1. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16: Electric Forces and Fields 16.1 Electric Charge 16.2 Electric Force 16.3 The Electric Field Chapter 16

2. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 1 Electric Charge Chapter 16 Objectives for Section 1 Electric Charge Understand the basic properties of electric charge. Differentiate between conductors and insulators. Distinguish between charging by contact, charging by induction, and charging by polarization.

3. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Properties of Electric Charge Atoms are the source of electric charge. –Positively charged particles are called protons. –Uncharged particles are called neutrons. –Negatively charged particles are called electrons. Friction Rods Electrons from animal fur are transferred to atoms in ebonite (hard rubber). Ebonite acquires a net excess of electrons. Electrons from a glass rod will transfer to a silk cloth and give rise to an excess of electrons on the silk. Your hair becomes positively charged when you rub a balloon across it. Electric charge is conserved. The amount of positive charge acquired by your hair equals the amount of negative charge acquired by the balloon.

4. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Electric Charge Section 1 Electric Charge

5. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Properties of Electric Charge There are two kinds of electric charge. –like charges repel –unlike charges attract The magnitude of electrical forces between two charged bodies often exceeds the gravitational attraction between the bodies. Electric charge is conserved.

6. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Properties of Electric Charge Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge. +e, +2e, +3e, … The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton. e = 1.602 176 x 10 –19 C Charge is measured in coulombs (C). -1.0 C contains 6.2 x10 18 electrons.

7. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 The Milikan Experiment (1909) Section 1 Electric Charge

8. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Milikan’s Oil Drop Experiment Section 1 Electric Charge

9. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Transfer of Electric Charge An electrical conductor is a material in which charges can move freely. Most metals are conductors. An electrical insulator, or nonconductor, is a material in which charges cannot move freely. (Electrons are tightly bound to the atom.) Nonmetallic materials, such as glass, rubber, and wood are good insulators.

10. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Transfer of Electric Charge Insulators and conductors can be charged by contact. Conductors can be charged by induction. Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor.

11. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Charging by Induction Section 1 Electric Charge

12. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Transfer of Electric Charge A surface charge can be induced on insulators by polarization. With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.

13. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 2 Electric Force Chapter 16 Section 2 Electric Force Objectives Calculate electric force using Coulomb’s law. Compare electric force with gravitational force. Apply the superposition principle to find the resultant force on a charge and to find the position at which the net force on a charge is zero.

14. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Coulomb’s Law (Charles Coulomb, 1780’s) Two charges near one another exert a force on one another called the electric force. Coulomb’s law describes this force. Coulomb’s law states that the electric force is proportional to the magnitude of each charge and inversely proportional to the square of the distance between them. Section 2 Electric Force k c = 8.99 x 10 9 N. m 2 /C 2 when F in N, r in m, and q in C

15. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Coulomb’s Law Electric force is a vector. When the charges (q 1 and q 2 ) are alike they repel each other, and when they are opposite they attract each other. The resultant force on a charge is the vector sum of the individual forces on that charge. Adding forces this way is an example of the principle of superposition. When a body is in equilibrium, the net external force acting on that body is zero. A charged particle can be positioned such that the net electric force on the charge is zero. Set Coulomb’s Law forces equal and solve for the distance between either charge and the equilibrium position. (Practice C) Section 2 Electric Force

16. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Superposition Principle Section 2 Electric Force

17. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem The Superposition Principle Consider three point charges at the corners of a triangle, as shown at right, where q 1 = 6.00  10 –9 C, q 2 = –2.00  10 –9 C, and q 3 = 5.00  10 –9 C. Find the magnitude and direction of the resultant force on q 3. Section 2 Electric Force

18. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem, continued The Superposition Principle 1. Define the problem, and identify the known variables. Given: q 1 = +6.00  10 –9 Cr 2,1 = 3.00 m q 2 = –2.00  10 –9 Cr 3,2 = 4.00 m q 3 = +5.00  10 –9 Cr 3,1 = 5.00 m  = 37.0º Unknown:F 3,tot = ? Diagram: F el = k C q1q2r2q1q2r2

19. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem, continued The Superposition Principle Tip: According to the superposition principle, the resultant force on the charge q 3 is the vector sum of the forces exerted by q 1 and q 2 on q 3. First, find the force exerted on q 3 by each, and then add these two forces together vectorially to get the resultant force on q 3. 2. Determine the direction of the forces by analyzing the charges. The force F 3,1 is repulsive because q 1 and q 3 have the same sign. The force F 3,2 is attractive because q 2 and q 3 have opposite signs.

20. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem 3. Calculate the magnitudes of the forces with Coulomb’s law. 2

21. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem, 4. Find the x and y components of each force. At this point, the direction each component must be taken into account. F 3,1 : F x = (F 3,1 )(cos 37.0º) = (1.08  10 –8 N)(cos 37.0º) F x = 8.63  10 –9 N F y = (F 3,1 )(sin 37.0º) = (1.08  10 –8 N)(sin 37.0º) F y = 6.50  10 –9 N F 3,2 :F x = –F 3,2 = –5.62  10 –9 N F y = 0 N

22. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem 5. Calculate the magnitude of the total force acting in both directions. F x,tot = 8.63  10 –9 N – 5.62  10 –9 N = 3.01  10 –9 N F y,tot = 6.50  10 –9 N + 0 N = 6.50  10 –9 N

23. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem 6. Use the Pythagorean theorem to find the magni- tude of the resultant force.

24. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 2 Electric Force Sample Problem 7. Use a suitable trigonometric function to find the direction of the resultant force. In this case, you can use the inverse tangent function:

25. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Coulomb’s Law The Coulomb force is a field force. A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects. Gravitational attraction is also a field force. F G = G Chapter 16 Section 2 Electric Force m1m2r2m1m2r2 F el = k C q1q2r2q1q2r2

26. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Section 3 The Electric Field Chapter 16 Objectives Calculate electric field strength. Draw and interpret electric field lines. Identify the four properties associated with a conductor in electrostatic equilibrium.

27. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Strength (Intensity) An electric field is a region where an electric force on a test charge ( small positive charge ) can be detected. Q exerts an electric field on q 0. The SI units of the electric field, E, are newtons per coulomb (N/C). The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge (the direction a + test charge accelerates). E = F elec q 0 Q q0q0 Q in the figure above is (+) because it is exerting a force of repulsion on the test charge.

28. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Electric Fields and Test Charges Section 3 The Electric Field

29. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Strength Electric field strength (intensity) depends on charge and distance. An electric field exists in the region around a charged object. Electric Field Strength Due to a Point Charge F elec = k C qq 0 r 2 E = F elec q 0 and Q q0q0 Q

30. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Strength Direction When q is (+) the electric field is directed outward radially from q. When q is (-) the electric field is directed toward q.

31. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Lines Electric field lines are used to analyze electric fields and show strength and direction. –The number of electric field lines is proportional to the electric field strength. –Electric field lines are tangent to the electric field vector at any point.

32. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Rules for Drawing Electric Field Lines Section 3 The Electric Field

33. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Rules for Sketching Fields Created by Several Charges Section 3 The Electric Field

34. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field 4 Properties of Conductors in Electrostatic Equilibrium (Distribution of Static Charges) 1.The electric field is zero everywhere inside the conductor (otherwise charges would move and it would not be at equilibrium). 2.Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface (because the excess charges repel each other).

35. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field 4 Properties of Conductors in Electrostatic Equilibrium (Distribution of Static Charges) 3.The electric field just outside a charged conductor is perpendicular to the conductor’s surface. 4.On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.

36. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Discharging Effects of Points The charge density is greatest at the point of greatest curvature. St. Elmo’s Fire A corona discharge or brush is a slow leakage of charge that occurs when the electric field intensity is great enough to produce ionization at sharp projections or corners. Spark Discharge Dry air can ionize at atmospheric pressure when a difference of 30,000 J/C/cm (V/cm) exists between two charged surfaces.

37. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Calculating Net Electric Field Section 3 The Electric Field

38. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Sample Problem Electric Field Strength A charge q 1 = +7.00 µC is at the origin, and a charge q 2 = –5.00 µC is on the x- axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin.

39. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 1. Define the problem, and identify the known variables. Given: q 1 = +7.00 µC = 7.00  10 –6 Cr 1 = 0.400 m q 2 = –5.00 µC = –5.00  10 –6 Cr 2 = 0.500 m  = 53.1º Unknown: E at P (y = 0.400 m) Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors. Section 3 The Electric Field

40. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge. Section 3 The Electric Field

41. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 3. Analyze the signs of the charges. The field vector E 1 at P due to q 1 is directed vertically upward, as shown in the figure, because q 1 is positive. Likewise, the field vector E 2 at P due to q 2 is directed toward q 2 because q 2 is negative. Section 3 The Electric Field

42. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 4. Find the x and y components of each electric field vector. For E 1 : E x,1 = 0 N/C E y,1 = 3.93  10 5 N/C For E 2 : E x,2 = (1.80  10 5 N/C)(cos 53.1º) = 1.08  10 5 N/C E y,1 = (1.80  10 5 N/C)(sin 53.1º)= –1.44  10 5 N/C Section 3 The Electric Field

43. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 5. Calculate the total electric field strength in both directions. E x,tot = E x,1 + E x,2 = 0 N/C + 1.08  10 5 N/C = 1.08  10 5 N/C E y,tot = E y,1 + E y,2 = 3.93  10 5 N/C – 1.44  10 5 N/C = 2.49  10 5 N/C Section 3 The Electric Field

44. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector. Section 3 The Electric Field

45. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector. In this case, you can use the inverse tangent function: Section 3 The Electric Field

46. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Sample Problem, continued Electric Field Strength 8. Evaluate your answer. The electric field at point P is pointing away from the charge q 1, as expected, because q 1 is a positive charge and is larger than the negative charge q 2. Section 3 The Electric Field

47. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Charging By Induction

48. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Transfer of Electric Charge

49. Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Lines