Prof. Kwang-Chun Ho.

Transcription

1 Electromagnetic Field Theory [Chapter 4: Electrostatic Fields] Prof. Kwang-Chun Ho Tel: Fax:

2 Outline What is electrostatics? Coulomb s law and electric field intensity Electric fields due to continuous charge distributions Study electric flux density due to electric field Gauss s law and the first one of Maxwell s equations Procedure for applying Gauss s law to calculate the electric field Basic concept of electric potential known as voltage Relationship between electric field and potential What is an electric dipole and flux lines? Energy density in electrostatic fields Dept. Information and Communication Eng. 2

3 What is electrostatics? Electrostatic Phenomena: Leave glass rod which h has been rubbed with silk Bring another glass rod close to that one. Two rods separate further Bring a plastic rod close to that one. Two rods approach each other (Movement of charges) (Electric Forces) Dept. Information and Communication Eng. 3

4 What is electrostatics? How do we interpret the results? Explanations: There exist two kinds of charges Unlike charges attract; Like charges repel May exist electric fields due to their divergence around charges Definition of electrostatics: Study the effect of static (or time-invariant) electric fields, due to charges at rest Practical Applications: Oscilloscope, Ink-jet printers, almost all computer peripheral devices,... Dept. Information and Communication Eng. 4

5 Coulomb s Law Two fundamental laws governing electrostatic fields: Coulomb s law Experimental law formulated in 1785 by Charles Coulomb, who is French army engineer Gauss s law Law of electrical force: Force law that governed the interactions between two charge objects Using two small electrically charged spheres, he deduces the force is proportional p to the product of the two charges and followed the inverse square law of distance between them Q 1 Q 2 Run Animation! R Suspension head Fiber Dept. Information and Communication Eng. 5

6 Coulomb s Law QQ 1 2 F 2 R If we insert a proportionality constant, Coulombs law may be written as QQ 1 2 F k R 2 where the measured value k is o k 910 Nm /C m/f The permittivity (or dielectric constant) of medium is In air, it becomes 9 10 medium F/m 36 Q1 and Q2 r1 and r2 2 medium r o If point charges are located at points having position vector, then the force on due to Q is Q 1 Dept. Information and Communication Eng. 6

7 Coulomb s Law QQ 1 2 F 12 a 2 R 12 Q 4 0 R 1 R Q2 12 r2 r1 F21 F12 and 2 1 r 12 1 Here, R r r R and R12 ar 12 R Combining two equations, we have Some useful notes: The force on Q 1 due to Q 2 is F 21 F 12 F r2 QQ r r r 2 r 1 Q and must be point charges and at rest! Q 1 Q 2 Like charges repel; Unlike charges attract Dept. Information and Communication Eng. 7

8 Coulomb s Law Principle of superposition: If there are N charges Q1, Q2,, QN located, respectively, at points with position vectors rr 1, 2, r N, the force on a charge located at point Q is the vector r sum of the forces exerted by the charges Q1, Q2,, QN r 1 N r r 1 Q Qk r rk F 3 4 r r 2 0 k 1 r r Example 4.1: Consider point charges Q2(1mC) and Q (2mC) 3 located at (3,2,1) and (1,1,4), respectively Q 1 (10 nc) k F N F 2 F 1 r r r N Calculate the electric force on a charge located at (0,3,1) Dept. Information and Communication Eng. 8

9 Coulomb s Law Solution: Using the following equation F 1 F F 13 1 QQ QQ F 1 a 2 12 a R 12 R 13 we have F ax 7.149ay 0.637az mn Origin where F 12 r r 2 1 R12 r1 r2 0,3,13, 2, 1, R13 0,3,11, 1, 4, R12 3,1, 2 R13 1, 4, 3 a12, a13 R 914 R r 3 Dept. Information and Communication Eng. 9

10 Coulomb s Law Example 4.2: Calculate l the electric force on a charge +2Q Solution: y Q Q Q110 a 5 cm 7 C 2Q 1 2 2Q x Dept. Information and Communication Eng. 10 3

11 Visual EMT using MatLab Will run the MATLAB program, which computes the force between point charges Plots the position of each charge Displays the net force on each charge Run Coulomb.m! Dept. Information and Communication Eng. 11

12 Electric Field Intensity Physical Meaning: There exists forces exerted by the charge everywhere in space surrounding a charge How can we detect the forces? Answer: Place a positive test charge Q t at arbitrary point P, and measure the force acting on it Dept. Information and Communication Eng. 12

13 Electric Field Intensity Mathematical Meaning: The force per unit charge (i.e. the force acting on test charge of +1C) is called Electric Field Intensity, which yields F 1 QQ E t Q E a 2 R a 2 R Qt Qt 4 0R 4 0R Thus, a R Q Qrt r E a 2 R 3 4 0R 4 0 rt r r t For N point charges Q located r 1, Q2,, QN at rr 1, 2, r N, the electric field intensity at point r is N 1 Qk r rk E 3 4 k 1 r r 0 k Dept. Information and Communication Eng. 13

14 Electric Field Intensity Example 4.3: Find the total electric field intensity at P due to Q 1 and Q 2 Solution: r 2 r r2 E E1 E2 r r Q1r r1 Q2r r2 1 a r 4 r r 4 r r r a 2 E 1 E 2 E Dept. Information and Communication Eng. 14

15 Electric Field Intensity Example 4.4: Point charges 5 nc and -2 nc are located at (204) (2,0,4) and (-3,0,5), 305) respectively (a) Determine e e the force on a 1 nc point charge located at (1,-3,7) (b) Find the electric field at (1,-3,7) Solution: (a) E (b) Dept. Information and Communication Eng. 15

16 Electric Field Intensity How can we visualize the electric field intensity? Done by drawing continuous lines from the charge which are everywhere tangent to E These lines are called Electric Flux (or field) Lines The spacing of lines is inversely proportional to the strength of the field! Dept. Information and Communication Eng. 16

17 Electric Field Intensity Dept. Information and Communication Eng. 17

18 Electric Field Intensity Dept. Information and Communication Eng. 18

19 Electric Fields due to Charge Distributions Various charge distributions and charge elements: 2 3 where L C/m, S C/m, and v C/m represent the line charge, surface charge, and volume charge density, respectively. What is the Electric Fields due to these charge distributions? Dept. Information and Communication Eng. 19

20 Electric Fields due to Charge Distributions Consider a differential volume charge dq vdv: Electric field intensity at a point P due to that charge then is de( r ) dq 4 4 r r dv r r v r r 0 r r Integrating g over volume V,, we have 1 v r r Er ( ) 3 4 dv r r 0 V dq V dv v r r r r Origin Surface E( r) 1 r r ds E( r) S S r r 4 0 Line 1 r r L 3 4 L r r dl Dept. Information and Communication Eng. 20

21 Electric Fields due to Charge Distributions Example 4.5: Line Charge Consider a line charge with uniform charge density extending from A to B along z-axis Then, associated with element dl at 0,0, z is Cartesian: Cylindrical: Thus, where de de dl R L 40 R,, 0,0, a z zaz dl dz, R x y z z R xax yay z z az R zz tan sec x z L de y Dept. Information and Communication Eng. 21

22 Electric Fields due to Charge Distributions 2 Considering z z tan, dzsec d, and integrating from A to B, it becomes a zz a L z Using E 2 3/2 4 dz 2 sin 0 z z tan, cos sec cosa sina 1 L z sec d cos 0 sec 1 2 L cos sin 4 a az d 0 1 L sin2 sin1 cos2 cos1 4 E a a z 0 For infinite line charge, that is, 1 /2, 2 /2, it is L E a 2 0 Dept. Information and Communication Eng. 22

23 Electric Fields due to Charge Distributions Example 4.6: Surface Charge Consider a infinite it sheet of charge in the xy-plane l with Then, associated with element 1 is de z S R R y x Dept. Information and Communication Eng. 23

24 Electric Fields due to Charge Distributions a ha SdS R S z de d d /2 40 R 40 h Due to the symmetry y of charge distribution, the contribution along a cancels Thus, the total electric field intensity is / / S h d d Sh d E dez az az h h h 1 a 2 2 S S 2 2 1/2 z 0 0 h 0 Finally, we can note that if the charge is in the xy-plane, E has only a component normal to the sheet a z Dept. Information and Communication Eng. 24

25 Electric Fields due to Charge Distributions Practical Application Examples: [Configuration of A Real Capacitor] z E E E E E E E E E E E E 0 E E E 0 S S 2 az az Dept. Information and Communication Eng. 25

26 Electric Fields due to Charge Distributions [Configuration of Cathode Ray Oscilloscope] x V x cathode anode z y deflection plates V y screen V x V y V A controls the deflection of electron along x-axis controls the deflection of electron along y-axis Dept. Information and Communication Eng. 26

27 Electric Fields due to Charge Distributions [TV tube with electron-deflecting charged plates (orange)] Dept. Information and Communication Eng. 27

28 Electric Fields due to Charge Distributions [Configuration of Color Ink-Jet Printer] V 0 Nozzle vibrating at ultrasonic frequency sprays ink in the form of droplets These droplets acquire charge proportional to the character to be printed while passing through a set of charged plates Vertical displacement of an ink droplet is proportional to its charge Blank space between characters is achieved by having no charge, then the ink droplets are collected by gutter ( at old version) Dept. Information and Communication Eng. 28

29 Electric Fields due to Charge Distributions [Configuration of Microstrip Lines] r Electric fields Dept. Information and Communication Eng. 29

30 Electric Fields due to Charge Distributions [Microwave Oven]

31 Electric Fields due to Charge Distributions Example 4.7: A circular disk of radius a is uniformly 2 charged S [ C/ m ]. If the disk lies on the z=0 plane, (a) Show that at point (0,0,h) S h E 1 a z 0 (b) From this, derive the electric field due to an infinite sheet of charge on the z=0 0 plane. (c) If a h, show that E is similar to the field due to a point charge h a E a S r ds Dept. Information and Communication Eng. 31

32 Electric Fields due to Charge Distributions Solution (a) Consider an element of area ds of the disk The contribution due to ds dd is SdS r SdS a haz de /2 0 r 4 h Since the sum of the contribution along gives zero, we have E z h 4 h 2 a 2 a S hdd S d 2 2 3/ / h a h 1 S S h 1 E h 20 a h 0 E a z z Dept. Information and Communication Eng. 32

33 Electric Fields due to Charge Distributions (b) As S a, E az 2 0 (c) For very large h, using a binomial series we have 2 h 1 a a h a h 1 h 1 a 1 a h 2 h 2 2 1/2 Then, the electric field becomes similar to a field due to a point charge 2 2 a a Q E a a a S S 2 z 2 z 2 z 0h 0h 0h Dept. Information and Communication Eng. 33

34 Electric Fields due to Charge Distributions Example 4.8: A square plate described by z 0 carries a surface charge density (a) Find the total charge on the plate 2 x2, 2 y2, 2 12 y [ mc/ m ] (b) Find the electric field intensity at P(0,0,10) Solution: 2 2 (a) Q ds 12 y dxdyd S (b) Using E S (4) 2 ydy 192 mc 0 ds ( S S ) a ds r r r r 4 r r 0 0 S r r r r ds Dept. Information and Communication Eng. 34

35 Electric Fields due to Charge Distributions, in which r r (0,0,10) ( x, y,0) ( x, y,10) we have E (12 10 ) y ( x, y,10) dxdy 3/ x y (1 / 2) d( y ) ( ) a 2 z dx 2 2 3/2 2 0 x y ( ) az dx x 104 x x x 104 ( ) a z ln a [ 2 z MV x x By the symmetry, 2 and ydy 1/2d y f( x) 2 f x a ( ) dx 2 ln ( ) ( ) f x f x a / m] Dept. Information and Communication Eng. 35

36 Electric Fields due to Charge Distributions Example 4.9: Planes x 2 and y 3, respectively, carry charges 10 nc/m 2 and 15 nc/m If the line x 0, z 2 carries charge 10 nc/m, calculate E at P(1,1,-1) due to the three charge distributions Solution: The contributions to E at point P(1,1,-1) 1 1) due to the infinite sheets are 2 E 3 E 1 E 2 Dept. Information and Communication Eng. 36

37 Electric Fields due to Charge Distributions Since, the unit vector gets as Then, the electric field is R a The total field can be obtained by Dept. Information and Communication Eng. 37

38 Electric Fields due to Charge Distributions Volume Charge: Consider the volume charge distribution ib ti with v Then, it is not so easy to calculate the total electric field intensity, and the result can be obtained through tedious mathematical procedurere If so, what is an easier way to get the result? The answer is Gauss s law, which will be discussed later in details Dept. Information and Communication Eng. 38

39 Electric Fields due to Charge Distributions What is the electric flux line? When a test charge is at one point in electric field, it moves along a certain path by force acting on the test charge This path is called a line of force or electric flux line Electric flux lines between charges Dept. Information and Communication Eng. 39

40 Electric Fields due to Charge Distributions Example 4.10: The figure shows the electric field lines for a system of two point charges What are the relative magnitudes of the charges? What are the signs of the charges? In what regions of space is the electric field strong? or weak? Dept. Information and Communication Eng. 40

41 Electric Fields due to Charge Distributions Solution: There are 32 lines coming from the charge on the bottom, while there are 8 converging on that on the top. Thus, the one on the bottom is 4 times larger than the one on the top The one on the bottom is positive; field lines leave it. The one on the top is negative; field lines end on it The field is strong near both charges. It is strongest on a line connecting the charges. Few field lines are drawn there, but this is for clarity. The field is weakest to the upward of the top charge Dept. Information and Communication Eng. 41

42 Electric Flux Density Definition: Let s assume that a point charge is enclosed in an imaginary sphere, which is called Gaussian surface Electric flux lines pass perpendicularly and uniformly through the surface of sphere Then, the flux lines per unit area is called the electric flux density, which is defined as 2 D 0 E [C/m ] + D Dept. Information and Communication Eng. 42

43 Electric Flux Density Thus, when the orientations of the surface defined and the electric flux lines are different, the total electric flux through the surface S can be evaluate by D ds S ds D D ds ds ds Dept. Information and Communication Eng. 43

44 Gauss s s Law Definition: The total electric flux through any closed surface (Gaussian surface) is equal to the total charge enclosed by that surface. That is, Q D ds dv S V v ( 0) ( 0) Applying divergence theorem to the middle term in equation above D ds Ddv ( 0) ( 0) No enclosed S V charge! Gauss' law is a form of one of Maxwell's equations, the four fundamental equations for electricity and magnetism. Dept. Information and Communication Eng. 44

45 Applications of Gauss s s Law we have D v Gauss' law is a powerful tool for the calculation l of electric fields when they originate from charge distributions of sufficient symmetry y to apply ppy it Part of the power of Gauss' law in evaluating electric fields is that it applies to any surface It is often convenient to construct an imaginary surface called a Gaussian surface to take advantage of the symmetry of the physical situation Dept. Information and Communication Eng. 45

46 Applications of Gauss s s Law Here, Dashed-Lines are Gaussian Surfaces Dept. Information and Communication Eng. 46

47 Applications of Gauss s s Law Example 4.11: Point Charge Suppose a point charge is located at the origin Determine D at a point P Solution: Since D D is everywhere rar normal to the Gaussian surface applying Gauss s law gives 2 Q D ds D ds D 4 r P S r S r Thus, the electric flux density is Q D a 2 r 4 r Q D Dept. Information and Communication Eng. 47

48 Applications of Gauss s s Law Example 4.12: Infinite Line Charge Arranging Suppose the infinite line of uniform charge lies along z-axis Determine D at a point P Solution: Choose a cylindrical surface containing P to satisfy symmetry condition D D is constant on and normal to a the surface Thus, the electric flux density is l Q D ds L D S L a 2 r S D ds D 2 rll l L Line charge L C/m Dept. Information and Communication Eng. 48 P D

49 Applications of Gauss s s Law Example 4.13: Infinite Sheet of Charge Suppose the infinite sheet of uniform charge lies on z=0 0 S plane. Determine at a point P D Surface charge C/m 2 S D P D ds ds D D D x y z Dept. Information and Communication Eng. 49

50 Applications of Gauss s s Law Solution: Choose a cylindrical box, cutting symmetrically by the sheet of charge. Then, we have two surfaces parallel to the sheet D D a is normal to the surface z z S A Q D ds Dz ds ds Dz A A S top botton Note that D ds evaluated on the sides of the box is zero Thus, the electric flux density is D S az, z>0 2 S az,z<0 2 Dept. Information and Communication Eng. 50

51 Applications of Gauss s s Law Example 4.14: S Suppose two infinite sheets of uniform charge,as shown in figure. Determine D everywhere in space Solution: S S S S S S D D D D D D ( Between plates ) S Dept. Information and Communication Eng. 51

52 Applications of Gauss s s Law Example 4.15: Uniformly Charged Sphere Consider a sphere of radius R with a uniform charge Determine D everywhere Solution: Construct Gaussian surfaces for r R and r R, respectively For r R, the total charge enclosed by the surface of radius r is 4 Qenc vdv v r 3 And, the electric flux is D ds D ds D 4 r r S 3 r 2 D a r a n ds 3 v [C/m ] Gaussian surface r Dept. Information and Communication Eng. 52

53 Applications of Gauss s s Law Thus, using Gauss s law, we have Qenc, v r Dr 4r 3 r for 0 D D var r R 3 3 For r R, since total charge enclosed by the surface is 4 3 Qenc vdv v R, 3 from Gauss s s lawtheelectric electric flux density becomes 4 Qenc, v R Dr 4 r 3 3 R D for 2 var r a 3r 3 2 r v R 3r 3 v 2 r Dept. Information and Communication Eng. 53

54 Applications of Gauss s s Law Dept. Information and Communication Eng. 54

55 Electric Potential Similarity between gravitational and electrical potential energy : Work done by a L gravitational force is W mg L mgl L L G Then, the gravitational potential energy decreases, and is equal to the negative of work done as follows: UG mg L mgl Similarly, as a positive charge moves in the direction of an electric a L Q Q Dept. Information and Communication Eng. 55

56 Electric Potential field, it experiences an electric force and the work done by the electric field becomes W FL QEL QEL E Then, the positive charge loses the electric potential energy and is equal to the negative of the work UE QE L QEL (Potential energy is negative!) because it moves from a point of higher potential to a point of lower potential The electric potential difference between two points is defined as the change in potential energy of a charge Q divided by the charge U V E E L Q Dept. Information and Communication Eng. 56

57 Electric Potential Suppose a charge Q is moved from pt A to pt B through a region of space described by E electric field E The potential difference between points A and B, VB VA, is defined as the change in potential energy (final minus initial value) of a charge, Q, moved from A to B, A B divided by the charge U V AB V B V A Q E L E Dept. Information and Communication Eng. 57

58 Electric Potential where the potentials V at pt A and pt B are defined as the A, VB potential energy per unit charge: U If V AB 10 [V] 0, 10 J of work is required to move a 1 C of charge between two points that are at potential difference of 10 V Now, let s consider a charge moving from pt A to pt B along arbitrary path in electric field E Then, the potential energy in displacing the charge by is du Fdl QEdl E dl A V AB, E, A, B F Dept. Information and Communication Eng. 58 Q Q F dl B Q

59 Electric Potential And, the total potential energy in moving from pt A to pt B is B U Q E dl E A Thus, the potential difference between A and B is denoted by B U E VAB VB VA E dl Q A Is the potential difference independent of path taken? Consider the case of constant t vector field E Then, the potential difference at the direction along path A-B is B V V V E dl Eh AB B A A B h A dl r C E Dept. Information and Communication Eng. 59

60 Electric Potential For long way round along path A-C-B, it becomes C B C V V V E dl E dl E dl sin 0 Er sin Eh AB B A A C A Integral does not depend d upon the path chosen to move from A to B so that the integral is the same for BOTH paths Example 4.16: Assume that the electric field due to a point charge Q located at the origin, we have Q E a 2 r 4 0 r A r A r Q E dr dl r B B Dept. Information and Communication Eng. 60

61 Electric Potential Then, determine the potential difference between A and B Solution: r B Q Q 1 1 VAB VB VA a 2 r drar 40r 4 r 0 rb ra A Here, if VA 0 as A, the potential at any point due to a point charge Q located at the origin is r r r B Choose infinity as reference because the potential at infinity is supposed to be zero (Ground)! V Q 4 r 0 r E dl Dept. Information and Communication Eng. 61

62 Electric Potential For N point charges Q1, Q2,, QN located at r1, r2, r, the N potential ti at r is Q k N 1 V( r) (point charges) 4 r r 0 k 1 For continuous charge distributions, the potential at becomes 1 L V( r) (line charge) r dl 4 0 r r L 1 S r V( r) (surface charge) 4 ds 0 r r S 1 v r V( r) (volume charge) 4 dv r r 0 V k r Dept. Information and Communication Eng. 62

63 Electric Potential Example 4.17: Charge L is uniformly distributed within a rod. Find the electric potential on the perpendicular bisector of the charged rod Solution: Since dq L dy, the potential due to dq at a point P becomes 1 1 L dv dq dy r 40 x y Thus, the potential due to the rod is a 2 2 ln L dy L a x a V a x y 0 a x a Dept. Information and Communication Eng. 63 L

64 Electric Potential Example 4.18: v Charge is uniformly distributed ib t d within a spherical shell of radius a. Then, the electric field is 3 r for 0, for 2 3 a E 3 var r a E var r a r 0 0 Find the potential everywhere For r a, r 3 3 a v a v V r a dra v out 2 r r 3 0r 3 0r Vout () r For 0 r a, V () in r r E r v Vin r V out r a ar dra r a 3 a 0 2 v 2 r a Dept. Information and Communication Eng. 64

65 Electric Potential Practical Application: battery "D-cell" or "AA-cell" has a rating of 1.5 volts, which means that every charge moving from the negative side of the cell to the positive side will do 1.5 Joules worth of work The difference between the D-cell and the AA-cell is that the D-cell has more charges, so it will last longer As shown in figure, every negative charge that passes through the light bulb does 1.5 joules worth of work which makes it give off light Bulb D AA Dept. Information and Communication Eng. 65

66 Electric Potential Curiosity Question: How can a bird (like this bluebird) stand on a high voltage line without getting zapped? Answer: Because there is no difference in Voltage across his feet! When does the current flow? If there are voltage or potential difference, then the current starts to flow from high voltage to low voltage Dept. Information and Communication Eng. 66

67 Electric Potential Dept. Information and Communication Eng. 67

68 Electric Potential But when a small bird sits on the power line, both feet are on the same voltage line! (no potential difference and no current flow!) If one leg B of a chicken is on the ground and the other one A is on the power line, then there are potential difference between these two legs Therefore, there is a flow of charge and eventually the chicken will be barbecued, as shown in the figures Dept. Information and Communication Eng. 68

69 Relationship between E and V Since the potential difference is independent of the path taken, we have V V That is, Applying Stokes s theorem to the equation above, it becomes BA AB V V E dl 0 BA AB E dl E ds 0 E 0 It is called the second Maxwell s equation The vector field is said to be conservative Thus, the electrostatic field is a conservative field Dept. Information and Communication Eng. 69 E

70 Relationship between E and V Now, let s define the relationship between E and V, satisfying i the conservative property Consider a point charge Q in an electrostatic field Then, the potential ti difference between two points is dv du E / Q E dl Ecos dl E dl Rearranging g for, we have E L dv dl E L (V/m) If dl is x, y and z-axis, it becomes E dv, dv, dv x Ey E z d d d L dx dy dz V B VA dv Q dl V A E V B L Dept. Information and Communication Eng. 70

71 Relationship between E and V Then, the vector E at any point is given by Finally, we have E E a E a E a x x y y z z V V V a a a x y z x y z (V/m) E ax ay azv V x y z (V/m) Note that: Since the curl of gradient of scalar function is always zero V 0, the electrostatic field must be a conservative field Dept. Information and Communication Eng. 71

72 Electric Dipole and Flux Lines What is an electric dipole? Two point charges of equal magnitude but opposite sign are separated by a small distance Now, use the potential to calculate the E field of a dipole Remember how messy the direct calculation was? The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms given as Dept. Information and Communication Eng. 72

73 Electric Dipole and Flux Lines Q 1 1 Q r r V() r r r 0 rr Rewriting this for special case r d it becomes Q dcos V() r 2 4 r because r r dcos, r r r Calculating E in spherical coordinates, we have V 1 V E V ar a r r Qd 3 2cosar sina 4 r 0 r 0 2 Dept. Information and Communication Eng. 73

74 Electric Dipole and Flux Lines Electric flux line V () r 0 V() r 0 Equipotential surface V() r 0 Dept. Information and Communication Eng. 74

75 Electric Dipole Antenna Dept. Information and Communication Eng. 75

76 Electric Dipole Antenna Dept. Information and Communication Eng. 76

77 Electric Dipole and Flux Lines Equipotentials of a charged sphere: The electric field of the charged sphere has spherical symmetry The potential depends only on the distance from the center of B the sphere A An equipotential surface is a surface on which all points are at tthe same potential ti The electric field at every point on an equipotential surface is perpendicular to the surface Electric flux line Equipotential surface Dept. Information and Communication Eng. 77

78 Electric Dipole and Flux Lines Why?? Along the surface, there is NO change in V (it s an equipotential!) B V V E dl 0, B A E dl 0 (Orthogonal relation between two factors) Thus, no work is required to move a charge at a constant speed on an equipotential surface A Dept. Information and Communication Eng. 78

79 Electric Dipole and Flux Lines Equipotential lines on the surface of the human body reflect the electric dipole nature of the heart Dept. Information and Communication Eng. 79

80 Visual EMT using MatLab Draw the potential ti due to two opposite point charges in three-dimensional space, and confirm the equipotential lines from the resulting shape Run potential.m! Dept. Information and Communication Eng. 80

81 Energy Density in Electrostatic Fields In electrostatic fields, we have, in a mathematical sense, two types of field Vector fields (electric fields) and scalar fields (potential) However, when we inquire i more deeply into the workings of the universe, we find that the fundamental nature of reality is the presence of energy The electric field is primarily an energy density field The electric field surrounding a charge stores energy in space Electric potential does the pulling apart of the positive and negative layers of space and the pulled-apart-ness stores and transfers energy Dept. Information and Communication Eng. 81

82 Energy Density in Electrostatic Fields Consider a system with just two charges Assume V21 is the electric potential due to charge Q1 at a point P When a second charge Q 2 moves from infinity to P without acceleration, the potential energy is U QV Q 1 Since the potential at point P due to is V 1 Q 1 QQ, U r12 40 r12 Because the second charge moves from lower potential to higher potential! Now, add a third charge Q 3 to the system (That is, a third charge moves from infinity to a point without acceleration) Q 1 r 12 P Q 2 Dept. Information and Communication Eng. 82

83 Energy Density in Electrostatic Fields Then, since the potential at the point due to and Q is Q1 2 1 Q1 Q 2 V31 V32, P 40 r13 r23 Q 2 the potential ti energy is defined d as r r QQ 1 3 QQ 2 3 U31 U r13 r23 Q 1 The total potential energy required r 13 to assemble the charges is 1 QQ 1 2 QQ 1 3 QQ 2 3 UE U21 U31 U32 40 r12 r13 r23 1 Rewriting the potential energy as UE UE UE, we have 2 Q 3 Dept. Information and Communication Eng. 83

84 Energy Density in Electrostatic Fields U E 1 1 QQ 1 2 QQ 1 3 QQ QQ 1 2 QQ 1 3 QQ r 12 r 13 r r 12 r 13 r23 1 Q2 Q 3 Q1 Q 3 Q1 Q 2 Q1 Q2 Q3 2 40r12 40r13 40r12 40r23 40r13 40r23 1 QV QV QV where V 1, V 2 and V 3 are the total potentials at Q 1, Q 2 and Q 3 In general, if there are n point charges, it becomes U 1 n E QV k k 2 k 1 Dept. Information and Communication Eng. 84

85 Energy Density in Electrostatic Fields If, instead of point charge, it is a continuous charge distribution, ib ti the summation becomes integration ti U E vvdv D Vdv D Vdv D Edv vol vol vol vol DV VD D V V, D, VD0 2 3 r r r r large Developing further, we can define electrostatic energy density as 2 du 1 E 1 2 D ue D E 0E dv Dept. Information and Communication Eng. 85

86 Visual EMT using MatLab Draw the energy density due to one point charge with negative value in three-dimensional space Run energydensity.m! Dept. Information and Communication Eng. 86

87 Homework Assignments Problem 4.1: Let us illustrate the use of the vector form of 4 Coulomb s law by locating a charge of Q [C] at 4 M 1, 2, 3 and a charge of Q 2 10 [C] at N 2,0,5 in a vacuum. Find the force exerted on by Problem 4.2: Find E at P 1,1,1 caused by four identical charges located at and, as shown in Fig. P 4 3[nC] P, P, P, Q2 Q1 87

88 Homework Assignments Problem 4.3: Find the total charge contained in a 2cm length of the electron beam shown in Fig. Problem 4.4: Given that 3 12nC/m,1 2 v 0,otherwise D Determine everywhere Problem 4.5: Determine the electric field 2 due to the potential V z 1sin 88

89 Homework Assignments Problem 4.6: 9 nc/m Two dipoles with dipole moment 5a z nc/m and a z are located at points 0,0, 2 and 0,0,3, respectively. Find the potential at the origin. Problem 4.7: A spherical charge distribution is given by Find V everywhere. r o, r a v a 0, r a 89