# Chapter 22: The Electric Field. Read Chapter 22 Do Ch. 22 Questions 3, 5, 7, 9 Do Ch. 22 Problems 5, 19, 24

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1 Chapter : The Electric Field Read Chapter Do Ch. Questions 3, 5, 7, 9 Do Ch. Problems 5, 19, 4

2 The Electric Field Replaces action-at-a-distance Instead of Q 1 exerting a force directly on Q at a distance, we say: Q 1 creates a field exerts a force on Q. and then the field NOTE: Since force is a vector then the electric field must be a vector field! E F qe

3 Does the field really exist? It exists due to the finite speed of light (maximum speed of signal propagation) Electric charges in isolating shells Isolation removed (door open) Force is exerted after the source charge is isolated Interaction by the field rather than by charge

4 Field E is defined as the force that would be felt by a unit positive test charge E F / q 0 SI units for the electric field: newtons per coulomb.

5 Electric Field Lines We visualize the field by drawing field lines. These are defined by three properties: Lines point in the same direction as the field. Density of lines gives the magnitude of the field. Lines begin on + charges; end on charges.

6 Electric field created by a negatively charged metal sphere

7

8

9 Coulomb s Law for the Field Coulomb s law for the force on q due to Q: F k qq r rˆ qe Coulomb s law for the field E due to Q: E k Q r rˆ

10 Example 1 What is the electric field strength at a distance of 10 cm from a charge of μc? E kq r ( )( 10 (10 10 ) 6 ) N / C So a one-coulomb charge placed there would feel a force of 180,000 newtons.

11 Q.-1 A point charge Q is far from all other charges. At a distance of m from Q, the electric field is 0 N/C. What is the electric field at a distance of 4m from Q? 1. 5 N/C. 10 N/C 3. 0 N/C N/C N/C

12 Adding fields Principle of superposition Electric fields due to different sources combine vector addition to form the one true total field. Q 1 Q 3 Q E 3 E E 1 E E E E E 1 3

13 Example 1. Find the electric field on the x axis. y -Q Q x E y 0 d E x E 1 E kq x ( kq x d). Where will the field be zero? x d x x. 4d

14 The shell theorems for E In Chapter 13 we had the shell theorems for gravity In Chapter 1 (p. 567) the shell theorems for electrostatics were stated. In Chapter 3 (p. 618) they will be proven. But we can easily understand them now from our knowledge of electric field lines.

15 The shell theorems for gravity Given a uniform spherical shell of mass: (1) The field outside is the same as if all the mass were concentrated at the center. () The field inside the shell is zero. (These theorems for gravity are given in Chapter 14.) (Newton s headache!)

16 Prove true also for electric field Use our knowledge of electric field lines to draw the field due to a spherical shell of charge: There is no other way to draw lines which satisfy all 3 properties of electric field lines, and are also spherically symmetric. Notice that both shell theorems are obviously satisfied.

17 Fields at r are the same! r r Q spread over shell point Q at center PROOF: (1) Spherical symmetry () Fields far away must be equal

18 Useful result for spherical r symmetry Field outside a sphere of total charge Q is radially outward with magnitude E kq r

19 Q.- A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell s charge were Concentrated at the center.. Concentrated at the point closest to the particle. 3. Concentrated at the point opposite the particle. 4. Zero.

20 Electric Dipole The combination of two charges of equal but opposite sign is called a dipole. If the charges +q and q are separated by a distance d, then the dipole moment p is defined as a vector pointing from q to +q of magnitude p = qd. q p q

21 Electric Field Due to a Dipole

22 Torque on a Dipole in a Field d F ( sin ) qe d sin pe p E sin

23 Chapters, 3: The Electric Field NOTE! If you have urgent questions, please send me e- mail with Your name Section At

24 Review: Electric Fields Definition of electric field: F = qe SI unit: newton per coulomb (N/C) Coulomb s Law for a point charge Q: F = kqq/r OR E = kq/r Principle of superposition (vector addition): E net E E E 1 3

25 Electric Field Lines We visualize the field by drawing field lines. These are defined by three properties: Lines point in the same direction as the field. Density of lines gives the magnitude of the field. Lines begin on + charges; end on charges. From these properties it is easy to see that Field lines never cross

26 Coulomb s Law for the Field Coulomb s law for the force on q due to Q: F k qq r rˆ qe Coulomb s law for the field E due to Q: E k Q r rˆ

27 Example Three point charges are placed on the y axis as shown. Find the electric field at point P on the x axis. E 1 kq b E E 1 E E 3 E x E E 3 Q k a b ( E E3) x E cos Eb / a b E ( E b E E 3/ E3) x E1 4kQb /( a b ) kq / E 3 1 E3 E y 0

28 Q.-1 What is the SI unit for the electric field? 1. Newtons. Coulombs 3. Newtons per Coulomb 4. Newtons per meter 5. Coulombs per meter

29 Q.-1 What is the SI unit for the electric field? Field is force per unit charge: 1. Newtons. Coulombs 3. Newtons per Coulomb 4. Newtons per meter 5. Coulombs per meter F qe

30 Three charges (one + and two -) are placed on the x and y axes as shown. What is the approximate direction of the electric field at the origin? Will it be pointing toward point 1,, 3, or 4? E Q.- -q 1 3 +q -q 4

31 Q.- 1. (1). () 3. (3) 4. (4) -q 1 3 +q -q 4

32 Q.- Three charges (one + and two -) are placed on the x and y axes as shown. What is the approximate direction of the electric field at the origin? Will it be pointing toward point 1,, 3, or 4? Solution. Imagine a positive test charge placed at the origin. It will be atracted to the q charges and repelled by the +q charge. 3 +q -q E 1 -q 4

33 Linear charge distribution Linear charge density = charge per unit length If a rod of length.5 m has a uniform linear charge density λ = 3 C/m, then the total charge on the rod is (.5 m) (3 C/m) = 7.5 C. If a rod of length L carries a non-uniform linear charge density λ(x), then adding up all the charge produces an integral: b Q ( x) dq a b a dx

34 Example: Ch. Prob. 3 Find the electric field at point P. Let q / L. Then dq dr dr, de r k dq r dr k r dq E de k a a L al dr 1 k k r r a 1 a a 1 L

35 Suppose density is not uniform Still true that dq dr, dr de dq k r r dr k r dq al dr E de k ( r) r a

36 Field of a charged ring Uniform linear charge density so dq = λds and de = kdq/r By symmetry, E x =E y =0 and so E = E z = sum of all de z, and dez = cos θ de. E k cos dq k cos r Q r

37 Charged ring continued E k cos dq k cos r Q r But cos z / r so E kqz r 3 ( z kqz R ) 3/

38 Charged ring result E k cos dq Q k cos r r kqz 3 r kqz ) ( z R 3/ NOTE: This is a good example of a special result, which is the answer to an example problem, not a fundamental principle to be memorized. It is the process we are supposed to be learning, not the result!

39 NOTE This result for the field on the axis of a charged ring can be derived more easily in Chapter 4 using the idea of electric potential.

40 Electric Dipole The combination of two charges of equal but opposite sign is called a dipole. If the charges +q and q are separated by a distance d, then the dipole moment p is defined as a vector pointing from q to +q of magnitude p = qd. q p q

41 Electric Field Due to a Dipole

42 Torque on a Dipole in a Field d F ( sin ) qe d sin pe p E sin

43 Gauss s Law Gauss s Law is the first of the four Maxwell Equations which summarize all of electromagnetic theory. Gauss s Law gives us an alternative to Coulomb s Law for calculating the electric field due to a given distribution of charges.

44 Gauss s Law: The General Idea The net number of electric field lines which leave any volume of space is proportional to the net electric charge in that volume. E lines leaving V volume V E lines entering V

45 Flux The flux Φ of the field E through the surface S is defined as E da S The meaning of flux is just the number of field lines passing through the surface.

46 Best Statement of Gauss s Law The outward flux of the electric field through any closed surface equals the net enclosed charge divided by ε 0.

47 Gauss s Law: The Equation S E da Q / enc 0 S is any closed surface. Q enc is the net charge enclosed within S. da is an element of area on the surface of S. da is in the direction of the outward normal SI units

48 Chapters, 3: The Electric Field Previous Homework: Read Chapter Do Ch. Questions 3, 5, 7, 9 Do Ch. Problems 5, 19, 4, 34

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