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Electric Force & Field 2012

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IBD Topic 6

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Electric Force & Field 2012

  1. 1. Topic 6: Fields and Forces Topic 6.2 Electric Force and Field
  2. 2. Electric Field & Force  What do you know about electric forces?  What does an electric field look like?  What experience have you had with electric fields?
  3. 3. Electrification by Friction Electric charge, or `electricity', can come from batteries and generators. But some materials become charged when they are rubbed. Their charge is sometimes called electrostatic charge or `static electricity'. It causes sparks and crackles when you take off a pullover, and if you slide out of a car seat and touch the door, it may even give you a shock.
  4. 4. Two Types of Charge  Polythene and Perspex can be charged by rubbing them with a dry, woollen cloth.
  5. 5.  When two charged polythene rods are brought close together, they repel (try to push each other apart).  The same thing happens with two charged Perspex rods.
  6. 6.  However, a charged polythene rod and a charged Perspex rod attract each other.  Experiments like this suggest that there are two different and opposite types of electric charge.  These are called positive (+) charge and negative (-) charge:
  7. 7. Conservation of Charge  Where charges come from  Everything is made of tiny particles called atoms.  These have electric charge inside them.  There is a central nucleus made up of protons and neutrons.  Orbiting the nucleus are much lighter electrons
  8. 8.  Electrons have a negative (-) charge.  Protons have an equal positive (+) charge.  Neutrons have no charge.
  9. 9.  Normally, atoms have equal numbers of electrons and protons, so the net (overall) charge on a material is zero.  However, when two materials are rubbed together, electrons may be transferred from one to the other.  One material ends up with more electrons than normal and the other with less.  So one has a net negative charge, while the other is left with a net positive charge.
  10. 10.  Rubbing materials together does not make electric charge. It just separates charges that are already there.  Charge is always conserved in any action, the distribution of charge is changed.
  11. 11. Question  If a million electrons are transferred in one rub by a cloth, what are the resulting charges in Coulombs of (i) 3 rubs of polythene (ii) 5 rubs of Perspex (iii) the rods from (i) and (ii) are made to touch
  12. 12. Conductors and Insulators  When some materials gain charge, they lose it almost immediately. This is because electrons flow through them or the surrounding material until the balance of negative and positive charge is restored.
  13. 13. Conductors  Conductors are materials that let electrons pass through them.  Metals are the best electrical conductors.  Some of their electrons are so loosely held to their atoms that they can pass freely between them.  These free electrons also make metals good thermal conductors.  Most non-metals conduct charge poorly or not at all, although carbon (in the form of graphite) is an exception.
  14. 14. Insulators  Insulators are materials that hardly conduct at all.  Their electrons are tightly held to atoms and are not free to move - although they can be transferred by rubbing.  Insulators are easy to charge by rubbing because any electrons that get transferred tend to stay where they are.
  15. 15. Semiconductors  Semiconductors are `in-between' materials.  They are poor conductors when cold, but much better conductors when warm.
  16. 16. Electrostatic Induction  Attraction of uncharged objects  A charged object will attract any uncharged object close to it. For example, the charged screen of a TV will attract dust.
  17. 17.  The previous diagram shows what happens if a positively charged rod is brought near a small piece of aluminium foil.  Electrons in the foil are pulled towards the rod, which leaves the bottom of the foil with a net positive charge.  As a result, the top of the foil is attracted to the rod, while the bottom is repelled.  However, the attraction is stronger because the attracting charges are closer than the repelling ones.
  18. 18. Electroscope  In 1786 the Rev. Abraham Bennet introduced a device called a ‘gold leaf electroscope’.  It consists of two extremely thin gold leaves hanging parallel to each other connected to a conducting stem and cap.  The internal system is protected from air currents by a glass enclosure.
  19. 19. Electroscope  The outer case and stand system is separated from the inner system by an insulating plug.  At PAC we use a needle electrometer where the needle replaces the easily damaged gold leaf.
  20. 20. Using an Electroscope  An electroscope can be charged positively by induction with a negative rod
  21. 21. Using an Electroscope  The electroscope can test the sign of a charge.  The electroscope is first charged positively  by stroking the cap with a strongly + ively charged glass rod or  by induction using a -ively charged ebonite rod.
  22. 22. Using an Electroscope  The object whose charge is to be tested is slowly brought near to the cap of the electroscope.  If the rod is uncharged, then the charged electroscope will induce a charge on it, and since the nearer side of the rod is negatively charged.
  23. 23. Using an Electroscope
  24. 24. Using an Electroscope  In effect the rod will behave as a negatively charged one, and repel electrons to the needle, neutralising positive charge and decreasing needle divergence.
  25. 25. Using an Electroscope  In effect the rod will behave as a negatively charged one, and repel electrons to the needle, neutralising positive charge and decreasing needle divergence.
  26. 26. Using an Electroscope  It is easy to distinguish a + ively charged electroscope.
  27. 27. Using an Electroscope
  28. 28. Using an Electroscope
  29. 29. Using an Electroscope  To tell the difference between a neutrally charged rod and a negatively charged rod, we need to earth the electroscope.  An uncharged rod will cause no deflection while the negatively charged rod will cause the needle to diverge.  Repulsion is the only sure test for determining the sign of a charge.
  30. 30. Coulomb’s Law The force between two point charges is directly proportional to the product of the charges and inversely proportional to their distance apart squared. +1 +3 r
  31. 31. Equations  F  q1 q2 / r2  Or F = k q1 q2 / r2  Where k = 1/4πε0  ε0 is the permittivity of free space  Therefore Coulomb’s Law can be written as  F = q1 q2 / 4πε0r2
  32. 32. Applying Coulomb’s Law  To determine the net force on a charge due to two or more other charges, you must use vector addition.  Find the force and direction due to each of the other charges in turn  and then resolve these forces to get the resultant force.
  33. 33. Coulomb’s Law  Coulomb’s law states that the force acting between two charges q1 and q2 whose distances are separated by a distance d is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The force is along the line joining the centres of the charges.
  34. 34. Coulomb’s Law  𝑭 = 𝟏 𝟒𝝅𝜺𝜺 𝒐 𝒒 𝟏 𝒒 𝟐 𝒓 𝟐
  35. 35. Question 1 1. Find the force when (a) 2 protons are placed 1.00 mm apart (b) proton and electron are placed 1.00 x 10-9m apart (c) helium nuclei and an electron are 1.00 x 10-6m apart
  36. 36. Q1 (a) Solution  𝐹 = 1 4𝜋𝜀 𝑜 𝑞1 𝑞1 𝑟2  𝐹 = 8.99 x 109 1.6 x 10−19 x1 .6 x 10−19 (1 x 10−3) 2  F = 2.30 x 10−22 N repulsion
  37. 37. Q1 (b) Solution  𝐹 = 1 4𝜋𝜀 𝑜 𝑞1 𝑞1 𝑟2  𝐹 = 8.99 x 109 1.6 x 10−19 x1 .6 x 10−19 (1 x 10−9) 2  F = 2.30 x 10−10 N attraction
  38. 38. Q1 (c) Solution  𝐹 = 1 4𝜋𝜀 𝑜 𝑞1 𝑞1 𝑟2  𝐹 = 8.99 x 109 3.2 x 10−19 x1 .6 x 10−19 (1 x 10−6) 2  F = 4.60 x 10−16 N attraction
  39. 39. Question 2 2. Find the force acting on a +10.0 C charge placed 60.0 cm east of a +6.00 C charge and 40.0 cm north of a -8.00 C charge.
  40. 40. Q 2 Solution q2 = -8.00 μ C q = +10.0 μ C q1 = +6.00 μ C 60 x 10-2 m 40 x 10-2 m
  41. 41. Q 2 Solution  𝐹𝑞𝑞1 = 1 4𝜋𝜀 𝑜 𝑞𝑞1 𝑟2  𝐹𝑞𝑞1 = 8.99 x 109 10 x 10−6 x 8 x 10−6 (60 x 10−2) 2  𝐹𝑞𝑞1 = 1.49833333 N East  Don’t round yet
  42. 42. Q 2 Solution  𝐹𝑞𝑞2 = 1 4𝜋𝜀 𝑜 𝑞𝑞2 𝑟2  𝐹𝑞𝑞2 = 8.99 x 109 10 x 10−6 x 8 x 10−6 (40 x 10−2) 2  𝐹𝑞𝑞2 = 4.495 N South  Don’t round yet
  43. 43. Q 2 Solution q2 = -8.00 μ C q = +10.0 μ C q1 = +6.00 μ C 60 x 10-2 m 1.49833333 N 4.495 N Add forces vectorially
  44. 44. Q 2 Solution  𝐹 = √(4.495)2 + (1.498333)2  𝐹 = 6.74 N  Direction  Tan θ = 𝑂𝑝 𝐴𝑑𝑗 = 4.495 1.498333 = 3.0000  θ = 71.6o  𝐹 = 6.74 N at 18.4 oT
  45. 45. Electric Field  A resultant force changes motion.  Many everyday forces are pushes or pulls between bodies in contact.  In other cases forces arise between bodies that are separated from one another.  Electric, magnetic and gravitational effects involve such action-at-a-distance forces and to deal with them physicists find the idea of a field of force, or simply a field, useful.
  46. 46.  Fields of these three types have common features as well as important differences.  An electric field is a region where an electric charge experiences a force.  If a very small, positive point charge Q, the test charge, is placed at any point in an electric field and it experiences a force F,  then the field strength E (also called the E-field) at that point is defined by the equation  𝐸 = 𝐹 𝑞
  47. 47.  The magnitude of E is the force per unit charge and its direction is that of F (i.e. of the force which acts on a positive charge).  Field strength E is thus a vector.  If F is in newtons (N) and Q is in coulombs (C) then the unit of E is the newton per coulomb (N C-1).  A commoner but equivalent unit is the volt per metre (V m-1).
  48. 48.  To determine the net field strength on a charge due to two or more other charges, we must use vector addition.  Find the field strength and direction due to each of the other charges in turn  and then resolve these field strengths to get the resultant field strength  Remember that the direction of a field is the direction in which a positive charge would move
  49. 49. Uniform Electric Field  The field between two parallel plates can be calculated by
  50. 50. Field Patterns  An electric field can be represented and so visualized by electric field lines.  These are drawn so that  (1) the field line at a point (or the tangent to it if it is curved) gives the direction of E at that point, i.e. the direction in which a positive charge would accelerate,  and (2) the number of lines per unit cross-section area is proportional to E.  The field line is imaginary but the field it represents is real.
  51. 51. Positive Charge +
  52. 52. A hollow sphere
  53. 53. Games  Check your knowledge of charges & fields  http://phet.colorado.edu/en/simulation/char ges-and-fields  Play Electric Field Hockey  http://phet.colorado.edu/en/simulation/elec tric-hockey

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