Physics 2049: Electrodynamics General Physics II (PHY 2049) Electrodynamics is the Physics of our everyday life. General Physics II (PHY 2049) General Physics II (PHY 2049) Communication Exploration! 1 General Physics II (PHY 2049) Modern Electronics Syllabus Instructor: James Strohaber Course: General Physics II (Electrodynamics) PHYS2049 Textbook: Fundamentals of Physics, Halliday, Resnick and Walker, 9th ed Class website: http://strohaber.webs.com/ Office: Centennial Building Rm 142 Phone: (850) 599-3889 Syllabus email: [email protected] Class times: M/ W/ F: 11:15AM to 12:30AM Class Location: B L Perry Rm 303 Recitation time: ? Recitation Location: ? Office hours: by appointment 2 All Exams are Grade Policy Homework 15% Recitation and Quizzes 5% 3 Exams (16 each) 48% Final 32% •Closed book •In general I will try to give problems in which the answers will require a formula solutions with variables •Problems will be similar to those on homework, recitation and quizzes Exam Schedule Homework and Class assignments Homework: There will be weekly homework assignments given on Monday and due the following Monday Class Assignments: There will be an assignment each class period. You will work in groups of 4—5 people and you will turn in your work at the end of class. All mid-term exams will be from 11:15 to 12:30 September 22 Exam I October 20 Exam II November 17 Exam III TBA Final 3 WebAssign WebAssign 4414 WebAssign 9999 WebAssign 4 Chapter 21: Coulomb’s Law 1. 2. 3. 4. 5. Static Electricity Electric Charge Conductors and Insulators Coulomb’s Law Charge Quantization Charge Conservation Chapter 21: Coulomb’s Law Hierarchy of biological matter The charge carriers are the protons and electrons Cells Atoms Proteins Chapter 21: Coulomb’s Law Chapter 21: Coulomb’s Law Question: What is charge? Answer: We don’t know exactly. What we do know however is that • It comes in two types and some particles have a non zero amount of it • charge can be associated with a “Field” • charged particles can interact with each other through a field to produce a force • force due to charge interaction can be quantified (Coulomb’s Law) 5 Chapter 21: Coulomb’s Law Chapter 21: Coulomb’s Law Coulomb’s Law This repulsion and attraction gives rise to a force Coulomb’s Law Rule 1: Like charges repel each other …and this force can be quantified Rule 2: Opposite charges attract each other Chapter 21: Coulomb’s Law Coulomb’s Law Chapter 21: Coulomb’s Law Coulomb’s Law qq F = k 1 2 2 rˆ r The proportionality factor k = 8.99 ×109 N ⋅ m 2 / C2 k= qq F = k 1 2 2 rˆ r 1 4πε 0 ε 0 = 8.85 × 10−12 C2 / N×m 2 qq F = k 1 2 2 rˆ r q1 is the total charge particle 1 carries and is measured in Coulombs (C) q1 = Qe Here Q is a whole number and is the number of elementary charges. e = 1.6 ×10−19 C e is the elementary charge, and it is the charge that the electron and proton carries 6 Chapter 21: Coulomb’s Law Coulomb’s Law Chapter 21: Coulomb’s Law Coulomb’s Law (superposition principle) qq F = k 1 2 2 rˆ r qq F = k 1 2 2 rˆ r r is the separation distance between particle 1 of charge q1 and particle 2 of charge q2. q1q3 q1qn −1 qq q1q2 F1,net rˆ1,n −1 + k 12 n rˆ1,n all = k 2 rˆ1,2 + k 2 rˆ1,3 + ⋯ + k 2 r1,2 r1,2 r1,n −1 r1,n r Coulomb’s Law is also known as an inverse square law such as that due to gravity mm F = −G 1 2 2 rˆ r Chapter 21: Coulomb’s Law Coulomb’s Torsion Balance: Use to verify the inverse-square law. Fnet = F1,2 + F1,3 + ⋯ + F1, n −1 + F1, n Chapter 21: Coulomb’s Law How to use Coulomb’s Law qq F = k 1 2 2 rˆ r • In this case both charges are of the same sign. Therefore, the force is repulsive and the magnitude of Coulomb’s law says that the force is a positive value. • Force is a vector, and vectors have direction. In the image above there are two vectors. Which one is it? • This is the job of the unit vector at the end of Coulomb’s law to answer. 7 Chapter 21: Coulomb’s Law Coulomb’s Law q1 qq F = k 1 2 2 rˆ r r1 r Chapter 21: Coulomb’s Law Coulomb’s Law q2 r2 r −r rˆ = 2 1 r2 − r1 q2 r2 Because x2 > x1 and y2 > y1, the displacements Δx > 0 and Δy > 0 are both positive. The unit vector in Coulomb’s law is referring to the force vector on particle 2. x −x y −y rˆ = 2 1 iˆ + 2 1 ˆj r2 − r1 r2 − r1 What are x2 − x1 and r1 r x −x y −y rˆ = 2 1 iˆ + 2 1 ˆj r2 − r1 r2 − r1 Pick one of the charges and label it q1 then label the other as q2 r2 = r1 + r r = r2 − r1 q1 qq F = k 1 2 2 rˆ r We write this as the force on particle 2 from due to particle 1 y2 − y1 Chapter 21: Coulomb’s Law Coulomb’s Law qq F = k 1 2 2 rˆ r Pick one of the charges and label it q1 then label the other as q2 qq F2,1 = k 1 2 2 rˆ r They are displacements q1 r1 r1 = r2 + r r = r1 − r2 r Chapter 21: Coulomb’s Law Coulomb’s Law: Hydrogen atom q2 qe q p F = k 2 rˆ r r2 r k = 9 ×109 Nm 2 /C2 r −r rˆ = 1 2 r1 − r2 qe = −1.6 × 10−19 C x −x y −y rˆ = 1 2 iˆ + 1 2 ˆj r2 − r1 r2 − r1 q p = 1.6 × 10−19 C because x2 > x1 and y2 > y1, the displacements Δx < 0 and Δy < 0 are both negative. The unit vector in Coulomb’s law is referring to the force vector on particle 1. qq F1,2 = k 1 2 2 r rˆ r = 0.5 ×10−10 m This value of r is known as the Bohr radius 8 Chapter 21: Coulomb’s Law Newton’s Universal Law Coulomb’s Law: Hydrogen atom r −19 2 ( F = − 9 ×109 ) (1.6 ×10 ) ( 0.5 ×10 ) −10 2 me = 9.11× 10−31 kg m p = 1.67 ×10−27 kg "Mass of grain of sand: 10-10 kg" ( F = mg = 10 )( kg 9.8m / s r me m p F = −G 2 rˆ r G = 6.67 × 10−11 Nm 2 /kg 2 F = −92.16 × 10−9 N −10 Chapter 21: Coulomb’s Law 2 ) = 10 r = 0.5 ×10−10 m −9 N 100 grains of sand Chapter 21: Coulomb’s Law Newton’s Universal Law This value of r is known as the Bohr radius Chapter 21: Coulomb’s Law Coulomb’s Law: Example r me m p F = −G 2 rˆ r qe q p F = k 2 rˆ r q1 = q2 = e r d /2 q3 = q4 = −e / 2 ( 9.11×10 )(1.67 ×10 ) ) ( 0.5 ×10 ) −31 ( F = − 6.67 × 10 −11 F = −4.1×10−47 N −27 −10 2 Fg Fe = 4.45 × 10−40 Fnet = F1,2 + F1,3 + F1,4 ke 2 ke2 ke 2 Fnet = − 2 + 2 cos(θ ) + 2 cos(θ ) R 2r 2r ke 2 ke 2 Fnet = − 2 + 2 cos(θ ) R r 9 Chapter 21: Coulomb’s Law Coulomb’s Law: Example 2 ke ke Fnet = − 2 + 2 cos(θ ) R r cos(θ ) = R 2r r= Coulomb’s Law: Example r 2 Chapter 21: Coulomb’s Law d /2 ke2 4ke2 Fnet = − 2 + R d 2 + R2 ( 1 2 1 2 d + R 4 4 r d /2 3/2 ) R F = −∇V 1 0.5 ke 2 4ke2 R Fnet = − 2 + 3/2 R d 2 + R2 ( 4 1 V = ke − R d 2 + R2 ) 2 ( Chapter 21: Coulomb’s Law Coulomb’s Law q1 Particle 1 has a charge of 3 μC and is positioned at (1, 2)mm. Particle 2 has a charge of 1.5 μC and is at position (4, 3)mm. r1 q2 r ) 0 -0.5 0 1 2 3 4 5 6 7 8 9 Chapter 21: Coulomb’s Law Coulomb’s Law Particle 1 has a charge of 3 μC and is positioned at (1, 2). Particle 2 has a charge of 1.5 μC and is at position (4, 3). r2 1/2 q1 r1 r q2 r2 (a) What is the magnitude of the force between the two particles r2 − r1 = ( x1 − x2 )2 + ( y1 − y2 )2 = (1 − 4)2 + (2 − 3) 2 = 3.16mm qq F = k 1 2 2 = 9 × 109 r ( ( 3 ×10 )(1.5 ×10 ) = 4 ×10 N ) ( 3.16 ×10 ) −6 −6 3 −3 2 (b) Write in unit vector notation the unit vector for the force acting on particle 1 x −x y −y 1− 4 ˆ 2 − 3 ˆ rˆ = 1 2 iˆ + 1 2 ˆj = i+ j r2 − r1 r2 − r1 3.16 3.16 rˆ = −0.95iˆ − 0.32 ˆj 10 Chapter 21: Coulomb’s Law Coulomb’s Law q1 Particle 1 has a charge of 3 μC and is positioned at (1, 2). Particle 2 has a charge of 1.5 μC and is at position (4, 3). r1 r Chapter 21: Coulomb’s Law Coulomb’s Law q2 Find the electrostatic force on charge Q r r2 θ we can see that the y-components cancel θ Qq Qq Qq F = k 2 cos(θ ) + k 2 cos(θ ) = 2k 2 cos(θ ) r r r (c) What is the angle of the unit vector r? tan(θ ) = cos(θ ) = θ = arctan r= From the positive x axis Chapter 21: Coulomb’s Law Qq 1 2 2 d +x 4 3/2 x Electron in uniform circular motion r For small values of x <<d/2 3/2 F = 2k Coulomb’s Law Find the electrostatic force on charge Q Qq 1 2 d + x2 4 Chapter 21: Coulomb’s Law Coulomb’s Law 1 2 2 d +x 4 Qq F = 2k 3 x r 0.32 o = 18.6 0.95 −0.32 −0.95 180o + 18.6o = 198.6o F = 2k x r r θ x ≈ 16k Qq d3 θ ∑ F = ma a= v2 r x r −k qe q p r2 = −m v2 r Newton’s second law ∑ F = mxɺɺ −16k Qq x = mxɺɺ d3 k Qq ɺɺ x + 16 x = ɺɺ x + ω2x = 0 m d3 v= x = sin(ωt ) xɺ = ω cos(ω t ) 2 ke mr k Qq x = sin 16 t m d3 ɺɺ x = −ω sin(ωt ) 11 Chapter 21: Coulomb’s Law Chapter 21: Coulomb’s Law Coulomb’s Law Charge quantization Consider an infinite number of identical charges (each of charge q) placed along the x axis at distances a, 2a, 2a, 4a … from the origin. What is the force on a particle of charge Q at the origin? Millikan use the apparatus on the right to measure the charge on droplets of oil. ∞ Fnet = ∑F Q , qi i Fnet = Fnet = kQq a2 π 2k 6a 2 Qq1 Qqn Qq2 Fnet = lim k 2 + k 2 + ⋯ + k 2 r1 r2 rn n →∞ Qqn Qq2 Qq1 Fnet = lim k 2 + k 2 2 +⋯ + k 2 2 a 2 a n a n →∞ ∞ ∞ ∑n 1 2 = n =1 kQq a2 ζ ( s) = ζ (2) ∑n 1 s He discovered that the charges were always integer multiples of the fundamental charge e n =1 Riemann zeta function Qq total charge q1 = Qe Whole number elementary charge Chapter 21: Coulomb’s Law Coulomb’s Law The force measured between two equally charged objects separated by 0.001 m is 5 pN. How many elementary charges are there? q1 = q2 = q 5 × 10−12 = 8.99 ×1010 q2 0.0012 q = 7.457 ×10−15 C q = Qe Q = q / e = (7.457 ×10−15 C) / (1.6 × 10−19 C ) = 46610.65 12
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