Chapter 21. Electric Fields. Lecture 2. Dr. Armen Kocharian

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1 Chapter 21 Electric Fields Lecture 2 Dr. Armen Kocharian

2 Electric Field Introduction The electric force is a field force Field forces can act through space The effect is produced even with no physical contact between objects Faraday developed the concept of a field in terms of electric fields

3 Electric Field Definition An electric field exists in the region of space around a charged object This charged object is the source charge When another charged object, the test charge, enters this electric field, an electric force acts on it

4 Electric Field Definition, cont The electric field is defined as the electric force on the test charge per unit charge The electric field vector, E, at a point in space is defined as the electric force F acting on a positive test charge, q o placed at that point divided by the test charge: E = F e / q o

5 Electric Field, Notes E is the field produced by some charge or charge distribution, separate from the test charge The existence of an electric field is a property of the source charge The presence of the test charge is not necessary for the field to exist The test charge serves as a detector of the field

6 Relationship Between F and E F e = qe This is valid for a point charge only One of zero size For larger objects, the field may vary over the size of the object If q is positive, F and E are in the same direction If q is negative, F and E are in opposite directions

7 Electric Field Notes, Final The direction of E is that of the force on a positive test charge The SI units of E are N/C We can also say that an electric field exists at a point if a test charge at that point experiences an electric force

8 Electric Field, Vector Form Remember Coulomb s law, between the source and test charges, can be expressed as F e qq = ke r o r ˆ 2 Then, the electric field will be F kqq / r kq E = lim = r = rˆ q 0 2 e e o ˆ e 0 lim 2 qo qo r E F q e = = o k e q r r ˆ 2

9 More About Electric Field Direction a) q is positive, F is directed away from q b) The direction of E is also away from the positive source charge c) q is negative, F is directed toward q d) E is also toward the negative source charge

10 Superposition with Electric Fields At any point P, the total electric field due to a group of source charges equals the vector sum of electric fields of all the charges E = k e i q r i r ˆ 2 i i

11 Superposition Example Find the electric field due to q 1, E 1 Find the electric field due to q 2, E 2 E = E 1 + E 2 Remember, the fields add as vectors The direction of the individual fields is the direction of the force on a positive test charge

12 Superposition Example, cont Find E 1 6 ( 70. x10 C) 2 q1 9Nm 5 N E1 = k = x = x, r1 C C ( 040m. ) Find E 2, 2 6 q (. 2 9Nm 50x10 C) 5 N E2 = k = 899. x10 = 18. x10, r2 C ( 050m. ) C 3 5 cos θ =, =. N E1 39x10 j 5 C 4 sinθ = 5 5 =. N 5. N E2 11x10 i 1 4x10 j C C 5 E = E 1 + E 2 = + =. N 5 +. N E E1 E2 11x10 i 2 5x10 j C C

13 Electric Field Continuous Charge Distribution The distances between charges in a group of charges may be much smaller than the distance between the group and a point of interest In this situation, the system of charges can be modeled as continuous The system of closely spaced charges is equivalent to a total charge that is continuously distributed along some line, over some surface, or throughout some volume

Electric Field Continuous Charge Distribution, cont Procedure: Divide the charge distribution into small elements, each of which contains Δq

14 Electric Field Continuous Charge Distribution, cont Procedure: Divide the charge distribution into small elements, each of which contains Δq Calculate the electric field due to one of these elements at point P Evaluate the total field by summing the contributions of all the charge elements

15 Electric Field Continuous Charge Distribution, equations For the individual charge elements Δq Δ E= k r e r ˆ 2 Because the charge distribution is continuous Δqi dq E= k lim ˆ ˆ e r 0 2 i = k e r Δq 2 i r r i i

16 Charge Densities Volume charge density: when a charge is distributed evenly throughout a volume ρ = Q / V Surface charge density: when a charge is distributed evenly over a surface area σ = Q / A Linear charge density: when a charge is distributed along a line λ = Q / l

17 Amount of Charge in a Small Volume For the volume: dq = ρ dv For the surface: dq = σ da For the length element: dq = λ dl

18 Problem Solving Hints Units: when using the Coulomb constant, k e, the charges must be in C and the distances in m Calculating the electric field of point charges: use the superposition principle, find the fields due to the individual charges at the point of interest and then add them as vectors to find the resultant field

19 Problem Solving Hints, cont. Continuous charge distributions: the vector sums for evaluating the total electric field at some point must be replaced with vector integrals Divide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating over the entire charge distribution Symmetry: take advantage of any symmetry to simplify calculations

20 Example 2 (field of a ring of charge) x P Uniformly charged ring, total charge Q, radius a What is the electic field at a point P, a distance x, on the axis of the ring. How to solve Consider one little piece of the ring Find the electric field due to this piece Sum over all the pieces of the ring (VECTOR SUM!!)

21 de = electric field due to a small piece of the ring of length ds dq = charge of the small piece of the ring Since the circumference is 2πa, and the total charge is Q: dq = Q (ds/2πa)

22 z The next step is to look at the components Before we do that, let s think! We are on the axis of the ring There cannot be any net y or z components A net y or z component would break the azimuthal symmetry of the problem Let s just add up the x-components and forget about the rest!

23 What is going on with the y and z components? z dq ds The y (or z) component of the electric field caused by the element ds is always exactly cancelled by the electric field caused by the element ds' on the other side of the ring

24 Now we sum over the whole ring, i.e. we take the integral:

25 Time to think about the integral now. The integration is "over the ring" k is a constant of nature a is the ring-radius, a constant for a given ring x is the distance from the center of the ring of the point at which we want the E-field, x is also a constant = Q

26 Sanity check: do limiting cases make sense? What do we expect for x=0 and x? At x=0 expect E=0 Again, because of symmetry Our formula gives E=0 for x=0 As x, ring should look like a point. Then, should get E kq/x 2 As x, (x 2 +a 2 ) x 2 Then E kxq/x 3 = kq/x 2

27 Electrical Field of Charged Ring (cont) E = 4 o xq ( + ) πε x R

28 Example Charged Disk The ring has a radius R and a uniform charge density σ Choose dq as a ring of radius r The ring has a surface area 2πr dr

29 Electric Field Lines Field lines give us a means of representing the electric field pictorially The electric field vector E is tangent to the electric field line at each point The line has a direction that is the same as that of the electric field vector The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region

30 Another electric field example Electric field on axis of a uniformly charged disk? As we did last time: Calculate the field due to a small piece of the disk Add-up the contributions from all the small pieces Key question: What is the most convenient way of breaking up the disk into small pieces? Exploit result from last time: field due to a ring Break up the disk into a bunch of concentric rings

31 Back to the disk Strategy: Consider ring, at radius r, small width dr Compute electric field de due to this ring Add up the electric fields of all the rings that make up the disk Electric field due to ring, using prev. result: dq = charge of ring

32 dq = charge of ring Next question: what is dq? Total charge of the disk = Q dq/q = [Area of the ring] / [Area of the disk] Area of the disk = πr 2 width of the ring Area of ring = 2πr dr dq = Q(2r/R 2 )dr circumference of the ring

33 Surface charge density: σ = Q/Area = Q/(πR 2 ) Q = πσr 2 Now use k=1/(4πε 0 )

34 Now we sum over the rings we integrate: What are we integrating over? Integrate over r, from r=0 to r=r

35 A curious result (counterintuitive?) Consider infinitely large disk, i.e., R x 2 +R 2, the 2 nd term in parenthesis 0 E σ /(2ε 0 ) Constant, independent of x!! The electric field of an infinitely large, uniformly charged plane is perpendicular to the plane, and constant in magnitude E=σ/(2ε 0 ) In the limit, this holds also for a finite plane provided the distance from the plane is small compared to the size of the plane

36 Another example (Problem ) Find electric field at any point on x-axis Brute force approach: - find the field due to a small piece of the disk - sum over all of the pieces (integrate) There is a better way: - use previous results + a trick!

37 Previous result: How can we apply it to: Imagine two rings with no holes Radius R 2, charge density +σ Radius R 1, charge density -σ The sum of the fields due to these two rings will be the same as the field of the ring with the hole!

38 Field due to ring of radius R 2, surface charge density +σ: Field due to ring of radius R 1, surface charge density -σ: Total field is the sum of the two:

39 Total electrical field at the center of semicircle = ( ) = cos θ = 0 N/C E E E E E x i x i i y i i i Δq Δq = = 4πε r 4πε R i 0 The linear charge density on the rod is λ = Q/L, where L is the rod s length Δq = λ Δs = (Q/L)Δs E Q = cosθ iδθ 4πε LR ( Q/ L) Δs Q = cosθ = cosθ Δ 2 2 x i i i 4πε 0R 4πε 0LR i E x 0 i s

40 Total electrical field at the center of semicircle We note that the arc length Δs is related to the small angle Δθ by Δs = RΔθ, With Δθ dθ, the sum becomes an integral over all angles forming the rod. θ varies from Δθ = π/2 to θ = +π/2. So we finally arrive at E x Q π /2 Q π /2 2Q = cosθdθ sinθ 4πε LR = 4πε LR = 4πε LR π /2 π /

41 Electric Field Lines, General The density of lines through surface A is greater than through surface B The magnitude of the electric field is greater on surface A than B The lines at different locations point in different directions This indicates the field is non-uniform

42 Electric Field Lines, Positive Point Charge The field lines radiate outward in all directions In three dimensions, the distribution is spherical The lines are directed away from the source charge A positive test charge would be repelled away from the positive source charge

43 Electric Field Lines, Negative Point Charge The field lines radiate inward in all directions The lines are directed toward the source charge A positive test charge would be attracted toward the negative source charge

44 Electric Field Lines Dipole The charges are equal and opposite The number of field lines leaving the positive charge equals the number of lines terminating on the negative charge

45 Electric Field Lines Like Charges The charges are equal and positive The same number of lines leave each charge since they are equal in magnitude At a great distance, the field is approximately equal to that of a single charge of 2q

46 Electric Field Lines, Unequal Charges The positive charge is twice the magnitude of the negative charge Two lines leave the positive charge for each line that terminates on the negative charge At a great distance, the field would be approximately the same as that due to a single charge of +q

47 Electric Field Lines Rules for Drawing The lines must begin on a positive charge and terminate on a negative charge In the case of an excess of one type of charge, some lines will begin or end infinitely far away The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge No two field lines can cross

48 Motion of Charged Particles When a charged particle is placed in an electric field, it experiences an electrical force If this is the only force on the particle, it must be the net force The net force will cause the particle to accelerate according to Newton s second law

49 Motion of Particles, cont F e = qe = ma If E is uniform, then a is constant If the particle has a positive charge, its acceleration is in the direction of the field If the particle has a negative charge, its acceleration is in the direction opposite the electric field Since the acceleration is constant, the kinematic equations can be used

50 Accelerating Positive Charge Uniform field constant force Use kinematics eqs. For displacement 1 xf = xi + vit + at 2 2, Positive charge in a uniform electric field For velocity vf = vi + at, ( ) 2 2 v = v + 2a x x f i f i

51 Accelerating Positive Charge, cont a v i = F m = = x = i q E m 0,, x f 2 at qe = = 2 2m t 2, Positive charge in a uniform electric field Work and Kinetic Energy 2 2 qe v f = 2 a x f = x f, m 2 mv f m 2 q E KE = = x f = qex 2 2 m W = Δ K = q E x. f f

52 Electron in a Uniform Field, Example The electron is projected horizontally into a uniform electric field The electron undergoes a downward acceleration It is negative, so the acceleration is opposite E Its motion is parabolic while between the plates

53 Electron in an Uniform Field, Example, cont a v v y fx fy F ee = =, m m = v = const, ix ee = ayt = t m Projectile motion of electron in a uniform electric field x y fx f = v t, ix 2 at qe = = 2 2m t at qe xfx y = = y = Ax 2 2 2mvix qe A = 2 2mv 2 f f fx ix

54 Motion in Nonuniform Field

55 The Cathode Ray Tube (CRT) A CRT is commonly used to obtain a visual display of electronic information in oscilloscopes, radar systems, televisions, etc. The CRT is a vacuum tube in which a beam of electrons is accelerated and deflected under the influence of electric or magnetic fields

56 Cathode Ray Tube (CRT), cont The electrons are deflected in various directions by two sets of plates The placing of charge on the plates creates the electric field between the plates and allows the beam to be steered

Electric Field Lines (last time) Visualization of the electric field Lines drawn parallel to the E-direction With arrows pointing in the direction of E Start

57 Electric Field Lines (last time) Visualization of the electric field Lines drawn parallel to the E-direction With arrows pointing in the direction of E Start on +ve charge, end on ve charge High density of lines strong field Uniqueness of E-field lines never cross Otherwise would have two directions at crossing point

58 Are these field-lines pattern correct? from A. OK B. OK C. No D. No E. No Direction of arrows is wrong Density of lines suggests that field is stronger on one side of the charge Direction of arrows is wrong

Are these OK field lines? No. Lines cannot cross What are the signs of charges A & B? A: -ve B: +ve (lines start on +ve charge, end on ve) Several locations are labeled.

59 Are these OK field lines? No. Lines cannot cross What are the signs of charges A & B? A: -ve B: +ve (lines start on +ve charge, end on ve) Several locations are labeled. Rank them in order of electric field strength, from smallest to largest DAECB or perhaps DAEBC high density of lines high E What are the signs of charges A through I? A. + B. - C. + D. - E. - F. + G. + H. + I. +

60 Rank magnitude of charges in each sketch A > B Density of lines to the left of A > density to right of B D > C F > E > G I > H > J

61 Chapter 23 Quiz questions

62 At the position of the dot, the electric field points 1. Left. 2. Down. 3. Right. 4. Up. 5. The electric field is zero.

63 At the position of the dot, the electric field points 1. Left. 2. Down. 3. Right. 4. Up. 5. The electric field is zero.

64 A piece of plastic is uniformly charged with surface charge density 1. The plastic is then broken into a large piece with surface charge density 2 and a small piece with surface charge density 3. Rank in order, from largest to smallest, the surface charge densities 1 to 3. 1P > η 2 > η 3 2. η 1 > η 2 = η 3 3. η 1 = η 2 = η 3 4. η 2 = η 3 > η 1 5. η 3 > η 2 > η 1

65 A piece of plastic is uniformly charged with surface charge density 1. The plastic is then broken into a large piece with surface charge density 2 and a small piece with surface charge density 3. Rank in order, from largest to smallest, the surface charge densities 1 to η 1 > η 2 > η 3 2. η 1 > η 2 = η 3 3. η 1 = η 2 = η 3 4. η 2 = η 3 > η 1 5. η 3 > η 2 > η 1

66 Which of the following actions will increase the electric field strength at the position of the dot? 1. Make the rod longer without changing the charge. 2. Make the rod shorter without changing the charge. 3. Make the rod fatter without changing the charge. 4. Make the rod narrower without changing the charge. 5. Remove charge from the rod.

67 Which of the following actions will increase the electric field strength at the position of the dot? 1. Make the rod longer without changing the charge. 2. Make the rod shorter without changing the charge. 3. Make the rod fatter without changing the charge. 4. Make the rod narrower without changing the charge. 5. Remove charge from the rod.

68 Which electric field is responsible for the trajectory of the proton? (1) (2) (3) (4) (5)

69 Which electric field is responsible for the trajectory of the proton? (1) (2) (3) (4) (5)

Chapter 23. Electric Fields

Chapter 23. Electric Fields Chapter 23 Electric Fields Electric Charges There are two kinds of electric charges Called positive and negative Negative charges are the type possessed by electrons Positive charges are the type possessed