5. Electric field (theoretical approach) and Gauss s law

Transcription

1 5. Electric field (theoretical approach) and Gauss s law

2 Announcement: Lab schedule will be posted later today

3 I went to the tutorial session yesterday A. yes B. No C. I don t remember

4 The tutorial session A. Was very useful B. Was useful C. Neutral D. Not useful E. Not at all useful

5 I am in 1. Mechanical 2. Architecture 3. Materials 4. Mining 5. Chemical 6. Computer 7. Physics 8. Math 9. Civil 10. Other

6 The electric field of one charge is (you can answer more than one) A. E = qrƹ r 2 qrƹ B. E = k 0 C. E = 1 r 2 qrƹ 4πε 0 r 2 r q Ƹ D. E = k 0 r q Ԧr E. E = k 0 r 3

7 The electric field at r of one charge at r i is (you can answer more than one) A. E = 1 qrƹ 4πε 0 r 2 B. E(Ԧr) = 1 qrƹ 4πε 0 r 2 C. E = 1 q r i 4πε 0 r 2 i D. E(Ԧr) = 1 q( Ԧr r i ) 4πε 0 Ԧr r 2 i E. E(Ԧr) = 1 q( Ԧr r i ) 4πε 0 Ԧr r 3 i

8 The electric field at r of charge q a at a and charge q b at b is (you can answer more than one) A. E = 1 q a + 1 q b 4πε 0 a 2 4πε 0 b 2 B. E = 1 q a a + 1 q b b 4πε 0 a 2 4πε 0 b 2 C. E Ԧr = 1 D. E Ԧr = 1 4πε 0 ( ( q a Ԧr a + q b Ԧr b 4πε 0 Ԧr a 3 Ԧr b 3 ) q Ԧr a q Ԧr b Ԧr a 3 + Ԧr b 3 )

9 The electric field at r of N charges q i at r i is (you can answer more than one) A. E Ԧr = σn q i Ԧr r i i=1 4πε 0 Ԧr r 3 i B. E Ԧr = 1 N q σ i Ԧr r i 4πε i=1 0 Ԧr r 3 i C. E Ԧr = q σn 4πε i=1 0 Ԧr r i Ԧr r i 3 D. E Ԧr = 1 N q σ i r i 4πε i=1 0 r 3 i

10 The electric field at r of N charges q i at r i and ρ r i is the uniform charge density of charge q i of volume V i (you can answer more than one) A. σn ρ r i dv i Ԧr r i i=1 4πε 0 Ԧr r 3 i B. σn q i Ԧr r i i=1 4πε 0 Ԧr r 3 i C. σn ρ r i V i Ԧr r i i=1 4πε 0 Ԧr r 3 i N ρ Ԧr V D. σ i Ԧr r i i=1 4πε 0 Ԧr r 3 i

11 Charge Densities Volume charge density: when a charge is distributed evenly throughout a volume ρ Q / V with units C/m 3 Surface charge density: when a charge is distributed evenly over a surface area σ Q / A with units C/m 2 Linear charge density: when a charge is distributed along a line λ Q / l with units C/m Section 23.5

12 Amount of Charge in a Small Volume If the charge is nonuniformly distributed over a volume, surface, or line, the amount of charge, dq, is given by For the volume: dq = ρ dv For the surface: dq = σ da For the length element: dq = λ dl Section 23.5

13 The electric field at r of N charges q i at r i and ρ r i is the uniform charge density of charge q i of volume dv i (you can answer more than one) A. σn ρ r i dv i Ԧr r i i=1 4πε 0 Ԧr r 3 i B. σn q i Ԧr r i i=1 4πε 0 Ԧr r 3 i C. σn ρ r i V i Ԧr r i i=1 4πε 0 Ԧr r 3 i N ρ Ԧr V D. σ i Ԧr r i i=1 4πε 0 Ԧr r 3 i

14 The electric field of a continuous charge distribution ρ Ԧr where dq = ρ Ԧr dv is (you can answer more than one) A. σn ρ r i dv i Ԧr r i i=1 4πε 0 Ԧr r 3 i B. V dv i ρ r i Ԧr r i 4πε 0 Ԧr r i 3 C. V dv ρ r Ԧr r 4πε 0 Ԧr r 3 D. dq 1 4πε 0 r 2

15 Team Scores Points Team Points Team 4.74 Architecture 3.55 Computer 3.39 Mechanical 3.1 Physics 2.79 Civil 2.68 Chemical 2.49 Math 2.35 Materials 2.31 Other 1.39 Mining

16 Electric Field Continuous Charge Distribution The distances between charges in a group of charges may be much smaller than the distance between the group and a point of interest. In this situation, the system of charges can be modeled as continuous. The system of closely spaced charges is equivalent to a total charge that is continuously distributed along some line, over some surface, or throughout some volume. Section 23.5

17 Electric Field Continuous Charge Distribution, cont Procedure: Divide the charge distribution into small elements, each of which contains Δq. Calculate the electric field due to one of these elements at point P. Evaluate the total field by summing the contributions of all the charge elements. Section 23.5

18 Electric Field Continuous Charge Distribution, equations For the individual charge elements q E k r e r ˆ 2 Because the charge distribution is continuous q dq E k lim rˆ k rˆ i e 0 2 i e q 2 i i ri r Section 23.5

19 Example 23.7: Electric Field Due to a Charged Rod

20 dq = dx, so de = k e [dq/(x 2 )] = k e [( dx)/(x 2 )] And E = k e [(dx)/(x 2 )] (limits x = a to x = l) E = k e [(1/a) 1/(a + l)]

21 Problem-Solving Strategy Conceptualize Establish a mental representation of the problem. Image the electric field produced by the charges or charge distribution. Categorize Individual charge? Group of individual charges? Continuous distribution of charges? Section 23.5

22 Problem-Solving Strategy, cont Analyze Analyzing a group of individual charges: Use the superposition principle, find the fields due to the individual charges at the point of interest and then add them as vectors to find the resultant field. Be careful with the manipulation of vector quantities. Analyzing a continuous charge distribution: The vector sums for evaluating the total electric field at some point must be replaced with vector integrals. Divide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating over the entire charge distribution. Symmetry: Take advantage of any symmetry to simplify calculations. Section 23.5

23 Problem Solving Hints, final Finalize Check to see if the electric field expression is consistent with your mental representation. Check to see if the solution reflects any symmetry present. Image varying parameters to see if the mathematical result changes in a reasonable way. Section 23.5

24 Gauss Law Theory attracts practice as the magnet attracts iron. It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment. C.F. Gauss Introduction

25 Gauss Law Theory attracts practice as the magnet attracts iron. It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment. C.F. Gauss 4πk q E e q ε o Introduction

26 Gauss Law Gauss Law can be used as an alternative procedure for calculating electric fields. Gauss Law is based on the inverse-square behavior of the electric force between point charges. It is convenient for calculating the electric field of highly symmetric charge distributions. Gauss Law is important in understanding and verifying the properties of conductors in electrostatic equilibrium. Introduction

27 Electric Flux Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field. Φ E = EA Units: N m 2 / C Section 24.1

28 What is the electric flux through a 45deg tilted flat surface infinitely large A. The same B. 30% less C. 30% more

29 Electric Flux, General Area The electric flux is proportional to the number of electric field lines penetrating some surface. The field lines may make some angle θ with the perpendicular to the surface. Then Φ E = EA cos θ Section 24.1

30 Electric Flux, Interpreting the Equation The flux is a maximum when the surface is perpendicular to the field. θ = 0 The flux is zero when the surface is parallel to the field. θ = 90 If the field varies over the surface, Φ = EA cos θ is valid for only a small element of the area. Section 24.1

31 Electric Flux, General In the more general case, look at a small area element. E A cosθ E A E i i i i i In general, this becomes lim E A E i i A 0 i E E da surface The surface integral means the integral must be evaluated over the surface in question. In general, the value of the flux will depend both on the field pattern and on the surface. Section 24.1

32 Participant Leaders Points Participant Points Participant 11 Bouchoutrouch-Ku, Tarik 11 Duan, Lin Pei 11 Ellis, Jacob 11 Teng, Yuan-Po 10 Fricker, Alexander