Chapter 16 Preview Objectives Properties of Electric Charge Section 1 Electric Charge Preview Objectives Properties of Electric Charge Transfer of Electric Charge

Chapter 16 Section 1 Electric Charge Objectives Understand the basic properties of electric charge. Differentiate between conductors and insulators. Distinguish between charging by contact, charging by induction, and charging by polarization.

Properties of Electric Charge 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.

Chapter 16 Electric Charge Section 1 Electric Charge Click below to watch the Visual Concept. Visual Concept

Properties of Electric Charge, continued 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 = 1.602 176 x 10–19 C

The Milikan Experiment Chapter 16 Section 1 Electric Charge The Milikan Experiment

Milikan’s Oil Drop Experiment Chapter 16 Section 1 Electric Charge Milikan’s Oil Drop Experiment Click below to watch the Visual Concept. Visual Concept

Transfer of Electric Charge 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.

Transfer of Electric Charge, continued 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.

Chapter 16 Charging by Induction Visual Concepts Click below to watch the Visual Concept. Visual Concept

Transfer of Electric Charge, continued 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.

Chapter 16 Preview Objectives Coulomb’s Law Sample Problem Section 2 Electric Force Preview Objectives Coulomb’s Law Sample Problem

Chapter 16 Objectives Calculate electric force using Coulomb’s law. 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.

Chapter 16 Section 2 Electric Force 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.

Coulomb’s Law, continued Chapter 16 Section 2 Electric Force 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.

Superposition Principle Chapter 16 Section 2 Electric Force Superposition Principle Click below to watch the Visual Concept. Visual Concept

Chapter 16 Sample Problem Section 2 Electric Force Sample Problem The Superposition Principle Consider three point charges at the corners of a triangle, as shown at right, where q1 = 6.00  10–9 C, q2 = –2.00  10–9 C, and q3 = 5.00  10–9 C. Find the magnitude and direction of the resultant force on q3.

Sample Problem, continued Chapter 16 Section 2 Electric Force Sample Problem, continued The Superposition Principle 1. Define the problem, and identify the known variables. Given: q1 = +6.00  10–9 C r2,1 = 3.00 m q2 = –2.00  10–9 C r3,2 = 4.00 m q3 = +5.00  10–9 C r3,1 = 5.00 m  = 37.0º Unknown: F3,tot = ? Diagram:

Sample Problem, continued Chapter 16 Section 2 Electric Force Sample Problem, continued The Superposition Principle Tip: According to the superposition principle, the resultant force on the charge q3 is the vector sum of the forces exerted by q1 and q2 on q3. First, find the force exerted on q3 by each, and then add these two forces together vectorially to get the resultant force on q3. 2. Determine the direction of the forces by analyzing the charges. The force F3,1 is repulsive because q1 and q3 have the same sign. The force F3,2 is attractive because q2 and q3 have opposite signs.

Sample Problem, continued Chapter 16 Section 2 Electric Force Sample Problem, continued The Superposition Principle 3. Calculate the magnitudes of the forces with Coulomb’s law.

Sample Problem, continued 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. F3,1: Fx = (F3,1)(cos 37.0º) = (1.08  10–8 N)(cos 37.0º) Fx = 8.63  10–9 N Fy = (F3,1)(sin 37.0º) = (1.08  10–8 N)(sin 37.0º) Fy = 6.50  10–9 N F3,2: Fx = –F3,2 = –5.62  10–9 N Fy = 0 N

Sample Problem, continued Chapter 16 Section 2 Electric Force Sample Problem, continued The Superposition Principle 5. Calculate the magnitude of the total force acting in both directions. Fx,tot = 8.63  10–9 N – 5.62  10–9 N = 3.01  10–9 N Fy,tot = 6.50  10–9 N + 0 N = 6.50  10–9 N

Sample Problem, continued 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.

Sample Problem, continued 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:

Coulomb’s Law, continued Chapter 16 Section 2 Electric Force 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 Preview Objectives Electric Field Strength Sample Problem Section 3 The Electric Field Preview Objectives Electric Field Strength Sample Problem Electric Field Lines Conductors in Electrostatic Equilibrium

Chapter 16 Objectives Calculate electric field strength. Section 3 The Electric Field Objectives Calculate electric field strength. Draw and interpret electric field lines. Identify the four properties associated with a conductor in electrostatic equilibrium.

Electric Field Strength 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.

Electric Fields and Test Charges Chapter 16 Section 3 The Electric Field Electric Fields and Test Charges Click below to watch the Visual Concept. Visual Concept

Electric Field Strength, continued 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

Calculating Net Electric Field Chapter 16 Section 3 The Electric Field Calculating Net Electric Field Click below to watch the Visual Concept. Visual Concept

Chapter 16 Sample Problem Electric Field Strength Section 3 The Electric Field Sample Problem Electric Field Strength A charge q1 = +7.00 µC is at the origin, and a charge q2 = –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.

Sample Problem, continued Chapter 16 Section 3 The Electric Field Sample Problem, continued Electric Field Strength 1. Define the problem, and identify the known variables. Given: q1 = +7.00 µC = 7.00  10–6 C r1 = 0.400 m q2 = –5.00 µC = –5.00  10–6 C r2 = 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.

Sample Problem, continued Chapter 16 Section 3 The Electric Field 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.

Sample Problem, continued Chapter 16 Section 3 The Electric Field Sample Problem, continued Electric Field Strength 3. Analyze the signs of the charges. The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.

Sample Problem, continued Chapter 16 Section 3 The Electric Field Sample Problem, continued Electric Field Strength 4. Find the x and y components of each electric field vector. For E1: Ex,1 = 0 N/C Ey,1 = 3.93  105 N/C For E2: Ex,2= (1.80  105 N/C)(cos 53.1º) = 1.08  105 N/C Ey,1= (1.80  105 N/C)(sin 53.1º)= –1.44  105 N/C

Sample Problem, continued Chapter 16 Section 3 The Electric Field Sample Problem, continued Electric Field Strength 5. Calculate the total electric field strength in both directions. Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08  105 N/C = 1.08  105 N/C Ey,tot = Ey,1 + Ey,2 = 3.93  105 N/C – 1.44  105 N/C = 2.49  105 N/C

Sample Problem, continued Chapter 16 Section 3 The Electric Field Sample Problem, continued Electric Field Strength 6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector.

Sample Problem, continued Chapter 16 Section 3 The Electric Field 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:

Sample Problem, continued Chapter 16 Section 3 The Electric Field Sample Problem, continued Electric Field Strength 8. Evaluate your answer. The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.

Chapter 16 Electric Field Lines 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.

Rules for Drawing Electric Field Lines Chapter 16 Section 3 The Electric Field Rules for Drawing Electric Field Lines Click below to watch the Visual Concept. Visual Concept

Rules for Sketching Fields Created by Several Charges Chapter 16 Section 3 The Electric Field Rules for Sketching Fields Created by Several Charges Click below to watch the Visual Concept. Visual Concept

Conductors in Electrostatic Equilibrium 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.