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Week of February 17 05Electric Field 1 Lecture 04 The Electric Field Chapter 22 - HRW.

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Presentation on theme: "Week of February 17 05Electric Field 1 Lecture 04 The Electric Field Chapter 22 - HRW."— Presentation transcript:

1 Week of February 17 05Electric Field 1 Lecture 04 The Electric Field Chapter 22 - HRW

2 Electric Field2 Week of February 17 05 Physics 2049 News WebAssign was due today Another one is posted for Friday You should be reading chapter 22; The Electric Field.  This is a very important concept.  It is a little “mathy” There will be a QUIZ on Friday.  Material from chapters 21-22.  Studying Works!

3 Electric Field3 Week of February 17 05 This is WAR You are fighting the enemy on the planet Mongo. The evil emperor Ming’s forces are behind a strange green haze. You aim your blaster and fire … but …… Ming the merciless this guy is MEAN !

4 Electric Field4 Week of February 17 05 Nothing Happens! The Green thing is a Force Field! The Force may not be with you ….

5 Electric Field5 Week of February 17 05 Side View The FORCE FIELD Force Position o |Force| Big!

6 Electric Field6 Week of February 17 05 Properties of a FORCE FIELD It is a property of the position in space. There is a cause but that cause may not be known. The force on an object is usually proportional to some property of an object which is placed into the field.

7 Electric Field7 Week of February 17 05 EXAMPLE: The Gravitational Field That We Live In. m M mgmg MgMg

8 Electric Field8 Week of February 17 05 The gravitational field: g The gravitational field strength is defined as the Force per unit mass that the field creates on an object This becomes g =( F /m)=(m g /m)= g The field strength is a VECTOR. For this case, the gravitational field is constant.  magnitude=g (9.8 m/s)  direction= down

9 Electric Field9 Week of February 17 05 Final Comment on Gravitational Field: Even though we know what is causing the force, we really don’t usually think about it.

10 Electric Field10 Week of February 17 05 Newton’s Law of Gravitation R M Earth m

11 Electric Field11 Week of February 17 05 The Calculation

12 Electric Field12 Week of February 17 05 Not quite correct …. Earth and the Moon (in background), seen from space)

13 Electric Field13 Week of February 17 05 More better … F Earth M Earth m Moon F moon mgmg

14 Electric Field14 Week of February 17 05 To be more precise … g is caused by  Earth (MAJOR)  moon (small)  Sun (smaller yet)  Mongo (extremely teeny tiny) g is therefore a function of position on the Earth and even on the time of the year or day.

15 Electric Field15 Week of February 17 05 The Electric Field E In a SIMILAR WAY  We DEFINE the ELECTRIC FIELD STRENGTH AS BEING THE FORCE PER UNIT CHARGE.  Place a charge q at a point in space.  Measure (or sense) the force on the charge – F  Calculate the E lectric Field by dividing the Force by the charge,

16 Electric Field16 Week of February 17 05

17 Electric Field17 Week of February 17 05 Electric Field Near a Charge

18 Electric Field18 Week of February 17 05 Two (+) Charges

19 Electric Field19 Week of February 17 05 Two Opposite Charges

20 Electric Field20 Week of February 17 05 A First Calculation Q r q A Charge The spot where we want to know the Electric Field Place a “test charge at the point and measure the Force on it.

21 Electric Field21 Week of February 17 05 Doing it Q r q A Charge The spot where we want to know the Electric Field F

22 Electric Field22 Week of February 17 05 General-

23 Electric Field23 Week of February 17 05 Continuous Charge Distribution

24 Electric Field24 Week of February 17 05 ymmetry

25 Electric Field25 Week of February 17 05 Let’s Do it Real Time Concept – Charge per unit length  dq=  ds

26 Electric Field26 Week of February 17 05 The math Why?

27 Electric Field27 Week of February 17 05 A Harder Problem A line of charge  =charge/length setup dx L r   x dEdE dE y

28 Electric Field28 Week of February 17 05 (standard integral)

29 Electric Field29 Week of February 17 05 Completing the Math 1/r dependence

30 Electric Field30 Week of February 17 05 Dare we project this?? Point Charge goes as 1/r 2 Infinite line of charge goes as 1/r 1 Could it be possible that the field of an infinite plane of charge could go as 1/r 0 ? A constant??

31 Electric Field31 Week of February 17 05 The Geometry Define surface charge density  =charge/unit-area dq=  dA dA=2  rdr (z 2 +r 2 ) 1/2 dq=  x dA = 2  rdr

32 Electric Field32 Week of February 17 05 (z 2 +r 2 ) 1/2 

33 Electric Field33 Week of February 17 05 (z 2 +r 2 ) 1/2 Final Result

34 Electric Field34 Week of February 17 05 Look at the “Field Lines”

35 Electric Field35 Week of February 17 05 What did we learn in this chapter?? FIELD We introduced the concept of the Electric FIELD.  We may not know what causes the field. (The evil Emperor Ming)  If we know where all the charges are we can CALCULATE E.  E is a VECTOR.  The equation for E is the same as for the force on a charge from Coulomb’s Law but divided by the “q of the test charge”.

36 Electric Field36 Week of February 17 05 What else did we learn in this chapter? We introduced continuous distributions of charge rather than individual discrete charges. Instead of adding the individual charges we must INTEGRATE the (dq)s. There are three kinds of continuously distributed charges.

37 Electric Field37 Week of February 17 05 Kinds of continuously distributed charges Line of charge   or sometimes = the charge per unit length.  dq=  ds (ds= differential of length along the line) Area   = charge per unit area  dq=  dA  dA = dxdy (rectangular coordinates)  dA= 2  rdr for elemental ring of charge Volume   =charge per unit volume  dq=  dV  dV=dxdydz or 4  r 2 dr or some other expressions we will look at later.

38 Electric Field38 Week of February 17 05 The Sphere dq r thk=dr dq=  dV=  x surface area x thickness =  x 4  r 2 x dr

39 Electric Field39 Week of February 17 05 Summary (Note: I left off the unit vectors in the last equation set, but be aware that they should be there.)

40 Electric Field40 Week of February 17 05 To be remembered … If the ELECTRIC FIELD at a point is E, then E = F /q (This is the definition!) Using some advanced mathematics we can derive from this equation, the fact that:

41 Electric Field41 Week of February 17 05 Example:

42 Electric Field42 Week of February 17 05 Solution

43 Electric Field43 Week of February 17 05 In the Figure, particle 1 of charge q 1 = -9.00q and particle 2 of charge q 2 = +2.00q are fixed to an x axis. (a) As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero? [1.89] L (b) Plot the strength of the electric field as a function of position (z). q 1 = -9qq 2 =+2q

44 Electric Field44 Week of February 17 05 Let’s do it backwards…

45 Electric Field45 Week of February 17 05 EXCEL aFirst TermSecond TermSum -30.13-0.88 -2.90.13-1.07-0.94 -2.80.14-1.15-1.01 -2.70.15-1.23-1.09 -2.60.15-1.33-1.18 -2.50.16-1.44-1.28 -2.40.17-1.56-1.39 -2.30.18-1.70-1.52 -2.20.20-1.86-1.66 -2.10.21-2.04-1.83 -20.22-2.25-2.03 -1.90.24-2.49-2.26 -1.80.26-2.78-2.52 ETC ….

46 Electric Field46 Week of February 17 05 alpha=1.89 ??

47 Electric Field47 Week of February 17 05 The mystery solved!!! BE CAREFULL!

48 Electric Field48 Week of February 17 05 In the Figure, the four particles are fixed in place and have charges q 1 = q 2 = +5e, q 3 = +3e, and q 4 = -12e. Distance d = 9.0 mm. What is the magnitude of the net electric field at point P due to the particles?

49 Electric Field49 Week of February 17 05

50 Electric Field50 Week of February 17 05 Figure 22-34 shows two charged particles on an x axis, q = -3.20 10 -19 C at x = -4.20 m and q = +3.20 10 -19 C at x = +4.20 m. (a) What is the magnitude of the net electric field produced at point P at y = -5.60 m? [7.05e-11] N/C (b) What is its direction? [180]° (counterclockwise from the positive x axis)

51 Electric Field51 Week of February 17 05 Figure 22-40 shows two parallel nonconducting rings arranged with their central axes along a common line. Ring 1 has uniform charge q 1 and radius R; ring 2 has uniform charge q 2 and the same radius R. The rings are separated by a distance d = 3.00R. The net electric field at point P on the common line, at distance R from ring 1, is zero. What is the ratio q 1 /q 2 ? [0.506]

52 Electric Field52 Week of February 17 05 In the Figure, eight charged particles form a square array; charge q = +e and distance d = 1.8 cm. What are the magnitude and direction of the net electric field at the center?


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