Download
electric charge and the electric field ch 26 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Electric Charge and the Electric Field (Ch 26) PowerPoint Presentation
Download Presentation
Electric Charge and the Electric Field (Ch 26)

Electric Charge and the Electric Field (Ch 26)

268 Vues Download Presentation
Télécharger la présentation

Electric Charge and the Electric Field (Ch 26)

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Electric Charge and the Electric Field (Ch 26) Four Fundamental Forces • Electromagnetic • Gravity • Strong Nuclear Force • Weak Nuclear Force

  2. Electric Charge • Electricity – Elektron (amber) • Amber (tree resin) gets a charge when rubbed with cloth • Ben Franklin • Positive = Charge on a rubber rod • Negative = Charge on an amber/plastic rod

  3. Electric Charge • Law of conservation of charge – Charge is never created or destroyed Cloth - - - - - - + + + + + Rubber rod

  4. Insulators/Conductors

  5. (Discuss movement of the electrons)

  6. Conduction • Transfer of charge by touching • Electrons can flow between substances • Can produce a net charge on the substance

  7. Induction • Transfer of charge without touching • Electrons migrate within the substance to cause a separation of charge • Net charge on substance is still zero

  8. Induction: Grounding • Only way to produce a net charge by induction • Earth can easily absorb or donate electrons • Electrons can leave a substance, then break the ground

  9. Electroscope

  10. Charging the electroscope by induction by conduction (charge separation) (net charge)

  11. Using the electroscope • Can be used to detect the charge on a substance • (Must charge electroscope first)

  12. Coulomb’s Law • Coloumb experimented to determine magnitude of electromagnetic force • Measured the forces between charged spheres (angle of deflection).

  13. F = k Q1Q2 r2 F = Force k = 9.0 X 109 N-m/C2 (proportionality constant) Q = Charge (C) r = Radius (m) Varies with inverse-square of the radius(distance)

  14. The Coulomb • 1 Coulomb = 1 Ampere•second • Unit of charge • Point charges – small objects (charge doesn’t get distributed much) • Elementary Charge • Charge on the electron and proton • Quantized (can’t have ½ an electron) • e = 1.602 X 10-19 C

  15. Determine the magnitude of the force between a proton and an electron in a hydrogen atom. Assume the distance from the electron to the nucleus is 0.53 X 10-10 m. (8.2 X 10-8 N) + -

  16. Calculate the force between an electron and the three protons in a Li atom if the distance is about 1.3 X 10-10 m. (3.9 X 10-8 N)

  17. Three charged particles are arranged in a straight line as shown in the diagram. Calculate the net force on particle 3. (-1.5 to the left) 0.30 m 0.20 m - - + Q1= -8.0 mC Q2= +3.0 mC Q3= -4.0 mC

  18. F = k Q1Q2 r2 F31 = (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(8.0X10-6 C) (0.50 m)2 F31 = 1.2 N (Repulsive to the right) F32 = (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(3.0X10-6 C) (0.20 m)2 F32 = 2.7 N (Attractive to the left)

  19. Fnet = F31 -F32 Fnet = 1.2 N – 2.7 N = -1.5 to the left

  20. Three charges are in a line. The first is +2.00 mC. The second is -2.50 mC at 25 cm. The third charge is +2.00 mC at the 40 cm mark. • Calculate the net force on the center charge. • Will the center charge move to the left or right?

  21. Three charges are in a line. The first is +3.00 mC. The second is -2.00 mC at 5 cm. Where should a the third charge (+4.0 mC) be placed so the middle charge does not move?

  22. Three charges are in a line. The first is -10.00 mC. The second is -15.00 mC at 100 cm. Where should a middle charge (+20.0 mC) be placed so it does not move?

  23. Coulomb’s Law: Ex 4 What is the resultant force on charge q3if the charges are arranged as shown below. The magnitudes of the charges are: q1 = +6.00 X 10-9 C q2 = -2.00 X 10-9 C q3 = +5.00 X 10-9 C

  24. 4.00 m q2 - + q3 37o 3.00 m 5.00 m q1 +

  25. First calculate the forces on q3 separately: F13= k Q1Q3 r2 F13 = (9.0 X 109 N-m2/C2)(6.00 X 10-9 C)(5.00 X 10-9 C) (5.00m )2 F13 = 1.08 X 10-8 N

  26. F23 = k Q2Q3 r2 F23 = (9.0 X 109 N-m2/C2)(2.00 X 10-9 C)(5.00 X 10-9 C) (4.00m )2 F23 = 5.62 X 10-9 N

  27. F13 = 1.08 X 10-8 N F23 = 5.62 X 10-9 N 37o q2 - + q3 37o q1 +

  28. F13x = F13cos37o = (1.08 X 10-8 N)cos37o F13x = 8.63 X 10-9 N F13y = F13sin37o = (1.08 X 10-8 N)sin37o F13y = 6.50 X 10-9 N Fx = F23 + F13x Fx = -5.62 X 10-9 N + 8.63 X 10-9 N = 3.01 X 10-9 N Fy= F13y = 6.50 X 10-9 N

  29. FR Fy Fx= 3.01 X 10-9 N Fy= 6.50 X 10-9 N FR = \/ (3.01 X 10-9 N)2 + (6.50 X 10-9 N)2 FR = 7.16 X 10-9 N sin q = Fy/FR sin q = (6.50 X 10-9 N)/ (7.16 X 10-9 N) q = 64.7o q Fx -

  30. Coulomb’s Law: Ex 5 Calculate the net electrostatic force on charge Q3 as shown in the diagram: Q3= +65 mC + 60 cm 30 cm 30o + - Q2= +50 mC Q1= -86 mC

  31. First calculate the forces on Q3 separately: F13= k Q1Q3 r2 F13 = (9.0 X 109 N-m2/C2)(65 X 10-6 C)(86 X 10-6 C) (0.60 m )2 F13 = 140 N F23 = (9.0 X 109 N-m2/C2)(65 X 10-6 C)(50 X 10-6 C) (0.30 m )2 F23 = 330 N

  32. F23 Q3 + F13 30o + - Q2 Q1

  33. F13x = F13cos30o = (140 N)cos30o F13x = 120N F13y = -F13sin30o = (140 N)sin30o F13y = -70N Fx = F13x = 120 N F7 = 330 N - 70N = 260 N

  34. FR Fy Fx= 120 N Fy= 260N FR = \/ (120 N)2 + (330N)2 FR = 290 N sin q = Fy/FR sin q = (260N)/ (290N) q = 64o q Fx -

  35. A small plastic bead has a mass of 15 mg and a charge of -10 nC (nano = 10-9). A glass rod of charge + 10 nC is held 1.0 cm above the bead. • Calculate the electric field strength of the rod at the position of the bead. • Calculate the force on the bead. • Will the bead leap off the table?

  36. Calculus Example 1 Calculate the force on charge q from the charged rod shown below. The charge per unit length of the rod is: l = Q/l

  37. F = kqQ x2 dF = kqdQdQ = ldx x2 dF = kqldx x2 F = kql∫ dx (from “a” to “a+l”) x2 F = -kql1 a+ l x a

  38. F = -kql1 a+ l x a F = -kql1 - l a+1 a F = kqll a(a+1)

  39. Calculus Ex 2 A total positive charge of Q is evenly distributed on a semicircular ring of radius R. Calculate the force felt by the charge q at the center of the semicircle.

  40. F = kqQ R2 However, x-components cancel Fy = Fsinq Fy = kqQ sinq R2 Break into a small unit of force dFy = kqdQ sinq R2

  41. l = Q/pR dQ = lds s = Rq dQ = l Rdq dFy = kq lR sinq dq R2

  42. Fy = kq l - cosq p R 0 Fy = 2kq l R

  43. Electric Field • Contact forces • Friction • Pushes and pulls • Forces at a distance • Gravity • Electromagnetism • Field – Invisible lines that extend from a body

  44. A positive test charge would be repelled by the field + Proton +

  45. A positive test charge would be attracted by the field + Electron -

  46. Opposite charges attract

  47. Like Charges repel

  48. E in terms of a test charge E = F q • Vector quantity • Force that the test charge q would feel. The smaller the charge, the larger the Force

  49. E in terms of a point charge E = kQ r2 • Vector quantity • Point charge is the source charge producing the electric field k = 1/ 4pe0 e0 =8.85 X 10-12 C2/Nm2 (permittivity constant)