Chapter 25. Electric Potential

Transcription

1 Chapter 25 Electric Potential

2 Electric Potential Electromagnetism has been connected to the study of forces in previous chapters. In this chapter, electromagnetism will be linked to energy. By using an energy approach, problems could be solved that were insoluble using forces. The concept of potential energy is of great value in the study of electricity. Introduction 2

3 Electric Potential Because the electrostatic force is conservative, electrostatic phenomena can be conveniently described in terms of an electric potential energy. This will enable the definition of electric potential. 3

4 Electrical Potential Energy When a test charge is placed in an electric field, it experiences a force. Fe q 0 E The force is conservative. If the test charge is moved in the field by some external agent, the work done by the field is the negative of the work done by the external agent. Section

5 Electrical Potential Energy, cont ds is an infinitesimal displacement vector that is oriented tangent to a path through space. The path may be straight or curved and the integral performed along this path is called either a path integral or a line integral. The work done within the charge-field system by the electric field on the charge is Fds q Eds 0 5

6 Electric Potential Energy, final As this work is done by the field, the potential energy of the charge-field system is changed by U q Eds 0 For a finite displacement of the charge from A to B, the change in potential energy of the system is B A 0 A Because the force is conservative, the line integral does not depend on the path taken by the charge. B U U U q E d s (25.1) Section

7 Electric Potential The potential energy per unit charge, U/q 0, is the electric potential. The potential is characteristic of the field only. The potential energy is characteristic of the charge-field system. The potential is independent of the value of q 0. The potential has a value at every point in an electric field. The electric potential is V U q 0 (25.2) Section

8 Electric Potential, cont. The potential is a scalar quantity. Since energy is a scalar As a charged particle moves in an electric field, it will experience a change in potential. U B V d (25.3) q E s A 0 The infinitesimal displacement is interpreted as the displacement between two points in space rather than the displacement of a point charge. The difference in potential is the meaningful quantity. Section

9 Electric Potential, final We often take the value of the potential to be zero at some convenient point in the field. Electric potential is a scalar characteristic of an electric field, independent of any charges that may be placed in the field. The potential difference between two points exists solely because of a source charge and depends on the source charge distribution. For a potential energy to exist, there must be a system of two or more charges. The potential energy belongs to the system and changes only if a charge is moved relative to the rest of the system. Section

10 Work and Electric Potential Assume a charge moves in an electric field without any change in its kinetic energy. The work performed on the charge is W = U = q V (25.4) Units: 1 V 1 J/C V is a volt. It takes one joule of work to move a 1-coulomb charge through a potential difference of 1 volt. Section

11 Work and Electric Potential In addition, 1 N/C = 1 V/m This indicates we can interpret the electric field as a measure of the rate of change of the electric potential with respect to position. 11

12 Voltage Electric potential is described by many terms. The most common term is voltage. A voltage applied to a device or across a device is the same as the potential difference across the device. The voltage is not something that moves through a device. Section

13 Electron-Volts Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt. One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt. 1 ev = J (25.5) Section

14 Quick Quiz 25.1 In Figure 25.1, two points A and B are located within a region in which there is an electric field. (i) How would you describe the potential difference ΔV = V B V A? (a) It is positive. (b) It is negative. (c) It is zero. (ii) A negative charge is placed at A and then moved to B. How would you describe the change in potential energy of the charge field system for this process? Choose from the same possibilities. Answer: (i) (b) (ii) (a) 14

15 Potential Difference in a Uniform Field The equations for electric potential between two points A and B can be simplified if the electric field is uniform: The displacement points from A to B and is parallel to the field lines. The negative sign indicates that the electric potential at point B is lower than at point A. Electric field lines always point in the direction of decreasing electric potential. B B B A E s s A A V V V d E d Ed (25.6) Section

16 Energy and the Direction of Electric Field When the electric field is directed downward, point B is at a lower potential than point A. When a positive test charge moves from A to B, the chargefield system loses potential energy. Electric field lines always point in the direction of decreasing electric potential. Section

17 If you can't see the image above, please install Shockwave Flash Player. If this active figure can t auto-play, please click right button, then click play. 25.2: Electric Potential and Gravitational Potential A charge q and a mass m are released in a uniform electric field and gravitation field, respectively. Both objects accelerate from release point A to final point B in the field. The animated bar charts show kinetic and potential energies of the two objectfield systems during the motion. The readout boxes display the initial and final electric potential energies (EPE) and the initial and final gravitational potential energies (GPE), as well as the potential difference between points A and B in the electric field, all in SI units. 17

18 More About Directions A system consisting of a positive charge and an electric field loses electric potential energy when the charge moves in the direction of the field. An electric field does work on a positive charge when the charge moves in the direction of the electric field. The charged particle gains kinetic energy and the potential energy of the charge-field system decreases by an equal amount. Another example of Conservation of Energy Section

19 Directions, cont. If q 0 is negative, then U is positive. A system consisting of a negative charge and an electric field gains potential energy when the charge moves in the direction of the field. In order for a negative charge to move in the direction of the field, an external agent must do positive work on the charge. Section

20 Equipotentials Point B is at a lower potential than point A. Points A and C are at the same potential. All points in a plane perpendicular to a uniform electric field are at the same electric potential. The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential. Section

21 Charged Particle in a Uniform Field, Example A positive charge is released from rest and moves in the direction of the electric field. The change in potential is negative. The change in potential energy is negative. The force and acceleration are in the direction of the field. Conservation of Energy can be used to find its speed. Section

22 Quick Quiz 25.2 The labeled points in Figure 25.4 are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves from A to B, from B to C, from C to D, and from D to E Answer: B to C, C to D, A to B, D to E 22

23 Example 25.1 A battery has a specified potential difference V between its terminals and establishes that potential difference between conductors attached to the terminals. A 12-V battery is connected between two parallel plates as shown in Figure The separation between the plates is d = 0.30 cm, and we assume the electric field between the plates to be uniform. (This assumption is reasonable if the plate separation is small relative to the plate dimensions and we do not consider locations near the plate edges.) Find the magnitude of the electric field between the plates. Solution: E V B V d A 12V V/m m 23

24 Example

25 Example 25.2 A proton is released from rest at point A in a uniform electric field that has a magnitude of V/m (Fig. 25.6). The proton under-goes a displacement of magnitude d = 0.50 m to point in the direction of B. Find the speed of the proton after completing the displacement. E 25

26 Example 25.2 Solution: V Ed K U 0 1 mv 2 2 2eV v m v 4 4 ( V/m)(0.50m) V 0 ev (1.610 C)( V) kg m/s 26

27 Potential and Point Charges An isolated positive point charge produces a field directed radially outward. The potential difference between points A and B will be 1 1 VBVA keq r B r Α (25.10) Section

28 Potential and Point Charges, cont. The electric potential is independent of the path between points A and B. It is customary to choose a reference potential of V = 0 at r A =. Then the potential due to a point charge at some point r is V q ke r (25.11) Section

29 Electric Potential of a Point Charge The electric potential in the plane around a single point charge is shown. The red line shows the 1/r nature of the potential. Section

30 Electric Potential with Multiple Charges The electric potential due to several point charges is the sum of the potentials due to each individual charge. This is another example of the superposition principle. The sum is the algebraic sum V = 0 at r = V k e i q r i i (25.12) Section

31 Electric Potential of a Dipole The graph shows the potential (y-axis) of an electric dipole. The steep slope between the charges represents the strong electric field in this region. Section

32 Potential Energy of Multiple Charges The potential energy of the system is U k e qq r (25.13) Section

33 Potential Energy of Multiple Charges If the two charges are the same sign, U is positive and work must be done to bring the charges together. If the two charges have opposite signs, U is negative and work is done to keep the charges apart. 33

34 If you can't see the image above, please install Shockwave Flash Player. If this active figure can t auto-play, please click right button, then click play. 25.9: Electric Potential and Potential Energy In this animation, you can move the negative (blue) charge in the upper left square or the point P in the upper right square. The animation moves both in tandem, but the two squares represent two separate experiments. 34

35 U with Multiple Charges, final If there are more than two charges, then find U for each pair of charges and add them. For three charges: U qq q q q q ke r12 r13 r23 (25.14) The result is independent of the order of the charges. Section

36 Quick Quiz 25.3 In Active Figure 25.9a, take q 1 to be a negative source charge and q 2 to be the test charge. (i) If q 2 is initially positive and is changed to a charge of the same magnitude but negative, what happens to the potential at the position of q 2 due to q 1? (a) It increases. (b) It decreases. (c) It remains the same. (ii) When q 2 is changed from positive to negative, what happens to the potential energy of the two-charge system? Choose from the same possibilities. Answer: (i) (c) (ii) (a) 36

37 Example 25.3 As shown in Figure 25.11a, a charge q l = 2.00 mc is located at the origin and a charge q 2 = 6.00 mc is located at (0, 3.00) m. 37

38 Example 25.3 (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. Solution: V V p p k e q r q r C C ( N m /C ) V 4.00 m 5.00 m 38

39 Example 25.3 (B) Find the change in potential energy of the system of two charges plus a third charge q 3 = 3.00 mc as the latter charge moves from infinity to point P (Fig b). Solution: U f q V 3 p U U f U i q V 3 p 0 ( C)( V) J 39

40 Finding E From V Assume, to start, that the field has only an x component. E x dv dx (25.16) Similar statements would apply to the y and z components. Equipotential surfaces must always be perpendicular to the electric field lines passing through them. Section

41 E and V for an Infinite Sheet of Charge The equipotential lines are the dashed blue lines. The electric field lines are the brown lines. The equipotential lines are everywhere perpendicular to the field lines. Section

42 E and V for a Point Charge The equipotential lines are the dashed blue lines. The electric field lines are the brown lines. The electric field is radial. dv Er dr The equipotential lines are (25.17) everywhere perpendicular to the field lines. Section

43 E and V for a Dipole The equipotential lines are the dashed blue lines. The electric field lines are the brown lines. The equipotential lines are everywhere perpendicular to the field lines. Section

44 Electric Field from Potential, General In general, the electric potential is a function of all three dimensions. Given V(x, y, z) you can find E x, E y and E z as partial derivatives: V V V Ex Ey Ez x y z (15.18) Section

45 Quick Quiz 25.4 In a certain region of space, the electric potential is zero everywhere along the x axis. (i) From this information, you can conclude that the x component of the electric field in this region is (a) zero, (b) in the positive x direction, or (c) in the negative x direction. (ii) Suppose the electric potential is +2 V everywhere along the x axis. From the same choices, what can you conclude about the x component of the electric field now? Answer: (i) (a) (ii) (a) 45

46 Example 25.4 An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a as shown in Figure 25.l3. The dipole is along the x axis P and is centered at the origin. 46

47 Example 25.4 (A) Calculate the electric potential at point P on the y axis. Solution: q i q q Vp k e ke i r i a y a y (B) Calculate the electric potential at point R on the positive x axis. Solution: q q q 2k qa V k k r x a x a x a i R e e i i e

48 Example 25.4 (C) Calculate V and E x at a point on the x axis far from the dipole. Solution: 2keqa 2keqa VR lim( ) ( x a) xa x a x dv d 2keqa Ex 2 dx dx x d 1 4keqa 2 keqa ( x a) 2 3 dx x x 48

49 Electric Potential for a Continuous Charge Distribution Method 1: The charge distribution is known. Consider a small charge element dq Treat it as a point charge. The potential at some point due to this charge element is dq dv ke r (25.19) Section

50 V for a Continuous Charge Distribution, cont. To find the total potential, you need to integrate to include the contributions from all the elements. dq V k e r This value for V uses the reference of V = 0 when P is infinitely far away from the charge distributions. (25.20) 50

51 V for a Continuous Charge Distribution, final If the electric field is already known from other considerations, the potential can be calculated using the original approach: B V E d s A If the charge distribution has sufficient symmetry, first find the field from Gauss Law and then find the potential difference between any two points, Choose V = 0 at some convenient point Section

52 Problem-Solving Strategies Conceptualize Think about the individual charges or the charge distribution. Imagine the type of potential that would be created. Appeal to any symmetry in the arrangement of the charges. Categorize Group of individual charges or a continuous distribution? The answer will determine the procedure to follow in the analysis step. Section

53 Problem-Solving Strategies, 2 Analyze General Scalar quantity, so no components Use algebraic sum in the superposition principle Keep track of signs Only changes in electric potential are significant Define V = 0 at a point infinitely far away from the charges. If the charge distribution extends to infinity, then choose some other arbitrary point as a reference point. Section

54 Problem-Solving Strategies, 3 Analyze, cont If a group of individual charges is given Use the superposition principle and the algebraic sum. If a continuous charge distribution is given Use integrals for evaluating the total potential at some point. Each element of the charge distribution is treated as a point charge. Section

55 Problem-Solving Strategies, 4 If the electric field is given Start with the definition of the electric potential. Find the field from Gauss Law (or some other process) if needed. 55

56 Problem-Solving Strategies, final Finalize Check to see if the expression for the electric potential is consistent with your mental representation. Does the final expression reflect any symmetry? Image varying parameters to see if the mathematical results change in a reasonable way. Section

57 V for a Uniformly Charged Ring P is located on the perpendicular central axis of the uniformly charged ring. The symmetry of the situation means that all the charges on the ring are the same distance from point P. The ring has a radius a and a total charge Q. Section

58 V for a Uniformly Charged Ring The potential and the field are given by V k E x e dq kq e r a x kx e Q 3/ a x 2 2 (25.21) (25.22) 58

59 V for a Uniformly Charged Disk The ring has a radius R and surface charge density of s. P is along the perpendicular central axis of the disk. P is on the cental axis of the disk, symmetry indicates that all points in a given ring are the same distance from P. Section

60 V for a Uniformly Charged Disk The potential and the field are given by V 2kes R x x E x 2kes 1 R x x 2 2 1/ 2 (25.23) (25.24) Section

61 V for a Finite Line of Charge A rod of line l has a total charge of Q and a linear charge density of l. There is no symmetry to use, but the geometry is simple. V 2 2 Q a ke ln a (25.25) Section

62 Example 25.5 (A) Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q. Solution: V V k e dq r k a 2 e x 2 k e a dq dq 2 2 x k Q a e 2 x 2 (25.21) 62

63 Example 25.5 (B) Find an expression for the magnitude of the electric field at point P. Solution: dv d Ex keq ( a x ) dx dx E x 2 2 1/ / 2 = keq 2 ( a x ) (2 x) kx ( a x ) e 2 2 3/ 2 Q (25.21) 63

64 Example 25.6 A uniformly charged disk has radius R and surface charge density s. 64

65 Cont (A) Find the electric potential at a point P along the perpendicular central axis of the disk. Solution: dq s da s (2rdr) 2srdr kedq ke 2srdr dv r x r x R 1/ 2rdr R 2 2 V kes k ( r x ) es 0 r x (B) Find the x component of the electric field at a point P along the perpendicular central axis of the disk. Solution: V E 2k s[( R x e dv dx 2 x 2 ) 1/ 2 x] 2k e 2 2 1/ 2 s[1 ( R x x ) ] 2 2rdr (25.23) (25.24) 65

66 Example 25.7 A rod of length l, located along the x axis has a total charge Q and a uniform linear charge density l. Find the electric potential at a point P located on the y axis a distance a from the origin (Fig. 25.l7). 66

67 Example 25.7 Solution: dq ldx dv ke ke r 2 2 a x ldx V k 0 e 2 2 a x ldx Q V kel k a x x a x V k Q [ln( 2 a 2 ) ln a] e e ln( ) Q a ke ln a (25.25) 67

68 V Due to a Charged Conductor Consider two points on the surface of the charged conductor as shown. E is always perpendicular to the displacement. Therefore, ds Eds 0 Therefore, the potential difference between A and B is also zero. Section

69 V Due to a Charged Conductor, cont. V is constant everywhere on the surface of a charged conductor in equilibrium. V = 0 between any two points on the surface The surface of any charged conductor in electrostatic equilibrium is an equipotential surface. Every point on the surface of a charge conductor in equilibrium is at the same electric potential. Because the electric field is zero inside the conductor, we conclude that the electric potential is constant everywhere inside the conductor and equal to the value at the surface. Section

70 Irregularly Shaped Objects The charge density is high where the radius of curvature is small. And low where the radius of curvature is large The electric field is large near the convex points having small radii of curvature and reaches very high values at sharp points. Section

71 Example 25.8 Two spherical conductors of radii r l and r 2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire as shown in Figure The charges on the spheres in equilibrium are q l and q 2, respectively, and they are uniformly charged. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres. 71

72 Example 25.8 Solution: q V ke k r (1) 1 2 e 1 r q E k and E k E E e 2 2 e 2 r1 r2 (2) q q q r r r q2 r1 q E r r r E r r r q 72

73 Cavity in a Conductor Assume an irregularly shaped cavity is inside a conductor. Assume no charges are inside the cavity. The electric field inside the conductor must be zero. Section

74 Cavity in a Conductor, cont The electric field inside does not depend on the charge distribution on the outside surface of the conductor. For all paths between A and B, V B V d A E s B A 0 A cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity. Section

75 Corona Discharge If the electric field near a conductor is sufficiently strong, electrons resulting from random ionizations of air molecules near the conductor accelerate away from their parent molecules. These electrons can ionize additional molecules near the conductor. This creates more free electrons. The corona discharge is the glow that results from the recombination of these free electrons with the ionized air molecules. The ionization and corona discharge are most likely to occur near very sharp points. 75

76 Millikan Oil-Drop Experiment Robert Millikan measured e, the magnitude of the elementary charge on the electron. He also demonstrated the quantized nature of this charge. Oil droplets pass through a small hole and are illuminated by a light. Section

77 Millikan Oil-Drop Experiment Experimental Set-Up Section

78 If you can't see the image above, please install Shockwave Flash Player. If this active figure can t auto-play, please click right button, then click play : The Millikan Oil-Drop Experiment This animation shows a simplified version of the apparatus used by Robert Millikan to measure the elementary charge on an electron. 78

79 Oil-Drop Experiment, 2 With no electric field between the plates, the gravitational force and the drag force (viscous) act on the electron. The drop reaches terminal velocity with F D mg Section

80 Oil-Drop Experiment, 3 An electric field is set up between the plates. The upper plate has a higher potential. The drop reaches a new terminal velocity when the electrical force equals the sum of the drag force and gravity. Section

81 Oil-Drop Experiment, final The drop can be raised and allowed to fall numerous times by turning the electric field on and off. After many experiments, Millikan determined: q = ne where n = 0, 1, 2, 3, e = C This yields conclusive evidence that charge is quantized. Section

82 Van de Graaff Generator Charge is delivered continuously to a highpotential electrode by means of a moving belt of insulating material. The high-voltage electrode is a hollow metal dome mounted on an insulated column. Large potentials can be developed by repeated trips of the belt. Protons accelerated through such large potentials receive enough energy to initiate nuclear reactions. Section

83 Electrostatic Precipitator An application of electrical discharge in gases is the electrostatic precipitator. It removes particulate matter from combustible gases. The air to be cleaned enters the duct and moves near the wire. Section

84 Electrostatic Precipitator As the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles become charged. Most of the dirt particles are negatively charged and are drawn to the walls by the electric field. 84

85 Electrostatic Precipitator 85