + (E2)x
22. Exercise: Electric Field Due to Two Charges Charges q1 and q2 are located on the x axis, at distances a and b, respectively, from the origin as shown in Figure. (A) Find the components of the net electric field at the point P, which is at position (0, y). (B) Evaluate the electric field at point P in the special case |q1| = |q2| and a = b.
23. Exercise: Electric Field Due to Two Charges (B) Evaluate the electric field at point P in the special case |q1| = |q2| and a = b.
24. Charge Density q If a charge Q is uniformly distributed throughout a volume V, the volume charge density ρ is defined by q If a charge Q is uniformly distributed on a surface of area A, the surface charge density σ (Greek letter sigma) is defined by q If a charge Q is uniformly distributed along a line of length l, the linear charge density λ is defined by ρ, σ, and λ have units of C/m3, C/m2, and C/m, respectively If the charge is nonuniformly distributed over a volume, surface, or line, the amount of charge dq
25. Electric Field due to a Charge Rod
26. Electric Field of a Uniform Ring of Charge
27. Electric Field of a Uniformly Charged Disk
28. Motion of a Charged Particle in a Uniform Electric Field q When a particle of charge q and mass m is placed in an electric field E, the electric force exerted on the charge is qE. If the only field exerts on the particle, it must be the net force and causes the particle to accelerate q The acceleration of the particle q If E is uniform (that is, constant in magnitude and direction), and the particle is free to move, the electric force on the particle is constant and we can apply the particle under constant acceleration model to the motion of the particle.
29. An Accelerating Positive Charge q 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. 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 in terms of energy
30. An Accelerating Positive Charge (B) Find the speed of the particle at (B) by modeling it as a non-isolated system in terms of energy
31. An Accelerated Electron (A) Find the acceleration of the electron while it is in the electric field.
32. An Accelerated Electron (C) Assuming the vertical position of the electron as it enters the field is yi = 0, what is its vertical position when it leaves the field?
33. Electric Flux q Considering an electric field that is uniform in both magnitude and direction as shown in Figure 24.1 q The field lines penetrate a rectangular surface of area whose plane is oriented perpendicular to the field. q The number of lines per unit area (in other words, the line density) is proportional to the magnitude of the electric field. q The total number of lines penetrating the surface is proportional to the product EA (Electric field x Area). q This product of the magnitude of the electric field and surface area perpendicular to the field is called the electric flux.
34. Electric Flux q Let’s now define how the electric flux through a surface • ! = E⊥A = EAcosθ • θ is the angle between the normal to a surface at a particular and the electric field passing through that point on the surface q The electric flux measures the amount of electric field passing through a surface A if the normal to A is tilted an angle θ with respect to the electric field
35. Electric Flux q If the electric field is non- uniform, or if the surface is non-uniform, then we can calculate the electric flux from each little piece where I can approximate thing as being uniform and then add all of these pieces together
36. Symmetrical Shape
37. Gaussian Surface q A Gaussian surface is defined as a completely enclosed surface q To be completely enclosed, it of course has to be 3-dimensional but will most often be represented by 2- dimensional projection q Usually we try to pick the Gaussian surfaces to match the observed symmetry of the charge distribution
38. Gauss’s Law q “The net flux through any closed surface surrounding a point charge q is given by q/ε0 and is independent of the shape of that surface”.
39. A Spherically Symmetric Charge Distribution q An insulating solid sphere of radius a has a uniform volume charge density ρ and carries a total positive charge Q (Fig. 24.10). (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.
40. A Spherically Symmetric Charge Distribution (B) Find the magnitude of the electric field at a point inside the sphere.
41. A Cylindrically Symmetric Charge Distribution q Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length λ
42. A Plane of Charge q Find the electric field due to an infinite plane of positive charge with uniform surface charge density σ.
43. A Sphere Inside a Spherical Shell q A solid insulating sphere of radius a carries a net positive charge Q uniformly distributed throughout its volume. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and carries a net charge -2Q. Using Gauss’s law, find the electric field in the regions labeled (1), (2), (3), and (4) in Figure and the charge distribution on the shell when the entire system is in electrostatic equilibrium.
44. Electric Potential
45. Electric Potential and Potential Difference q For an infinitesimal displacement ds of a point charge q immersed in an electric field, the work done within the charge–field system by the electric field on the charge is q Internal work done in a system is equal to the negative of the charge q displaced, the electric potential energy (charge-field is changed by a small amount) q The change in electric potential energy of the system when the charge is displaced at finite distance:
46. Electric Potential and Potential Difference q Let’s now consider the situation in which an external agent moves the charge in the field. If the agent moves the 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 = ΔU. q The work done by an external agent in moving a charge q through an electric field at constant velocity is q The SI unit of both electric potential and potential difference is Voltage: q The SI unit of electric field (N/C) can also be expressed in volts per meter
47. Electric Potential due to a Dipole An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a as shown in Figure. The dipole is along the x axis and is centered at the origin. (A) Calculate the electric potential at point P on the y axis. (B) Calculate the electric potential at point R on the positive x axis (C)Calculate V and Ex at a point on the x axis far from the dipole
48. Electric Potential due to a Dipole (B) Calculate the electric potential at point R on the positive x axis
49. Electric Potential due to a Uniformly Charged Ring (A) Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q (B) Find an expression for the magnitude of the electric field at point P.
50. Electric Potential due to a Uniformly Charged Ring (B) Find an expression for the magnitude of the electric field at point P.
51. Electric Potential due to a Uniformly Charged Disk A uniformly charged disk has radius R and surface charge density σ. (A) Find the electric potential at a point P along the perpendicular central axis of the disk (B) Find the x component of the electric field at a point P along the perpendicular central axis of the disk
52. Electric Potential due to a Uniformly Charged Disk Find the electric potential at a point P along the perpendicular central axis of the disk
53. Electric Potential due to a Finite Line of Charge
54. Two Connected Charged Spheres Two spherical conductors of radii r1 and r2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire as shown in Figure. The charges on the spheres in equilibrium are q1 and q2, respectively, and they are uniformly charged. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres.
55. Capacitance and Dielectrics
56. Calculating Capacitance q Capacitance of an isolated charged sphere q Capacitance of parallel-plate
57. The Cylindrical Capacitor
58. The Spherical Capacitor
59. Energy Stored in a Charged Capacitor q Suppose q is the charge on the capacitor at some instant during the charging process. At the same instant, the potential difference across the capacitor is ΔV = q/C. q The work necessary to transfer an increment of charge dq from the plate carrying charge -q to the plate carrying charge q (which is at the higher electric potential) is The work required to transfer the charge dq is the area of the tan rectangle in Figure 26.11 q The work done in charging the capacitor appears as electric potential energy UE stored in the capacitor.
60. Electrical Dipole q We have known electric dipoles before, either induced for permanent. q Now we can calculate the electric field from a dipole
61. Electric Dipole q Along the x-axis, the general formula for the electric field from a dipole is This is in opposite direction of the dipole moment, and half of the magnitude of the field along the y-axis
62. Microscopic Model of Current q Considering the current in a cylindrical conductor of cross-sectional area A (Fig. 27.2) • Δx: conductor length • Volume ≡ AΔx • n: the number of mobile charge carriers per unit volume q The total charge ΔQ in this segment The number of carriers v : drift velocity in the segment d
63. Drift Speed in a Copper Wire
64. Ohm’s Law and Resistance q A current density and an electric field are established in a conductor whenever a potential difference is maintained across the conductor. In some materials, the current density is proportional to the electric field: (σ: conductivity) If potential difference: q Ohm’s law: For many materials (including most The magnitude of potential metals), the ratio of the current difference: density to the electric field is a constant s that is independent of the R electric field producing the current.
65. The Resistance of Nichrome Wire q The radius of 22-gauge Nichrome wire is 0.32 mm. (A)Calculate the resistance per unit length of this wire. (B) If the potential difference of 10 V is maintained across a 1.0-m length of the Nichcrome, what is the current in the wire?
66. The Resistance of Nichrome Wire If the potential difference of 10 V is maintained across a 1.0-m length of the Nichcrome, what is the current in the wire?