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Physics 2112 Unit 2: Electric Fields Today’s Concepts: A) The Electric Field B) Continuous Charge Distributions Unit 2, Slide 1.

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Presentation on theme: "Physics 2112 Unit 2: Electric Fields Today’s Concepts: A) The Electric Field B) Continuous Charge Distributions Unit 2, Slide 1."— Presentation transcript:

1 Physics 2112 Unit 2: Electric Fields Today’s Concepts: A) The Electric Field B) Continuous Charge Distributions Unit 2, Slide 1

2 Fields Unit 2, Slide 2 MATH:  q1q1 2 What if I remove q 2 ? Is there anything at that point in space? q1q1 q2q2 F 12

3 If there are more than two charges present, the total force on any given charge is just the vector sum of the forces due to each of the other charges:  q 1  q 1  direction reversed Vector Field Unit 2, Slide 3 F 2,1 F 3,1 F 4,1 F1F1 q1q1 q2q2 q3q3 q4q4 F 2,1 F 3,1 F 4,1 F1F1 F 2,1 F 3,1 F 4,1 F1F1 F 2,1 F 3,1 F 4,1 F1F1 q1q1 q2q2 q3q3 q4q4 MATH: 

4 The electric field E at a point in space is simply the force per unit charge at that point. Electric field due to a point charged particle Superposition E2E2 E3E3 E4E4 E Field points toward negative and Away from positive charges. Field point in the direction of the force on a positive charge. “What exactly does the electric field that we calculate mean/represent? “ “What is the essence of an electric field? “ Electric Field Unit 2, Slide 4 q4q4 q2q2 q3q3

5 Two equal, but opposite charges are placed on the x axis. The positive charge is placed to the left of the origin and the negative charge is placed to the right, as shown in the figure above. What is the direction of the electric field at point A? A. A. Up B. B. Down C. C. Left D. D. Right E. E. Zero CheckPoint: Electric Fields1 Unit 2, Slide 5 A B x QQ QQ

6 Two equal, but opposite charges are placed on the x axis. The positive charge is placed to the left of the origin and the negative charge is placed to the right, as shown in the figure above. What is the direction of the electric field at point B? A. A. Up B. B. Down C. C. Left D. D. Right E. E. Zero CheckPoint: Electric Fields2 Electricity & Magnetism Lecture 2, Slide 6 A B x QQ QQ

7 Example 2.1 (Field from three charges) Electricity & Magnetism Lecture 2, Slide 7 Calculate E at point P. d P d qq qq qq1 2 3

8 CheckPoint: Magnitude of Field (2 Charges) Electricity & Magnetism Lecture 2, Slide 8 In which of the two cases shown below is the magnitude of the electric field at the point labeled A the largest? A. A. Case 1 B. B. Case 2 C. C. Equal QQ QQ QQ QQ A A Case 1 Case 2

9 _ _   _  Electricity & Magnetism Lecture 2, Slide 9

10 CheckPoint Results: Motion of Test Charge Electricity & Magnetism Lecture 2, Slide 10 A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2Q. How will the test charge move immediately after being released? A. A. To the left B. B. To the right C. C. Stay still D. D. Other

11 x q1q1 Q2Q2  0.4m,0  A charge of q 1 = +4uC is placed at the origin and another charge Q 2 = +10uC is placed 0.4m away. At what point on the line connected the two charges is the electric field zero? Example 2.2 (Zero Electric Field) Unit 2, Slide 11  0,0 

12  Q  L Summation becomes an integral (be careful with vector nature) “I don't understand the whole dq thing and lambda.” WHAT DOES THIS MEAN ? Integrate over all charges  dq  r is vector from dq to the point at which E is defined r dE Continuous Charge Distributions Electricity & Magnetism Lecture 2, Slide 12 Linear Example: charges pt for E dq   dx

13 Clicker Question: Charge Density Electricity & Magnetism Lecture 2, Slide 13 Linear  Q  L  Coulombs/meter Surface  Q  A  Coulombs/meter 2 Volume   Q  V  Coulombs/meter 3 What has more net charge?. A) A sphere w/ radius 2 meters and volume charge density  = 2 C/m 3 B) A sphere w/ radius 2 meters and surface charge density  = 2 C/m 2 C) Both A) and B) have the same net charge. “I would like to know more about the charge density.” Some Geometry

14 “Please go over infinite line charge.” Example 2.3 (line of charge) Electricity & Magnetism Lecture 2, Slide 14 Charge is uniformly distributed along the x -axis from the origin to x  a. The charge density is C  m. What is the x -component of the electric field at point P   x,y    0,h  ? Let’s do one slightly different. x y a h P x r dq   dx

15 Clicker Question: Calculation Electricity & Magnetism Lecture 2, Slide 15 A) B) C) D) E) What is ? Charge is uniformly distributed along the x -axis from the origin to x  a. The charge density is C  m. What is the x -component of the electric field at point P   x,y    0,h  ? x y a h P x r dq   dx We know:

16 Clicker Question: Calculation Electricity & Magnetism Lecture 2, Slide 16 We know: What is ? A) B) C) D) Charge is uniformly distributed along the x -axis from the origin to x  a. The charge density is C  m. What is the x -component of the electric field at point P   x,y    0,h  ? x a P x r 11  dq   dx y

17 We know: sin  2 DEPENDS ON x ! Clicker Question: Calculation Electricity & Magnetism Lecture 2, Slide 17 What is ? A) B) C) neither of the above Charge is uniformly distributed along the x -axis from the origin to x  a. The charge density is C  m. What is the x -component of the electric field at point P   x,y    0,h  ? x a P x r 11 22 dq   dx y

18 We know: Clicker Question: Calculation Electricity & Magnetism Lecture 2, Slide 18 Charge is uniformly distributed along the x -axis from the origin to x  a. The charge density is C  m. What is the x -component of the electric field at point P   x,y    a,h  ? What is x in terms of  ? A) B) C) D) x = h*tan   x = h*cos   x = h*sin   x = h / cos   x a P x r 11 22 dq   dx y

19 We know: Calculation Electricity & Magnetism Lecture 2, Slide 19 What is ? Charge is uniformly distributed along the x -axis from the origin to x  a. The charge density is C  m. What is the x -component of the electric field at point P   x,y    a,h  ? x = h*tan  x a P x r 11 22 dq   dx y dx = h*sec 2  d 

20 Exercise for student: Change variables: write  in terms of x Result: do integral with trig sub Observation Electricity & Magnetism Lecture 2, Slide 20 x a P x r 11 22 dq   dx y since E x < 0 make since?

21 We had: Back to the pre-lecture Electricity & Magnetism Lecture 2, Slide 21 “Please go over infinte line charge. How does R get outside the intergral?” x a P x r 11 22 dq   dx y How would this integral change if the line of charge were infinite in both directions? A) The limits would be  1 to –  1 B) The limits would be + to - C) The limits would be -  /2 to  /2 D) sin  would turn to tan  E) sin  would turn to cos  8 8

22 For an infinite line of charge, we had: Back to the pre-lecture Electricity & Magnetism Lecture 2, Slide 22 x a P x r 11 22  dq   dx y How would this integral change we wanted the y component instead of the x component? A) The limits would be  1 to –  1 B) The limits would be +/- infinity C) The limits would be -  to  D) sin  would turn to tan  E) sin  would turn to cos 

23 k in terms of fundamental constants Unit 2, Slide 23 Note

24 CheckPoint: Two Lines of Charge Electricity & Magnetism Lecture 2, Slide 24 Two infinite lines of charge are shown below. Both lines have identical charge densities +λ C/m. Point A is equidistant from both lines and Point B is located a above the top line as shown. How does E A, the magnitude of the electric field at point A, compare to E B, the magnitude of the electric field at point B? A. A.E A < E B B. B. E A = E B C. C. E A > E B

25 Example 2.4 (E-field above a ring of charge) Unit 2, Slide 25 a h P What is the electric field a distance h above the center of ring of uniform charge Q and radius a? y x

26 Example 2.5 (E-field above a disk) Unit 2, Slide 26 a h P What is the electric field a distance h above the center of disk of uniform charge Q and radius a? y x

27 dipoles Unit 2, Slide 27 q q d - + Dipole moment = p = qd Points from negative to positive. (opposite the electric field.) q qCl Na

28 Torque on dipole Unit 2, Slide 28 q q -+   = 2*(qE X d/2) = p X E dU= -dW =  d  = pE*sin  d   U  pEcos  Define U = when  =  /2 U  -p E 

29 Example 2.6 (Salt Dipole) Unit 2, Slide The two atoms in a salt (NaCl) molecule are separated by about 500pm. The molecule is placed in an electric field of strength 10N/C at an angle of 20 o. What is the torque on the molecule? What is the potential energy of the molecule? 20 o


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