Lecture 2 [Chapter 21] Tuesday, Jan 17th

Transcription

1 Lecture 2 [Chapter 21] Tuesday, Jan 17th

2 Administrative Items Assignments this week: read Ch 21 and Ch 22 in the textbook complete Pre-Lecture Ch22 HW assignment complete Ch 21 HW assignment [Pre-Lecture Ch 22 HW will be available today and due Thursday by 8am.] [Ch 21 HW should be available today and due Friday.] No Discussion Sections this week. First quiz will be next week in Discussion (Jan 25th).

3 Where were we? Coulomb s Law: F = 1 4πε0 q1q2 r 2 ^ r12

4 Electric forces are much stronger than gravity Compare gravitational and electric forces between an electron and a proton for a given separation r: F g = G m p m e r 2 F e F g = ke2 Gm p m e = F e = k q e q p r 2 = ke2 r 2 ( ) ( ) ( ) 2 ( )( ) F e F g The electric force is almost 40 orders of magnitude stronger than the gravitational force. So why don t we usually notice electric forces?

5 Electric forces are much stronger than gravity Compare gravitational and electric forces between an electron and a proton for a given separation r: F g = G m pm e r 2 F e F g = ke2 Gm p m e = F e = k q eq p r 2 = ke2 r 2 ( ) ( ) ( ) 2 ( )( ) F e F g The electric force is almost 40 order of magnitude stronger than the gravitational force. So why don t we usually notice electric forces? Gravity is always attractive. Ordinary matter contains almost exactly equal amount of positive and negative charges, and thus electric forces tend to cancel out.

6 Principle of Superposition: 3 or more charges Suppose we want to find the net force exerted by charge #1 AND #2 on charge #3. Charge #2 Charge #1 Charge #3 The principle of superposition says that we can just work out the force q 1 exerts on q 3 and add it vectorially to the force that q 2 exerts on q 3.

7 Principle of Superposition Example: What s total force on q3? Two pt. charges are located on the x-axis: q1 = 1.0 nc at x = +2.0 cm and q2 = -3.0 nc at x = +4.0 cm. What is the total force exerted by q1 and q2 on a charge q3 = 5.0 nc at x=0?

8 Principle of Superposition Example: What s total force on q3? Two pt. charges are located on the x-axis: q1 = 1.0 nc at x = +2.0 cm and q2 = -3.0 nc at x = +4.0 cm. What is the total force exerted by q1 and q2 on a charge q3 = 5.0 nc at x=0?

9 Principle of Superposition Example: What s total force on q3? Two pt. charges are located on the x-axis: q1 = 1.0 nc at x = +2.0 cm and q2 = -3.0 nc at x = +4.0 cm. What is the total force exerted by q1 and q2 on a charge q3 = 5.0 nc at x=0?

10 Principle of Superposition Example: What s total force on q3? Two pt. charges are located on the x-axis: q1 = 1.0 nc at x = +2.0 cm and q2 = -3.0 nc at x = +4.0 cm. What is the total force exerted by q1 and q2 on a charge q3 = 5.0 nc at x=0? F 1 3 = 112µN (in the negative x direction) F 1 2 = 84µN (in the positive x direction) F2 3 k e = 9 x 10 9 N. m 2 /C 2 Vector sum: (i.e. negative x direction)

11 Principle of Superposition: 3 or more charges Charge #1 Charge #2 Charge #3 Remember: add electrical forces using vector addition Split in x and y components for each force. Sum the x(y)-components together. Get the magnitude using Pythagorean Thm. and the direction angle using tanθ = y/x (where θ is measured counter-clockwise from the +x-axis).

12 Quiz: Superposition principle I Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charges #1 and #2 are positive (+q) and charge #3 is negative ( q). The net electric force that charges #2 and #3 exert on charge #1 is in A.the +x-direction B.the -x-direction C.the +y-direction D.the -y-direction E.None of the above y x Charge #1 +q Charge #2 +q q Charge #3

13 Quiz: Superposition principle I Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charges #1 and #2 are positive (+q) and charge #3 is negative ( q). The net electric force that charges #2 and #3 exert on charge #1 is in A.the +x-direction B.the -x-direction C.the +y-direction D.the -y-direction E.None of the above y x Charge #1 +q Charge #2 +q q Charge #3

14 Quiz: Superposition principle II Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charge #1 is positive (+q) and charges #2 and #3 are negative ( q). The net electric force that charges #2 and #3 exert on charge #1 is in A.the +x-direction B.the -x-direction C.the +y-direction D.the -y-direction E.None of the above y x Charge #1 +q Charge #2 q q Charge #3

15 Quiz: Superposition principle II Three point charges lie at the vertices of an equilateral triangle as shown. All three charges have the same magnitude, but charge #1 is positive (+q) and charges #2 and #3 are negative ( q). The net electric force that charges #2 and #3 exert on charge #1 is in A.the +x-direction B.the -x-direction C.the +y-direction D.the -y-direction E.None of the above y x Charge #1 +q Charge #2 q q Charge #3

16 Quiz: Electric Force I Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charge ball Q 0 on the line between the two charges such that the net force on Q 0 will be zero? +Q A B C D E +4Q R 2R 3R

17 Quiz: Electric Force I Two balls with charges +Q and +4Q are separated by 3R. Where should you place another charge ball Q 0 on the line between the two charges such that the net force on Q 0 will be zero? +Q A B C D E +4Q R 2R 3R F on Q 0 due to +Q is: F on Q 0 due to +4Q is: F = k Q 0 Q R 2 F = k Q 4Q 0 ( 2R) 2 Since +4Q is 4 times bigger than +Q, Q0 needs to be farther from +4Q. If Q 0 is 2x as far from +4Q to have net zero force - as the force depends on the distance squared in Coulomb s law.

18 Coulomb s law problem solving strategy Use Coulomb s law whenever you need to find the electric force between two (or more) charges. 1.Make a diagram of the location of all charges, their signs, and magnitudes 2.Setup the (x,y) coordinate system 3.Identify the particle(s) (eg Q) that you want to find the electric force on 4.For each particle that exerts a force on Q, find the magnitude using Coulomb s law 5.Draw the FBD of all forces acting on Q, showing the magnitude and direction of the force vectors. Direction will depend on whether the force is attractive or repulsive (i.e. the two charges involved) 6.Use the principle of superposition to find the resultant force on Q by adding the vectors in the FBD. You may need to break the force vectors into (x,y) components to do this. Tips: Use symmetries whenever possible to simplify the problem. Watch out for the units! r: meter q: Coulomb µc=10-6 C nc=10-9 C Check that your numerical answer makes sense!!!

19 Example: Electric force in a plane Two equal and positive charges q 1 =q 2 =2.0µC are located at x=0, y=0.3m and x=0, y=-0.3m, respectively. What are the magnitude and direction of the total electric force that q 1 and q 2 exert on a third charge Q=4.0µC at x=0.4m and y=0? Charge #1 y q 1 Q x q 2 Charge #2

20 Example: Electric force in a plane Two equal and positive charges q 1 =q 2 =2.0µC are located at x=0, y=0.3m and x=0, y=-0.3m, respectively. What are the magnitude and direction of the total electric force that q 1 and q 2 exert on a third charge Q=4.0µC at x=0.4m and y=0? y Charge #1 q 1 q 2 Charge #2 Q x First, look for symmetries: q 1 and q 2 have the same magnitude, the same charge, and are at the same distance from Q magnitude of the forces F 1 and F 2 will be equal. q 1 and q 2 are located symmetrically along the y-axis. y-components of F 1 and F 2 will cancelout. Force will be in +x direction.

21 Example: Electric force in a plane Two equal and positive charges q 1 =q 2 =2.0µC are located at x=0, y=0.3m and x=0, y=-0.3m, respectively. What are the magnitude and direction of the total electric force that q 1 and q 2 exert on a third charge Q=4.0µC at x=0.4m and y=0?

22 Example: Electric force in a plane Two equal and positive charges q 1 =q 2 =2.0µC are located at x=0, y=0.3m and x=0, y=-0.3m, respectively. What are the magnitude and direction of the total electric force that q 1 and q 2 exert on a third charge Q=4.0µC at x=0.4m and y=0?

23 Example: Electric force in a plane Two equal and positive charges q 1 =q 2 =2.0µC are located at x=0, y=0.3m and x=0, y=-0.3m, respectively. What are the magnitude and direction of the total electric force that q 1 and q 2 exert on a third charge Q=4.0µC at x=0.4m and y=0? x-component of F 1 : F 1 cosα x-component of F 2 : F 2 cosα Since F 1 =F 2, the total force on Q is: (y-components cancel)

24 Balloon Bending Water Trick (Balloon has just been rubbed on hair.)

25 Balloon Bending Water Trick (Balloon has just been rubbed on hair.)

26 How do balloons bend water? Electrons from hair give balloon extra negative (-) charge. The H2O molecule is a dipole : => has a positive (+) and negative (-) side. The (+) sides are attracted to the balloon. Molecules orient themselves w/ (+) side (where H s are) pointing towards balloon. While the (-) side is repelled, it s farther from the balloon, so the (+) tug wins.

27 Electric dipoles Nature is FULL of dipoles Many molecules are essentially dipoles A dipole is two equal magnitude, opposite sign point charges q, separated by a distance d. Dipole has no net charge. The electric field arises from the slight separation of the two opposite charges. Eg: Dipole moment of H 2 O is responsible for some of its important properties (i.e. almost a universal solvent) In water, Na + & Cl - (salt) can be pulled apart due to the H 2 0 dipole moment

28 Electric forces on neutral objects We ve seen that charged objects exert forces on each other. A charge object can also exert a force on an uncharged object. Induced charge effect: electric charges shift (polarization) leading to a stronger attractive force than repulsion force.

29 Electric Fields

30 Idea of Electric field q2 Force on q2 from q1: q1

31 Idea of Electric field q2 q3 Force on q2 from q1: Force on q3 from q1: q1

32 Idea of Electric field q2 q3 Force on q2 from q1: Force on q3 from q1: q1 Force on q4 from q1: q4

33 Idea of Electric field Force from q1 on any charge qi at position r: q1 qi

34 Idea of Electric field Force from q1 on any charge qi at position r: q1 qi Electric field: Depends only on the charge q1 and spatial location.

35 Idea of Electric field Force from q1 on any charge qi at position r: q1 qi!"!!" F = 0 q0 E

36 Idea of Electric field Force from any charge q on any other charge q0 at position r: q q!"!!" F = 0 q0 E Electric field from point charge q:

37 Electric Field vs. Electric Force Electric field: Exists in all points around a charged object When another charged object (charge q 0 ) enters this electric field, an electric force acts on it.!"!!" F = 0 q0 E

38 Electric Field vs. Electric Force Electric field: Exists in all points around a charged object When another charged object (charge q 0 ) enters this electric field, an electric force acts on it.!"! = q0 E!" F 0 A charge distribution creates an electric field, which exerts a force on any charge that is present in the field.

39 Quiz (show of hands, no clickers) What is the direction of the force felt by a negative charge in the presence of this electric field? E a) F b) F

40 Quiz What is the direction of the force felt by a negative charge in the presence of this electric field? E!"!!" F = 0 q0 E a) b) F F

41 Electric Field of a Point Charge

42 Quiz (show of hands, no clickers) Two point charges and a point P lie at the vertices of an equilateral triangle as shown. Both point charges have the same amount of charge (q), but are of opposite sign. There is nothing at point P. The net electric field that charges #1 and #2 produce at point P is in A.the +x-direction B.the -x-direction C.the +y-direction D.the -y-direction E.None of the above y P Charge #1 q x +q Charge #2

43 Quiz (show of hands, no clickers) Two point charges and a point P lie at the vertices of an equilateral triangle as shown. Both point charges have the same amount of charge (q), but are of opposite sign. There is nothing at point P. The net electric field that charges #1 and #2 produce at point P is in A.the +x-direction B.the -x-direction C.the +y-direction D.the -y-direction E.None of the above y P Charge #1 q x +q Charge #2

44 Example: Electric field of a point charge A point charge q=-8.0 nc is located at the origin. Find the electric field vector at (x=1.2m, y=-1.6m). -ve Need magnitude & direction r (1.2m, -1.6m)

45 Example: Electric field of a point charge A point charge q=-8.0 nc is located at the origin. Find the electric field vector at (x=1.2m, y=-1.6m). -ve Need magnitude & direction r (1.2 m) 2 + (-1.6 m) 2 (1.2m, -1.6m)

46 Example: Electric field of a point charge A point charge q=-8.0 nc is located at the origin. Find the electric field vector at (x=1.2m, y=-1.6m). -ve Need magnitude & direction E (1.2 m) 2 + (-1.6 m) 2 (1.2m, -1.6m)

47 Electric field: continuous charge distribution The system of closely spaced charges is equivalent to a total charge that is continuously distributed alone some line, over a surface, or throughout some volume. Procedure: 1. Divide the charge distribution into small elements, each of which contains dq 2. Calculate the electric field, de, due to one of these elements at point P 3. Evaluate the total field by summing the contributions of all the charge elements. de = 1 dq 4πε 0 r ˆr 2

48 Charge Densities Linear charge density: when a charge is distributed along a line λ Q / l with units C/m Surface charge density: when a charge is distributed evenly over a surface area σ Q / A with units C/m 2 Volume charge density: when a charge is distributed evenly throughout a volume ρ Q / V with units C/m 3 If the charge is uniformly distributed over a volume, surface, or line, the amount of charge, dq, is given by For the length element: dq = λ dl For the surface: dq = σ da For the volume: dq = ρ dv

49 Example: Ring of charge Suppose we have a total charge Q distributed uniformly around a thin ring of radius a. We want to find the electric field at point P, located at a distance x from the center of the ring, along the axis normal to the ring. dq ds = Q 2πa 1.Break charge distribution into small segments ds, with charge dq.

50 Example: Ring of charge Suppose we have a total charge Q distributed uniformly around a thin ring of radius a. We want to find the electric field at point P, located at a distance x from the center of the ring, along the axis normal to the ring. dq ds = Q 2πa 1.Break charge distribution into small segments ds, with charge dq. 2.Evaluate the electric field de due to dq at P. Because P is on the symmetry axis of the ring: We can see that de y for a segment at the top of the ring will cancel-out with de y for a segment at the bottom of the ring. Thus E y =0 and only E x contributes. Every point on the ring is at the same distance from point P: r = x 2 + a 2 de = 1 4πε 0 dq r 2 = 1 4πε 0 Q 2πa ds x 2 + a 2

51 Example: Ring of charge (cont.) de = 1 4πε 0 dq r 2 = 1 4πε 0 Q 2πa ds x 2 + a 2 The magnitude of de and the angle α are the same for every point on the ring, the x-component of de is the same for every point on the ring: de x = decosα de x = 1 4πε 0 Q 2πa ds x 2 + a 2 x x 2 + a 2

52 Example: Ring of charge (cont.) de = 1 4πε 0 dq r 2 = 1 4πε 0 Q 2πa ds x 2 + a 2 The magnitude of de and the angle α are the same for every point on the ring, the x- component of de is the same for every point on the ring: de x = decosα de x = 1 4πε 0 Q 2πa ds x 2 + a 2 x x 2 + a 2

53 Example: Ring of charge (cont.) de = 1 4πε 0 dq r 2 = 1 4πε 0 Q 2πa ds x 2 + a 2 The magnitude of de and the angle α are the same for every point on the ring, the x- component of de is the same for every point on the ring: de x = decosα de x = 1 4πε 0 Q 2πa ds x 2 + a 2 x x 2 + a 2 Now we can apply the principle of superposition and integrate over the whole ring: See book example 21.11: line of charge See book example 21.12: uniform charge disk E x = de x = E x = Q 4πε 0 1 4πε 0 Q 2πa x ( x 2 + a 2 ) 3/2 2πa x ds ( x 2 + a 2 ) 3/2 0

54 Example: Field of uniformly charged disk of radius R Think of many thin rings Field of each ring of radius a & charge q: σ Q / A with units C/m 2

55 Example: Field of uniformly charged disk of radius R q of each ring Field of each ring of radius a & charge q: σ Q / A with units C/m 2 at any radius r:

56 Example: Field of uniformly charged disk of radius R q of each ring (radius r) Field of each ring of radius a & charge q: σ Q / A with units C/m 2

57 Example: Field of uniformly charged disk of radius R σ Q / A with units C/m 2 Field of each ring of radius a & charge q: at any radius r:

58 Quiz Limit of an infinitely large plate (R > infinity)?

59 Quiz Limit of an infinitely large plate (R > infinity)? A. B. C.

60 Quiz Limit of an infinitely large plate (R > infinity)? A. B. C.

61 Example: Field of uniformly charged disk of radius R Limit of an infinitely large plate (R > infinity)

62 Example: Field of uniformly charged disk of radius R Limit of an infinitely large plate (R > infinity) If P is to the left of the plane, all the same except -ve sign:

63 Infinite plane sheet

64 2 infinite plane sheets

65 2 infinite plane sheets

66 2 infinite plane sheets

67 2 infinite plane sheets

68 2 infinite plane sheets The field between oppositely changed sheets is uniform as long as the sheets are much larger than their separation. Field outside is zero.

69 Example: Electron in a uniform field -e = C m e = kg E= N/C a) What is the acceleration of the electron? b) What speed & kinetic energy does the electron acquire while traveling 1cm? c) How much time does it take the electron to travel the distance

70 Example: Electron in a uniform field -e = C m e = kg E= N/C a) What is the acceleration of the electron?

71 Example: Electron in a uniform field -e = C m e = kg E= N/C c) How much time does it take the electron to travel the distance

72 Example: Electron in a uniform field 1 ev = the work done by electric field when a charge e is moved through a potential of 1 Volt 1 ev = Joule -e = C m e = kg E= N/C b) what is speed and KE of e after traveling 1cm?

73 Electric field lines: Visualization of E-Field

74 Electric field lines: Visualization of E-Field

75 Electric field lines: Visualization of E-Field An E-field line is an imaginary line or curve such as the direction of the E-field is tangent to the curve at every point. Number of field lines ending/starting on a charge is proportional to the magnitude of the charge. For a dipole (two equal and opposite charges), the field lines start on the positive charge and end on the negative charge. The electric field can only have one direction at any single point and thus never intersect. The density of the field lines near a particular point is an indication of the strength (magnitude) of the electric field there. Field lines close together where the field is strong Field lines are farther apart where the field is weak For a uniform E-field (eg 2 large parallel charged planes), the field lines are parallel and evenly space. The field as the same direction and the same magnitude everywhere.

76 Electric field lines: Visualization of E-Field