1 / 72

CHAPTER 18 & 19 ELECTRIC FIELD, ELECTRICAL ENERGY and CAPACITANCE

CHAPTER 18 & 19 ELECTRIC FIELD, ELECTRICAL ENERGY and CAPACITANCE. Michael Faraday developed the concept of electric field. A charge creates an electric field about it in all directions. When another charged object enters this electric field, it experiences an electric force.

mac
Télécharger la présentation

CHAPTER 18 & 19 ELECTRIC FIELD, ELECTRICAL ENERGY and CAPACITANCE

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. CHAPTER 18 & 19ELECTRIC FIELD,ELECTRICAL ENERGY and CAPACITANCE

  2. Michael Faraday developed the concept of electric field. A charge creates an electric field about it in all directions. When another charged object enters this electric field, it experiences an electric force.

  3. Suppose I have a +Q charge. I want to know the strength of the electric field (E) at a point which is a distance r from it.

  4. Electric field intensity (E) at a point which is a distance r from +Q is defined as the force experienced by a small positive charge +q when it is placed at that point. E +q Q r

  5. The direction of the electric field intensity at a point is the same as the direction in which a positive charge +q would move when placed at that point.

  6. Consider a negative charge –Q. The electric field intensity at a point which is at a distance r away is E +q Q r

  7. E +q Q r

  8. Electric field Lines are imaginary lines drawn in such a manner that their direction at any point is the same as the direction of the electric field at that point. The strength of the electric field is indicated by the spacing between the lines. The closer the electric field lines the stronger is the electric field.

  9. Consider a positive charge. The electric field lines around it will point away from the charge.

  10. Consider a negative charge. The electric field lines around it will point towards the charge.

  11. Field Lines for two positively charged objects

  12. Field Lines for a negatively and positively charged object

  13. Electric field lines for objects with unequal strengths http://dev.physicslab.org/asp/applets/pointcharges/default.asp

  14. What is the electric field intensity at a distance 2 m from a charge -12 x 10-6 C? Example

  15. A negative charge of 2 x 10-8 C experiences a force of 0.060 N to the right in an electric field. What is the magnitude of the electric field? Problem #1

  16. A positive test charge of 5 x 10-4 C is in an electric field that exerts a force of 2.5 x10-4 N on it. What is the magnitude of the electric field at the location of the test charge? Problem #2

  17. What charge exists on a test charge that experiences a force of 1.4 x 10-8 N at a point where the electric field intensity is 2.0 x 104 N/C? Problem #3

  18. Q Problem #4 Find the electric field intensity at a point P, located 6 mm to the left of a point charge of 8C. What are the magnitude and direction of the force on a –2nC charge placed at a point P. 6mm p 8 x 10-6 C

  19. Q 6mm p 8 x 10-6 C

  20. Q 6mm p 8 x 10-6 C

  21. B Problem #5 Determine the electric field intensity at the midpoint between a –60 C charge and a + 40 C charge. The charges are 70 mm apart in air. EA p A EB -60 x 10-6 C 40 x 10-6 C

  22. B EA p A EB -60 x 10-6 C 40 x 10-6 C

  23. B EA p A EB -60 x 10-6 C 40 x 10-6 C

  24. B A Problem #6 Find the electric field intensity at a point 30 mm to the right of a 16 x 10-9 C charge and a 40 mm to the left of a 9 x 10-9 C charge. EA p 30mm 40mm EB 16 x 10-9 C 9 x 10-9 C

  25. B A EA p 30mm 40mm EB 16 x 10-9 C 9 x 10-9 C

  26. B A EA p 30mm 40mm EB 16 x 10-9 C 9 x 10-9 C

  27. Problem #7 Two charges of +16 x 10-6 C and +8 x 10-6 C are 200 mm apart in air. At what point between the two charges is the field intensity equal to zero?

  28. q1=8x10-6 C q2=16x10-6 C

  29. A good conductor contains charges that are not bound to any atom and are free to move about within the material.When no net motion of charges occurs within a conductor, the conductor is said to be in electrostatic equillibrium. CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM

  30. 1.The electric field is zero everywhere inside the conductor. AN ISOLATED CONDUCTOR HAS THE FOLLOWING PROPERTIES 2. Any excess charge on an isolated conductor resides entirely on its surface.

  31. 3. The electric field just outside a charged conductor is perpendicular to the conductor’s surface. AN ISOLATED CONDUCTOR HAS THE FOLLOWING PROPERTIES 4. On an irregularly shaped conductor, the charge tends to accumulate at sharp points.

  32. CHAPTER 19 HOORAY!!!

  33. ENERGY & ELECTRIC POTENTIAL Point B Point A Consider a fixed positive charge placed at a point A and a fixed negative charge at point B. There is a force of attraction between the two charges.

  34. ENERGY & ELECTRIC POTENTIAL Point B Point A Point B Point C Therefore to bring the charge at B to a point C a distance (r) from B, will require work to be done against the force of attraction. Thus the charge at B will undergo a change in potential energy.

  35. ELECTRICAL POTENTIAL DIFFERENCE IS MEASURED IN VOLTSIT IS THE CHANGE IN POTENTIAL ENERGY PER UNIT CHARGE.

  36. This formula is used when dealing with charges

  37. This formula is used when working with parallel plates

  38. This formula is used when work on a charge is given

  39. Example The electric field intensity between the two charged parallel plates is 8000 N/C. The plates are 0.05 m apart. What is the potential difference between the two plates?

  40. Problem #1 A voltmeter reads 500 V when placed across two charged, parallel metal plates. The plates are 0.02 m apart. What is the electric field between them?

  41. Problem #2 What work is done on a 5 C charge that is raised by a potential of 1.5 V

  42. Problem #3 If 120 J of work is done to move one Coulomb of charge from a positive plate to a negative plate, what voltage difference exists between the plates?

  43. Problem #4 How much work is done to transfer 0.15 C of charge through a potential difference of 9 V?

  44. Problem #5 An electron is moved through a potential difference of 500 V. How much work is done on the electron?

  45. Problem #6 A force of 0.053 N is required to move a charge of 37 x 10-6 C a distance of 25 cm in an electric field. What is the size of the potential difference between the two points?

  46. Parallel Plates When two parallel plates are connected across a battery, the plates will become charged and an electric field will be established between them.

  47. The direction of an electric field is defined as the direction that a positive test charge would move. So in this case, the electric field would point from the positive plate to the negative plate. The field lines are parallel to each other and hence this type of electric field is uniform and is calculated with the equation E = V / d.

More Related