1 / 29

Physics 2112 Unit 2: Electric Fields

Physics 2112 Unit 2: Electric Fields. Today’s Concepts: A) The Electric Field B) Continuous Charge Distributions. Fields. What if I remove q 2 ? Is there anything at that point in space?. q 1. q 1. q 2. 2. F 12. MATH:. =. Vector Field. F 4,1. F 2,1. F 1. F 3,1.

arch
Download Presentation

Physics 2112 Unit 2: Electric Fields

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

  2. Fields What if I remove q2? Is there anything at that point in space? q1 q1 q2 2 F12 MATH: =

  3. Vector Field F4,1 F2,1 F1 F3,1 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: q2 F4,1 q2 F2,1 q1 F1 F1 F3,1 F1 q4 q1 F2,1 F2,1 F3,1 F3,1 q3 +q1-> -q1direction reversed F4,1 F4,1 q4 MATH: = q3

  4. Electric Field “What exactly does the electric field that we calculate mean/represent? “ “What is the essence of an 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 q2 E4 E2 Field points toward negative and Away from positive charges. Field point in the direction of the force on a positive charge. E E3 q4 q3

  5. CheckPoint: Electric Fields1 A B x +Q -Q 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? Up Down Left Right Zero

  6. CheckPoint: Electric Fields2 A B x +Q -Q 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? Up Down Left Right Zero

  7. Calculate E at point P. Example 2.1 (Field from three charges) +q -q +q 1 P d 2 d 3

  8. CheckPoint: Magnitude of Field (2 Charges) In which of the two cases shown below is the magnitude of the electric field at the point labeled A the largest? Case 1 Case 2 Equal -Q +Q +Q +Q A A Case 1 Case 2

  9. _ _ _ + + +

  10. CheckPoint Results: Motion of Test Charge 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? To the left To the right Stay still Other

  11. Example 2.2 (Zero Electric Field) q1 Q2 x (0,0) (0.4m,0) A charge of q1 = +4uC is placed at the origin and another charge Q2 = +10uC is placed 0.4m away. At what point on the line connected the two charges is the electric field zero?

  12. Continuous Charge Distributions Linear Example: dE r charges dq=l dx “I don't understand the whole dq thing and lambda.” Summation becomes an integral (be careful with vector nature) WHAT DOES THIS MEAN ? Integrate over all charges (dq) r is vector from dq to the point at which E is defined l=Q/L pt for E

  13. Linear (l =Q/L)Coulombs/meter Surface (s = Q/A)Coulombs/meter2 Volume (r=Q/V) Coulombs/meter3 Clicker Question: Charge Density Some Geometry “I would like to know more about the charge density.” What has more net charge?. A) A sphere w/ radius 2 meters and volume charge density r = 2 C/m3 B) A sphere w/ radius 2 meters and surface charge density s = 2 C/m2 C) Both A) and B) have the same net charge.

  14. Example 2.3 (line of charge) “Please go over infinite line charge.” y P r Let’s do one slightly different. h dq=l dx x Charge is uniformly distributed along the x-axis from the origin to x=a. The charge density is lC/m. What is the x-component of the electric field at point P:(x,y)=(0,h)? x a

  15. Clicker Question: Calculation y P Charge is uniformly distributed along the x-axis from the origin to x=a. The charge density is lC/m. What is the x-component of the electric field at point P:(x,y)=(0,h)? r h dq=l dx x x a We know: What is ? B) E) A) D) C)

  16. Clicker Question: Calculation Charge is uniformly distributed along the x-axis from the origin to x=a. The charge density is lC/m. What is the x-component of the electric field at point P:(x,y)=(0,h)? y P q2 r q1 x x a dq=l dx We know: What is ? B) A) D) C)

  17. Clicker Question: Calculation y Charge is uniformly distributed along the x-axis from the origin to x=a. The charge density is lC/m. What is the x-component of the electric field at point P:(x,y)=(0,h)? P q2 r q1 x x We know: a dq=l dx What is ? A) B) sinq2DEPENDS ON x! C) neither of the above

  18. Clicker Question: Calculation y Charge is uniformly distributed along the x-axis from the origin to x=a. The charge density is lC/m. What is the x-component of the electric field at point P:(x,y)=(a,h)? P q2 r q1 x x a dq=l dx We know: What is x in terms of Q ? A) B) x = h / cosQ2 D) x = h*sinQ2 x = h*cosQ2 x = h*tanQ2 C)

  19. Calculation y Charge is uniformly distributed along the x-axis from the origin to x=a. The charge density is lC/m. What is the x-component of the electric field at point P:(x,y)=(a,h)? P q2 r q1 x x a We know: dq=l dx x = h*tanQ dx = h*sec2Q dq What is ?

  20. Observation y P q2 r q1 since x x a dq=l dx Ex< 0 make since? Exercise for student: Change variables: write Qin terms of x Result: do integral with trig sub

  21. Back to the pre-lecture y “Please go over infinte line charge. How does R get outside the intergral?” P q2 r We had: q1 x x a dq=l dx How would this integral change if the line of charge were infinite in both directions? D) sinQwould turn to tanQ A) The limits would be Q1 to –Q1 8 8 B) The limits would be + to - E) sinQwould turn to cosQ C) The limits would be -p/2to p/2

  22. Back to the pre-lecture y q2 For an infinite line of charge, we had: P r q2 q1 x x a dq=l dx How would this integral change we wanted the y component instead of the x component? D) sinQwould turn to tanQ A) The limits would be Q1 to –Q1 B) The limits would be +/- infinity E) sinQwould turn to cosQ C) The limits would be -p/2to p/2

  23. k in terms of fundamental constants Note

  24. CheckPoint: Two Lines of Charge 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 EA, the magnitude of the electric field at point A, compare to EB, the magnitude of the electric field at point B? • EA < EB • EA = EB • EA > EB

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

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

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

  28. Torque on dipole • = 2*(qE X d/2) • = p X E Q q q + dU= -dW = tdQ = pE*sinQ*dQ DU= -pEcosQ Define U = when Q = p/2 U= -p E -

  29. Example 2.6 (Salt Dipole) + 20o - 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 20o. What is the torque on the molecule? What is the potential energy of the molecule?

More Related