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Physics 212 Lecture 2, Slide 1 Physics 212 Lecture 2 Today's Concept: The Electric Field Continuous Charge Distributions.

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Presentation on theme: "Physics 212 Lecture 2, Slide 1 Physics 212 Lecture 2 Today's Concept: The Electric Field Continuous Charge Distributions."— Presentation transcript:

1 Physics 212 Lecture 2, Slide 1 Physics 212 Lecture 2 Today's Concept: The Electric Field Continuous Charge Distributions

2 Music Who is the Artist? A)Bill Frisell B)Cindy Cashdollar C)Daniel Lanois D)Marc Ribot E)Tony Rice Great musician/album names. All these people were at the Ellnora Guitar Festival this year. Physics 212 Lecture 2

3 Our Comments 1.If your TA doesn’t show up – go to undergrad office 231/3 (or home page for telephone numbers of discussion and lab masters) 2.Please operate only your own clicker 3.Unfortunately we really can’t help with PreLecture, Checkpoint or homework physics questions by email – too many students: Office Hours, CARE (see link), your friends –Fridays 2:00 – 7:00 p.m. in 271 Loomis –Sundays 2:00 – 9:00 p.m. in 143 Loomis –Mondays 11:00 a.m. – 9:00 p.m. in 143 Loomis 4.Quite a bit of homework this week 5.Don’t panic! Physics 212 Lecture 2, Slide 3

4 Physics 212 Lecture 2, Slide 4 33 q1q1q1q1 q2q2q2q2 q3q3q3q3 q4q4q4q4 F 2,1 F 3,1 F 4,1 F1F1F1F1 F 2,1 F 3,1 F 4,1 F1F1F1F1 F 2,1 F 3,1 F 4,1 F1F1F1F1 F 2,1 F 3,1 F 4,1 F1F1F1F1 q1q1q1q1 q2q2q2q2 q3q3q3q3 q4q4q4q4 Coulomb’s Law (from last time) 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 reversedMATH:

5 Physics 212 Lecture 2, Slide 5 Electric Field 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 q2q2q2q2 q3q3q3q3 q4q4q4q4 E2E2E2E2 E3E3E3E3 E4E4E4E4 E 08 Field points toward negative and away from positive charges. Electric fields and also their use with lines of charge really confuse me! “ “Electric fields and also their use with lines of charge really confuse me! “ How fields add together“ “ How fields add together“

6 Physics 212 Lecture 2, Slide 6 simulation 09 Checkpoint 1 “Field directions were confusing.” Two equal, but opposite charges are place on the x axis. The positive charge is placed to the left of the origin and the negative charge to the right, as shown in the figure above. What is the direction of the electric field at point A? A. Up B. Down C. Left D. Right E. Zero What is the direction of the electric field at point B? A. Up B. Down C. Left D. Right E. Zero

7 In which of the two cases below is the magnitude of the electric field at the point labeled A the largest? A Case 1 B Case 2 C Same Physics 212 Lecture 2, Slide 7 12 Checkpoint 2EE “the two positive charges would cancel eachother out in the x direction since they both have fields that point away from them, while the charges in case one have fields that both point in the same direction.” “In two there is a component of the field pointing away.” “The coulomb forces exerted at point A are at right angles in both cases, so that the magnitude stays the same.”

8 Physics 212 Lecture 2, Slide 8 Two Charges Two charges q 1 and q 2 are fixed at points (-a,0) and (a,0) as shown. Together they produce an electric field at point (0,d) which is directed along the negative y-axis. x y q1q1q1q1 q2q2q2q2 (-a,0) (a,0) (0,d) Which of the following statements is true: a) Both charges are negative b) Both charges are positive c) The charges are opposite d) There is not enough information to tell how the charges are related 22

9 Physics 212 Lecture 2, Slide 9 -- ++ -+ 23

10 Physics 212 Lecture 2, Slide 10 12 Checkpoint 3 “ although the charge on the right is doubled, the force on q by the charge is inversely proportional to the square of the distance between the charges, so the force is effectively halved compared to the charge on the left. “ “radius gets squared so it moves to the right even though the second charge is stronger.” “The forces from each charge will have the same force because of their relative distances.” INTERESTING: statement is correct, but given in support of “to the left” !! A positive test charge q is released from rest at a 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. To the left B. To the right C. Stay still D. Other

11 Physics 212 Lecture 2, Slide 11 Example Calculate E at point P. 20 Just some more examples would be cool “Just some more examples would be cool” What is the direction of the electric field at point P, the unoccupied corner of the square? d +q +q -q P d (A) (B) (C) (D) Need to know d Need to know d & q (E)

12 Physics 212 Lecture 2, Slide 12 Continuous Charge Distributions Summation becomes an integral (be careful with vector nature) = Q/L 25 my head is spinning, what does the dQ mean? “ my head is spinning, what does the dQ mean? ” WHAT DOES THIS MEAN ?? Integrate over all charges (dq) r is vector from dq to the point at which E is defined pt for E Linear Example: charges dq = dx r dE

13 Physics 212 Lecture 2, Slide 13 Charge Density Linear ( =Q/L) Coulombs/meter Surface (  Q/A) Coulombs/meter 2 Volume (  = Q/ V) Coulombs/meter 3 What has more net charge?. A) A)A sphere w/ radius 2 meters and volume charge density  = 2 C/m 3 B) B)A sphere w/ radius 2 meters and surface charge density  = 2 C/m 2 C) C)Both A) and B) have the same net charge. 28 What exactly is charge density, how is it calculated? “What exactly is charge density, how is it calculated?” Some Geometry

14 Physics 212 Lecture 2, Slide 14 29 Checkpoint 4 Both lines have identical charge densities + C/m. Point A is equidistant from both lines and Point B is located 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. E A E B Two infinite lines of charge are shown below. “ cancel at A, sum at B“ “Both fields are parallel and both lines are parallel and identical.” “b is twice as far from the bottom one as the other, therefore smaller E.”

15 Physics 212 Lecture 2, Slide 15 Calculation Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? 33 We know: xya h P x What is ? (A) (B) (C) (D) (E) r dq= dx “I really think it would be nice to go over how to find the electric field with the infinite lines of charge with the integrals”

16 Physics 212 Lecture 2, Slide 16 Calculation Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? 33 We know: x y a hPx r dq= dx 1111 2222 What is ?(A) (B) (C) (D) 2222

17 Physics 212 Lecture 2, Slide 17 Calculation Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? 33 We know: x y a hPx r dq= dx 1111 2222 2222 What is ? (A) (B) (C) neither of the above cos  2 DEPENDS ON x !! “The concept I found most difficult was integrating to find electric field when the charge is distributed over a distance.”

18 Physics 212 Lecture 2, Slide 18 Calculation Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? 33 We know: x y a hPx r dq= dx 1111 2222 2222 What is ?(A) (B) (C) (D)

19 Physics 212 Lecture 2, Slide 19 Calculation Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? 33 We know: x y a hPx r dq= dx 1111 2222 2222 What is ?

20 Physics 212 Lecture 2, Slide 20 Observation Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? 33 x y a hPx r dq= dx 1111 2222 2222 Note that our result can be rewritten more simply in terms of  1. Exercise for student: Change variables: write x in terms of  Result: obtain simple integral in 

21 Physics 212 Lecture 2, Slide 21 Notes Preflight + Prelecture 3 due by 8:00 AM Tuesday Jan. 24 Homework 1 is due Tuesday Jan. 24 Labs start Monday Jan. 23 Discussion Quiz next week will be on Coulomb’s Law and E


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