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Chapter 16: Electric Forces and Fields

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## Chapter 16: Electric Forces and Fields

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**Chapter 16**Chapter 16: Electric Forces and Fields 16.1 Electric Charge 16.2 Electric Force 16.3 The Electric Field**Section 1 Electric Charge**Chapter 16 Objectives for Section 1 Electric Charge • Understandthe basic properties of electric charge. • Differentiatebetween conductors and insulators. • Distinguishbetween charging by contact, charging by induction, and charging by polarization.**Section 1 Electric Charge**Chapter 16 Properties of Electric Charge • Atoms are the source of electric charge. • Positively charged particles are calledprotons. • Uncharged particles are calledneutrons. • Negatively charged particles are calledelectrons. • 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.**Section 1 Electric Charge**Chapter 16 Electric Charge**Section 1 Electric Charge**Chapter 16 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.**Section 1 Electric Charge**Chapter 16 Properties of Electric Charge • Electric charge is quantized.That is, when an object is charged, its charge is always a multiple of afundamental 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 x1018 electrons.**Section 1 Electric Charge**Chapter 16 The Milikan Experiment (1909)**Section 1 Electric Charge**Chapter 16 Milikan’s Oil Drop Experiment**Section 1 Electric Charge**Chapter 16 Transfer of Electric Charge • Anelectrical conductoris a material in which charges can move freely. Most metals are conductors. • Anelectrical 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.**Section 1 Electric Charge**Chapter 16 Transfer of Electric Charge • Insulators and conductors can be charged by contact. • Conductors can be charged byinduction. • Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor.**Section 1 Electric Charge**Chapter 16 Charging by Induction**Section 1 Electric Charge**Chapter 16 Transfer of Electric Charge • A surface charge can be induced on insulators bypolarization. • With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.**Section 2 Electric Force**Chapter 16 Section 2 Electric Force Objectives • Calculateelectric force using Coulomb’s law. • Compareelectric force with gravitational force. • Applythe 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.**Section 2 Electric Force**Chapter 16 Coulomb’s Law (Charles Coulomb, 1780’s) • Two charges near one another exert a force on one another called theelectric 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. kc = 8.99 x 109 N.m2/C2 when F in N, r in m, and q in C**Section 2 Electric Force**Chapter 16 Coulomb’s Law • Electric force is a vector. When the charges (q1 and q2) 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**Chapter 16 Superposition Principle**Section 2 Electric Force**Chapter 16 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.**Section 2 Electric Force**Chapter 16 q1q2 r2 Sample Problem, continued Fel = kC 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 q = 37.0º Unknown:F3,tot= ? Diagram:**Section 2 Electric Force**Chapter 16 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.**Section 2 Electric Force**Chapter 16 Sample Problem 3. Calculate the magnitudes of the forces with Coulomb’s law. 2**Section 2 Electric Force**Chapter 16 Sample Problem, 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**Section 2 Electric Force**Chapter 16 Sample Problem 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**Section 2 Electric Force**Chapter 16 Sample Problem 6. Use the Pythagorean theorem to find the magni-tude of the resultant force.**Section 2 Electric Force**Chapter 16 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:**Section 2 Electric Force**Chapter 16 q1q2 r2 Coulomb’s Law Fel = kC • 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. FG = G m1m2 r2**Section 3 The Electric Field**Chapter 16 Objectives • Calculateelectric field strength. • Draw and interpretelectric field lines. • Identifythe four properties associated with a conductor in electrostatic equilibrium.**Section 3 The Electric Field**q0 Q E = Felec q0 Chapter 16 Electric Field Strength (Intensity) • Anelectric fieldis a region where an electric force on a test charge (small positive charge) can be detected. Q exerts an electric field on q0. • The SI units of the electric field, E, arenewtons 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). Q in the figure above is (+) because it is exerting a force of repulsion on the test charge.**Section 3 The Electric Field**Chapter 16 Electric Fields and Test Charges**Section 3 The Electric Field**Q q0 Felec =kC qq0 r2 E = Felec q0 Chapter 16 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 and Q**Section 3 The Electric Field**Chapter 16 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.**Section 3 The Electric Field**Chapter 16 Electric FieldLines • 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.**Section 3 The Electric Field**Chapter 16 Rules for Drawing Electric Field Lines**Section 3 The Electric Field**Chapter 16 Rules for Sketching Fields Created by Several Charges**Section 3 The Electric Field**Chapter 16 4 Properties of Conductors in Electrostatic Equilibrium (Distribution of Static Charges) • The electric field is zero everywhere inside the conductor (otherwise charges would move and it would not be at equilibrium). • Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface (because the excess charges repel each other).**Section 3 The Electric Field**Chapter 16 4 Properties of Conductors in Electrostatic Equilibrium (Distribution of Static Charges) • 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.**Section 3 The Electric Field**Chapter 16 Discharging Effects of Points 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. • The charge density is greatest at the point of greatest curvature. St. Elmo’s Fire 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.**Section 3 The Electric Field**Chapter 16 Calculating Net Electric Field**Section 3 The Electric Field**Chapter 16 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.**Section 3 The Electric Field**Chapter 16 Sample Problem, continued Electric Field Strength 1.Define the problem,andidentifytheknownvariables. 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.**Section 3 The Electric Field**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**Chapter 16 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 E2at P due to q2 is directed toward q2 because q2 is negative.**Section 3 The Electric Field**Chapter 16 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**Section 3 The Electric Field**Chapter 16 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**Section 3 The Electric Field**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**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**Chapter 16 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.**Section 1 Electric Charge**Chapter 16 Charging By Induction**Section 1 Electric Charge**Chapter 16 Transfer of Electric Charge**Section 3 The Electric Field**Chapter 16 Electric Field Lines