1 / 57

SACE Stage 2 Physics

SACE Stage 2 Physics. Electric Fields. Electric Fields. Unit of Charge. Electric charge ( q ) is measured in units of Coulomb ( C ) Definition The coulomb is defined as the amount of charge transferred by an electric current of 1 Ampere in 1 second. 1 coulomb = 1 ampere x 1 second.

pfannenstiel
Download Presentation

SACE Stage 2 Physics

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. SACE Stage 2 Physics Electric Fields

  2. Electric Fields • Unit of Charge Electric charge (q) is measured in units of Coulomb(C) Definition The coulomb is defined as the amount of charge transferred by an electric current of 1 Ampere in 1 second. 1 coulomb = 1 ampere x 1 second

  3. Electric Fields • Unit of Charge Comparison of Charge and Mass of Electrons and Protons The charge on an electron is exactly the same as the charge on a proton. However, a proton has a mass approximately 2000 times the mass of an electron. me qe = e (symbol for the charge on an electron) qe= -1.6 x 10-19 C mp=2000me qe= -1.6 x 10-19 C qp= +1.6 x 10-19 C

  4. Electric Fields • Size of the Coulomb The Coulomb is a very large unit of charge. Eg, consider the amount of charge contained in the electrons in 1 g of hydrogen atoms. 1 g contains Avogadro's number of particles (6.02 x 1023)  number of electrons (Ne) = 6.02 x 1023 Total Charge of electrons (qe) = Ne x charge on an electron (e) = (6.02 x 1023 ) x (- 1.60 x 10-19) = - 96320 Coulomb. Similarly the total charge of the protons is + 96320 Coulomb.

  5. Electric Fields • Size of the Coulomb Consider the number of electrons which need to be moved to produce a charge of 10-6 C. Hence to produce this charge it is necessary to shift 6.3 x1013 electrons from 1 g of hydrogen (for example). As a fraction, this means that we shift (6.3 x1013)/(6.02 x1023)  1 x10-10 or 10-8 %, ie, producing a charge of 10-6 C in the lab would involve moving 0.00000001 % of the electrons in 1 gram of hydrogen

  6. Electric Fields • Size of the Coulomb Hence the Coulomb unit is very large and therefore we use smaller units: eg 1 milli-Coulomb = 10-3 = 1 mC 1 micro-Coulomb = 10-6 = 1 C 1 nano-Coulomb = 10-9 = 1 nC 1 pico-Coulomb = 10-12 = 1 pC

  7. Electric Fields • Conservation of Charge The fundamental concept in the model of electricity used here is the concept of Conservation of Charge In any system the total amount of charge remains constant.

  8. Electric Fields • Conservation of Charge Pair production: high energy bundle of electromagnetic radiation (gamma ray photon) interacts with the nucleus of an atom knocking out a pair of sub-atomic particles, a proton and an electron. electron protons and electrons are produced (created) in equal numbers to conserve charge. heavy nucleus charge before = 0 charge after = e+ + e- = 0 proton

  9. F F Line joining charges 10 C 1 pC 10 C 1 pC Interaction Between Charges • Coulomb’s Law Newton's Third Law still applies. The size of the two forces is the same, but the directions of the forces exerted by each object on the other are opposite.

  10. Only one force acts on this object Interaction Between Charges • Coulomb’s Law NOTE : ONLY ONE FORCE ACTS ON EACH OBJECT, SO THERE IS A NET FORCE ON EACH OBJECT .

  11. v e- FC proton Interaction Between Charges • Coulomb’s Law This force causes an acceleration (eg centripetal acceleration) It is found that the force also depends on the charges and their distance apart.

  12. Interaction Between Charges • Coulomb’s Law Any two point charges have acting on them equal sized, oppositely directed forces acting along the line joining their centres. The magnitude of these equal sized forces is directly proportional to the product of the charges and inversely proportional to the square of their distance apart. The forces are attractive for unlike charges and repulsive for like charges. [Constant = 9 x 109 N m2 C-2 in a vacuum and can be assumed to be the same in air] constant

  13. Interaction Between Charges • Example Consider the repulsive force between the electrons in 1 g of hydrogen atoms placed on either side of the earth. Earth F F D = 12700 km 1 g of H 1 g of H

  14. Interaction Between Charges An equivalent mass of 50,000kg would be needed to produce the same force! Electric forces are much stronger than gravitational forces.

  15. R electron proton Ratio of Gravitational Forces and Electric forces Can find the ratio between Electric force and Gravitational force by considering an electron and a proton. (hydrogen atom, 1 proton, 1 electron)

  16. Ratio of Gravitational Forces and Electric forces Fe = Fe: Fg = 2.3x10-28r-2 1.01 x 10-67r-2 ratio of force independent of separation of particles. = 2.3 x 1039 Fe = 2.3 x 1039 Fg since qp = +e or = = 2.3x10-28r-2 Fg= = 1.01 x 10-67r-2

  17. F = Form of Constant in Coulomb's Law is known as the Permittivity constant of free space where = 9 x 109 N m2 C-2 Do not represent as ke, it is incorrect as it has its own meaning.

  18. C qC +5 C 4 cm 3 cm qA+5 C qB +5 C B 5 cm A Electric Force due to More than 1 Charge The Charge at C will experience forces due to the charge at A and the charge at B. To find the resultant field we need to add the forces due to A and B as vectors.

  19. Electric Force due to More than 1 Charge The Force on C Due to B: F = The Force on C Due to A: F = = = = 250 N along the line BC and away from B = 140.6 N Away from A

  20. FB= 250 N FResultant q FA = 140.6 N Electric Force due to More than 1 Charge ie. By VECTOR ADDITION: ie adding FA and FC – shown opposite. By Pythagoras’ Theorem

  21. Electric Force due to More than 1 Charge tan q = q = 60.6° ie. Fresultant = 286.8 N at 60.6° to the line joining A and C. (As shown)

  22. Electric Fields The field concept is needed to explain how charges exert "forces at a distance" on other charges. Field model for Electric Charges large force small force +qT test charge strong field +qT test charge weak field Source Charge

  23. F -qT +qT F Electric Fields An electric field exists in a region of space if electric charges experience forces in that space. The direction of the field is the direction of the force on POSITIVE CHARGES

  24. Example A Calculate the electric field strength at a point P 10 cm from a point charge of + 10 C.

  25. E E Electric Fields If we use lines to represent these electric fields, then the spacing of the lines represents the size of E and the direction of the lines represents the direction of E.

  26. Derivation of the Electric Field Strength at a point in an Electric Field Consider a small test charge qt placed a distance r from a source charge qs qt r qs

  27. Example A Calculate the electric field strength at a point P 10 cm from a point charge of + 10 C.

  28. Example B Calculate the force on an electron placed at point P. (Charge on an electron, qe = 1.6 x 10-19C.)

  29. Example B Calculate the force on an electron placed at point P. (Charge on an electron, qe = 1.6 x 10-19C.)

  30. Example C Calculate the acceleration of the electron.

  31. Example C Calculate the acceleration of the electron.

  32. +q +q NOT F F Incorrect correct Rules For Drawing Electric Field Lines • Electric field lines do not cross • begin on positive charges and end on negative charges • The direction of the field is the direction of the force on a positive test charge

  33. F Strong field Weak field Rules For Drawing Electric Field Lines • The number of field lines per unit area represents the strength of the field. • Field lines cut conductors at 90o. • No field exists in a conductor.

  34. - Electric Fields due to Point Charges

  35. Electric Fields in Nature In household wires 10-2 In radio waves 10-1 In the atmosphere 102 In sunlight 103 Under a thundercloud 104 In a lightning bolt 104 In an X-ray tube 106 At the electron in a hydrogen atom 6 x 1011 At the surface of a uranium nucleus 2 x 1021

  36. P 4 cm 3 cm qA = +5 C 5 cm qB = +5 C Electric Fields due to more than one charge Consider the electric field at a point P due to two positive charges of magnitude +5C positioned as shown below. At point P there are electric fields due to A and B. Any charge at P will experience forces due to the charge at A and the charge at B. To find the resultant field we need to add the fields due to A and B as vectors.

  37. Electric Fields due to more than one charge Electric field due to qA Electric field due to qB

  38. Electric Fields due to more than one charge Vector Diagram EB To calculate the length and direction of the resultant Electric Field, will need to use the cosine rule (if right angled triangle, use Pythagoras) and your trig ratios. Er EA

  39. - Electric Field due to point charges of equal magnitude but opposite sign:

  40. + Electric Field due to point charges of equal magnitude and the same sign:

  41. An Infinite Conducting Plate The electric field for an infinite conducting plate must be uniform. A positive charge leaving a positively charged surface would be equally repelled from all directions. The positive charge must move at right angles to the surface. A similar charge at a different position on the plate must also have the same force exerted on it from all directions so it too must move away from the plate at right angles.

  42. An Infinite Conducting Plate For a positively charged infinite plate:

  43. An Infinite Conducting Plate Two Parallel Infinite Conducting Plates Field due to Negative Plate Net Field here is ZERO! Field due to Positive Plate Negative Plate Positive Plate Net Field here is ZERO! Net field outside the plates must be zero as the uniform field from the infinite negative plate means the field strength must be the same everywhere, and the same applies to the positive plate ie. at a given point outside the plates the fields from each plate cancel by the principle of superposition.

  44. + _ Electric Fields and Conductors There cannot be a parallel component to the electric field at the surface of a conductor as this would imply that there is a force on the charges on the surface and as they are static charges and that the field is uniform at the surface, therefore parallel component = 0. There is no electric field inside the conducting material.

  45. + _ Electric Field Inside a Hollow Conductor A Hollow Spherical Uncharged Conductor Placed in the Region Between Two Oppositely Charged Parallel Plates: No field inside the hollow conductor

  46. Electric Fields and Sharp Points: The electric field around a Pear Shaped Object: Stronger field is in the vicinity of the sharpest curvature (at points of maximum curvature).

  47. P P Electric Fields and Sharp Points A test charge is placed at position P as shown below. If the net force at P is zero, then there must be a greater charge on the smaller sphere. Now this corresponds to the fact that if the charged object is pear shaped then considering the position P inside the object, hence net force on a test charge at P will only be zero if there are more charges at the sharper end of the object.

  48. Corona Discharge At the sharp end of a pair shaped conductor, the build up of charge can be large enough to create an electric field large enough to accelerate ionic particles in the air, which, when accelerating, collide with other particles in the air and they to, become ionised. This charges the air in the vicinity of the sharp end of the conductor to draw away any excess charge.

  49. Application: Photocopier and Laser Printers Photocopying Materials, such as selenium, that conduct better in light than the dark are known as photoconductive. If a selenium surface was positively charged in the dark, light will cause electrons to be drawn to it to neutralise the charge only where light was incident. Because selenium is not a good conductor the charge remains on the surface in a static position.

  50. The Process of Photocopying Important mechanical part of a photocopier is the cylindrical aluminium drum with a coating of Selenium. Aluminium is a good conductor of electricity and is therefore earthed. All processes occur as the drum rotates. • Charging the Drum • A highly charged wire (Corona Wire) at 5000V to 8000V extends the length of the drum and close to it. As the corona wire is highly charged, the corona effect allows positive charges to ‘jump’ on to the drum.

More Related