2. Electric Fields, Voltage, Electric Current, and Ohm’s Law ISAT 241 Fall 2003 David J. Lawrence

3. Properties of Electric Charges  Two kinds of charges. Unlike charges attract, while like charges repel each other.  The force between charges varies as the inverse square of their separation: F  1/r 2.  Charge is conserved. It is neither created nor destroyed, but is transferred.  Charge is quantized. It exists in discrete “packets”: q =  /  N e, where N is some integer.

4. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.2

5. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.2

6. Properties of Electric Charges  “Electric charge is conserved” means that objects become “charged” when charges (usually electrons) move from one neutral object to another.  This movement results in  a Net Positive charge on one object, and  a Net Negative charge on the other object.

7. Properties of Electric Charges  Neutral, uncharged matter contains as many positive charges as negative charges.  Net charge is caused by an excess (or shortage) of charged particles of one sign.  These particles are protons and electrons.

8. Properties of Electric Charges  Charge of an electron =  e =  1.6  10 -19 C  Charge of a proton =  e =  1.6  10 -19 C  “C” is the Coulomb.  Charge is Quantized! Total Charge = N  e = N  1.6  10 -19 C where N is the number of positive charges minus the number of negative charges. But, for large enough N, quantization is not evident.

9. Electrical Properties of Materials  Conductors: materials in which electric charges move freely, e.g., metals.  Insulators: materials that do not readily transport charge, e.g., most plastics, glasses, and ceramics.

10. Electrical Properties of Materials  Semiconductors: have properties somewhere between those of insulators and conductors, e.g., silicon, germanium, gallium arsenide, zinc oxide.  Superconductors: “perfect” conductors in which there is no “resistance” to the movement of charge, e.g., some metals and ceramics at low temperatures: tin, indium, YBa 2 Cu 3 O 7

11. Coulomb’s Law  The electric force between two charges is given by: (newtons, N)  Attractive if q 1 and q 2 have opposite sign.  Repulsive if q 1 and q 2 have same sign.  r = separation between the two charged particles.  k e = 9.0 x 10 9 Nm 2 /C 2 = Coulomb Constant.

13.  = electric force exerted by q 1 on q 2  r 12 = unit vector directed from q 1 to q 2 Coulomb’s Law  Force is a vector quantity. 

15. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.9

16. Gravitational Field  Consider the uniform gravitational field near the surface of the earth  If we have a = small “test mass” m o, the force on that mass is F g = m o g g y momo  We define the gravitational field to be Recall that g = | g | = 9.8 m/s 2

17. The Electric Field  The electric field vector E at a point in space is defined as the electric force F E acting on a positive “Test Charge” placed at that point, divided by the magnitude of the test charge q o. q >> q o qoqo FEFE q

18. The Electric Field q >> q o qoqo FEFE q  Units: ~newtons/coulomb, N/C

19. Serway & Jewett, Principles of Physics, 3 rd ed. See Figure 19.11

20. The Electric Field  In general, the electric force on a charge q o in an electric field E is given by + FEFE  FEFE E

21. The Electric Field u E is the electric field produced by q, not the field produced by q o. u Direction of E = direction of F E (q o > 0). u q o << |q| u We say that an electric field exists at some point if a test charge placed there experiences an electric force.

22. The Electric Field u For this situation, Coulomb’s law gives: F E = |F E | = k e (|q||q o |/r 2 ) u Therefore, the electric field at the position of q o due to the charge q is given by: E = |E| = |F E |/q o = k e (|q|/r 2 ) q >> q o qoqo E q | q | >> q o qoqo E q

23. Gravitational Field Lines u Consider the uniform gravitational field near the surface of the earth = g u If we have a small “test mass” m o, the force on that mass is F g = m o g u We can use gravitational field lines as an aid for visualizing gravitational field patterns. g y momo Recall that g = | g | = 9.8 m/s 2

24. Electric Field Lines u An aid for visualizing electric field patterns. u Point in the same direction as the electric field vector, E, at any point. u E is large when the field lines are close together, E is small when the lines are far apart.

25. Electric Field Lines u The lines begin on positive charges and terminate on negative charges, or at infinity in the case of excess charge. u The number of lines leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. u No two field lines can cross. u E is in the direction that a positive test charge will tend to go.

26. Electric Field Lines u The lines begin on positive charges and terminate on negative charges, or at infinity in the case of excess charge. +

27. Electric Field Lines u The lines terminate on negative charges. -

28. Electric Field Lines u More examples Field lines cannot cross!

29. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.17

30. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.18

31. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.19

32. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.20 See the discussion about this figure on page 683 in your book.

33. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.21 See Example 19.6 on page 684 in your book.

34. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 19.22 See Example 19.7 on page 685 in your book.

35. Work Done by a Constant Force (Review) u Fluffy exerts a constant force of 12N to drag her dinner a distance of 3m across the kitchen floor. u How much work does Fluffy do?

36. Work Done by a Constant Force (Review)  u Ingeborg exerts a constant force of 12N to drag her dinner a distance of 3 m across the kitchen floor.   = 30 o u How much work does Ingeborg do?

37. Similar to Serway & Jewett, Principles of Physics, 3 rd ed. Figure 6.1 See page 179 in your book.

38. Work Done by a Force (Review) u Is there a general expression that will give us the work done, whether the force is constant or not? u Yes! u Assume that the object that is being moved is displaced along the x-axis from x i to x f. u Refer to Figure 6.7 and Equation 6.11 on p. 184. = area under graph of F x from x i to x f

39. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 6.7

40. Gravitational Field u Consider the uniform gravitational field near the surface of the earth = g u Recall that g = | g | = 9.8 m/s 2 g ybyb yaya y b momo a momo d Suppose we allow a “test mass” m o to fall from a to b, a distance d.

41. Gravitational Field u How much work is done by the gravitational field when the test mass falls? g ybyb yaya y b momo a momo d Suppose we allow a “test mass” m o to fall from a to b, a distance d.

42. Electric Field u A uniform electric field can be produced in the space between two parallel metal plates. u The plates are connected to a battery. E

43. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 20.3

44. Electric Field u How much work is done by the electric field in moving a positive test charge (q o ) from a to b? E d a qoqo b qoqo

45. Electric Field u Recall that F E = q o E  Magnitude of displacement = d E d a qoqo b qoqo

46. Potential Difference = Voltage u Definition g The Potential Difference or Voltage between points a and b is always given by = (work done by E to move test chg. from a to b) (test charge) g This definition is true whether E is uniform or not.

47. Potential Difference = Voltage u For the special case of parallel metal plates connected to a battery -- g The Potential Difference between points a and b is given by gThis is also called the Voltage between points a and b. gRemember, E is assumed to be uniform.

48. Potential Difference = Voltage u We need units! u Potential Difference between points a & b  Voltage between points a & b

49. Potential Difference = Voltage u More units! u Recall that for a uniform electric field so In your book’s notation: Where d is positive when the displacement is in the same direction as the field lines are pointing.

50. Potential Difference = Voltage u In the general case = a “path integral” or “line integral” u Therefore

51. Potential Difference = Voltage u If E, F E, and the displacement are all along the x-axis, this doesn’t look quite so imposing! u So

52. Potential Difference = Voltage u What about the uniform E case? E d a qoqo b qoqo

53. Serway & Jewett, Principles of Physics, 3 rd ed. Figure 20.3 See Example 20.1 on page 714 in your book.

54. Voltage & Electric Field B A 12V d u If the separation between the plates is d = 0.3 cm, find the magnitude of the electric field between the plates. E

55. Voltage & Electric Field  Solution: Recall that for a uniform electric field, the voltage (or electric potential difference) between two points is given by V A  V B  V AB  E d where d is the distance between the two points.  E  V AB / d u In our case, we know that the voltage between the two plates is just the battery voltage, so V AB = 12 V. The two plates are separated by d = 0.3cm   E  12V/(0.3cm x 10 -2 m/cm) = 4000 V/m

56. Voltage and Electric Potential u Sometimes, physicists talk about the “electric potential” at some location, e.g., V A = “electric potential at point A” V B = “electric potential at point B”  Electric potential really needs to be measured with respect to some reference point. For example, the reference could be “ground” (the earth) or some distant point in space (“at  ”), so to be precise we could say V A = electric potential difference (voltage) between point A and ground, etc.

57. Voltage and Electric Potential u If V A = electric potential difference (voltage) between point A and ground, and V B = electric potential difference (voltage) between point B and ground, u Then the voltage (electric potential difference) between points A & B can be written V AB = V A  V B and V BA = V B  V A  If A is more + than B  V AB > 0 V BA < 0  If B is more + than A  V BA > 0 V AB < 0

58. Electric Field Lines and Electric Potential u Electric field lines always point in the direction of decreasing electric potential. u Page 713 In your book’s notation: Where d is positive when the displacement is in the same direction as the field lines are pointing. So d is negative when the displacement is in the opposite direction as the field lines are pointing.

59. Gravitational Field -- P.E. u Consider the uniform gravitational field near the surface of the earth = g u Recall that g = | g | = 9.8 m/s 2 g ybyb yaya y b m a m d Suppose we lift a “test mass” m from a to b, a distance d, against the field g.

60. Gravitational Field -- P.E. u Gravitational Potential Energy: U g = mgy where y is the height. u The change in the gravitational P.E. as we lift the mass is:  P.E. g =  U g = U gb  U ga = mgy b  mgy a = mg d +++ positive u If instead we let the mass fall from b to a:  P.E. g =  U g = U ga  U gb = mgy a  mgy b =  mg d --- negative

61. Electric Potential Energy  If a particle with charge q moves through a potential difference  V = V final  V initial, then the change in electric potential energy of the particle is given by  P.E. E =  U E = q  V or U E final  U E initial = q  V final  V initial )

62. Electric Potential Energy u Repeating Note that: Electric Potential   Electric Potential Energy but they are related (by the above equation)

63. Electric Potential Energy u Consider a uniform electric field = E in an environment without gravity. u V ba > 0 “point b is more positive than point a” E ybyb yaya y b qoqo a qoqo d Suppose we move a “test charge” q o from a to b, a distance d, against the field E (q o is positive).

64. Electric Potential Energy u The change in electric P.E. of the test charge when we move it is:  P.E. E =  U E = q o  V U Eb  U Ea = q o V ba  V ba = E d  U Eb  U Ea = q o E d = = the work we do in moving the charge

65. Energy -- Units u Recall that the SI unit of energy is the Joule (J). u Another common unit of energy is the electron volt (eV), which is the energy that an electron (or proton) gains or loses by moving through a potential difference of 1 V.  1 eV = 1.602  10 -19 J  Example: electron in beam of CRT has speed of 5  10 7 m/s  KE = 0.5mv 2 = 1.1  10 -15 J = 7.1  10 3 eV  electron must be accelerated from rest through potl. diff. of 7.1  10 3 V in order to reach this speed.

66. Electric Current u Consider a bar of material in which positive charges are moving from left to right: imaginary surface I u Electric current is the rate at which charge passes through the surface, I avg =  Q/  t, and the instantaneous current is I = dQ/dt.

67. Electric Current u SI unit of charge: Coulomb (C) u SI unit of current: Ampere (1A= 1C/s) u A current of 1 ampere is equivalent to 1 Coulomb of charge passing through the surface each second.

68. Electric Current u By definition, the direction of the current is in the direction that positive charges would tend to move if free to do so, i.e., to the right in this example. u In ionic solutions (e.g., salt water) positive charges (Na + ions) really do move. In metals the moving charges are negative, so their motion is opposite to the conventional current. u In either case, the direction of the current is in the direction of the electric field.

69. Electric Current u Na + ions moving through salt water u Electrons moving through copper wire EI E I

70. Electric Current u The electric current in a conductor is given by where n = number of mobile charged particles (“carriers”) per unit volume q = charge on each carrier v d = “drift speed” (average speed) of each carrier A = cross-sectional area of conductor  In a metal, the carriers have charge q  e.

71. Electric Current u The average velocity of electrons moving through a wire is ordinarily very small ~ 10 -4 m/s. u It takes over one hour for an electron to travel 1 m!!! E I

72. Ohm’s Law u For metals, when a voltage (potential difference) V ba is applied across the ends of a bar, the current through the bar is frequently proportional to the voltage. area A VbVb VaVa E I H The voltage across the bar is denoted: V ba = V b  V a.

73. Ohm’s Law u This relationship is called Ohm’s Law. u The quantity R is called the resistance of the conductor.  R has SI units of volts per ampere. One volt per ampere is defined as the Ohm ( . 1  =1V/A. u Ohm’s Law is not always valid!!

74. Ohm’s Law u The resistance can be expressed as where is the length of the bar (m) A is the cross-sectional area of the bar (m 2 ) , “Rho”, is a property of the material called the resistivity. SI units of ohm-meters (  -m). area A VbVb VaVa E I

75. Ohm’s Law u The inverse of resistivity is called conductivity: u So we can write

76. Resistance and Temperature u The resistivity of a conductor varies with temperature (approximately linearly) as where  resistivity at temperature T ( o C)  o  resistivity at some reference temperature T o (usually 20 o C)  “temperature coefficient of resistivity”. u Variation of resistance with T is given by

77. Electrical Power u The power transferred to any device carrying current I (amperes) and having a voltage (potential difference) V (volts) across it is P = VI u Recall that power is the rate at which energy is transferred or the rate at which work is done. u Units: W (Watt) = J/s

78. Electrical Power u Since a resistor obeys Ohm’s Law V = IR, we can express the power dissipated in a resistor in several alternative ways: