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 Electric charge is conserved. –Positively charged particles are called protons. –Uncharged particles are called neutrons. –Negatively charged particles are called electrons.

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 1 Electric Charge Properties of Electric Charge, continued Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge. Charge is measured in coulombs (C). The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton. e = x 10 –19 C

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

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

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. An electrical insulator is a material in which charges cannot move freely.

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

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Coulomb’s Law Two charges near one another exert a force on one another called the electric force. Coulomb’s law states that the electric force is propor- tional to the magnitude of each charge and inversely proportional to the square of the distance between them. Section 2 Electric Force

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Coulomb’s Law, continued 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. Section 2 Electric Force

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

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

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 =  10 –9 Cr 2,1 = 3.00 m q 2 = –2.00  10 –9 Cr 3,2 = 4.00 m q 3 =  10 –9 Cr 3,1 = 5.00 m  = 37.0º Unknown:F 3,tot = ? Diagram:

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 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.

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

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

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

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Coulomb’s Law, continued 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. Chapter 16 Section 2 Electric Force

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Strength An electric field is a region where an electric force on a test charge can be detected. 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.

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Electric Field Strength, continued Electric field strength depends on charge and distance. An electric field exists in the region around a charged object. Electric Field Strength Due to a Point Charge

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

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 = µC is at the origin, and a charge q 2 = –5.00 µC is on the x- axis m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis m from the origin.

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 = µC = 7.00  10 –6 Cr 1 = m q 2 = –5.00 µC = –5.00  10 –6 Cr 2 = m  = 53.1º Unknown: E at P (y = 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

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

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

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

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  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

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

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

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

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

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

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

Copyright © by Holt, Rinehart and Winston. All rights reserved. ResourcesChapter menu Chapter 16 Section 3 The Electric Field Conductors in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor. Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface. The electric field just outside a charged conductor is perpendicular to the conductor’s surface. On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.