Electricity and Magnetism Electric Fields

Transcription

1 Electricit and Magnetism Electric Fields Lana Sheridan De Anza College Sept 29, 2015

2 Last time Coulomb s law force from man charges current electric field charges and conductors

3 Warm Up Questions ar more work an and Which phsi-ofields bde : Spherical the following could be the charge on the particle hidden theofquestion mark? Gaussian hpothetical surface ssian surface, alculations of ution. For exsphere with a we discuss in? ace to the E n a Gaussian (A) 0 C Fig A spherical Gaussian ited example, (B) 1 C surface. If the electric field vectors ard from (C) the are Cof uniform magnitude and point l tells us that (D) 1 µc radiall outward at all surface points, a particle or ou can conclude that a net positive 1 Figure from Hallida, Resnick, Walker

4 Warm Up Questions ar more work an and Which phsi-ofields bde : Spherical the following could be the charge on the particle hidden theofquestion mark? Gaussian hpothetical surface ssian surface, alculations of ution. For exsphere with a we discuss in? ace to the E n a Gaussian (A) 0 C Fig A spherical Gaussian ited example, (B) 1 C surface. If the electric field vectors ard from (C) the are Cof uniform magnitude and point l tells us that (D) 1 µc radiall outward at all surface points, a particle or ou can conclude that a net positive 1 Figure from Hallida, Resnick, Walker

5 Warm Up Questions Which expression relating force to electric field is correct? (A) F = m 0 E (B) E = q 0 F (C) F = q 0 E (D) F = E

6 Warm Up Questions Which expression relating force to electric field is correct? (A) F = m 0 E (B) E = q 0 F (C) F = q 0 E (D) F = E

7 Warm Up Questions What are the units of electric field? (A) Nm (B) N/C (C) Nm 2 /C 2 (D) C/N

8 Warm Up Questions What are the units of electric field? (A) Nm (B) N/C (C) Nm 2 /C 2 (D) C/N

9 41 (a) What equal positive charges would have to be placed on Earth and on the Moon to neutralize their gravitational attraction? (b) Wh don t ou need to know the lunar distance to solve this prob- 42 (c) & 43 How man kilograms of hdrogen ions (that is, protons) pg 573lem? would be needed to provide the positive charge calculated in (a)? Homework Questions 42 In Fig , two tin conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L.Assume that u is so small that tan u can be replaced b its approximate equal, sin u.(a) Show that x q 2 L 2 0 mg 1/3 gives the equilibrium separation x of the balls. (b) If L 120 cm, m 10 g, and x 5.0 cm, what is q? 43 (a) Explain what happens to the balls of roblem 42 if one of them is discharged (loses its charge q to, sa, the ground). (b) Find the new equilibrium separation x,using the given values of L and m and the computed value of q. 44 SSM How far apart must two protons be if the magnitude of the q L θθ x L Fig roblems 42 and 43. q tude of th 50 Figu length L, at a distan are attach respective fixed sphe rod is hor the rod ex zontal and 51 A ch cross-sect

10 Overview field due to a point charge field from multiple point charges electric fields of charge distribution

11 Field from a oint Charge We want an expression for the electric field from a point charge, q. Using Coulomb s Law the force on the test particle is F 0 = k qq 0ˆr. r 2 E = F ( ) 1 k q q 0 = q 0 q 0 r 2 ˆr The field at a displacement r from a charge q is: E = k q r 2 ˆr

12 Field from a oint Charge The field at a displacement r from a charge q is: This is a vector field: charge and a nearb negative point charge that are equal in magnitude.the charges attract each other.the pattern of field lines and the electric field it represents have rotational smmetr about an axis passing through both charges in the plane of the page.the electric field vector at one point is shown; the vector is tangent to the field line through the point. E = k q r 2 ˆr The direction of is d toward the point charg F : The direction of is directl awa from th Because there is gives the field at eve point charge is shown We can quickl fin charge. If we place a p from Eq. 21-7, the net f E : Therefore, from Eq. 2 Fig The electric field vectors at various points around a positive point charge. Here E : i is the electri Equation 22-4 shows fields as well as to ele

13 Field from man charges The field is just the force divide b the charge. So, what is the force from man charges?

14 Field from man charges The field is just the force divide b the charge. So, what is the force from man charges? F net! F net,0 = F 1 0 F F n 0

15 Field from man charges The field is just the force divide b the charge. So, what is the force from man charges? F net! F net,0 = F 1 0 F F n 0 E net = F net q 0 Total electric field: E net = E 1 E 2... E n

16 Question about field from point charges 22-4 THE ELECTRIC FIELD DUE TO A OINT C Consider a proton p and an electron e on an x axis. x S e R p on p and an electron e on on of the electric field due to the electron at (a) point S and ction of What the net iselectric the direction field at (c) of point the electric R and (d) field point due S? to the electron at point S and point R? (A) leftward at S, leftward at R Sample roblem (B) leftward at S, rightward at R Net electric field due to three charged particles (C) rightward at S, leftward at R particles (D) with rightward charges at qs, 1 2Q, rightward at R E each a distance d from the origin Q 4 0 d and E produced at the origin? We next must find the orientations o 1 Figure from Hallida, Resnick, Walker, field page vectors 583. at the origin. Because q 1 is

17 Question about field from point charges 22-4 THE ELECTRIC FIELD DUE TO A OINT C Consider a proton p and an electron e on an x axis. x S e R p on p and an electron e on on of the electric field due to the electron at (a) point S and ction of What the net iselectric the direction field at (c) of point the electric R and (d) field point due S? to the electron at point S and point R? (A) leftward at S, leftward at R Sample roblem (B) leftward at S, rightward at R Net electric field due to three charged particles (C) rightward at S, leftward at R particles (D) with rightward charges at qs, 1 2Q, rightward at R E each a distance d from the origin Q 4 0 d and E produced at the origin? We next must find the orientations o 1 Figure from Hallida, Resnick, Walker, field page vectors 583. at the origin. Because q 1 is

18 Question about field from point charges 22-4 THE ELECTRIC FIELD DUE TO A OINT C Consider a proton p and an electron e on an x axis. x S e R p on p and an electron e on on of the electric field due to the electron at (a) point S and ction of What the net iselectric the direction field at (c) of point the net R and electric (d) point fields? at point S and point R? (A) leftward at S, leftward at R Sample roblem (B) leftward at S, rightward at R Net electric field due to three charged particles (C) rightward at S, leftward at R particles (D) with rightward charges at qs, 1 2Q, rightward at R E each a distance d from the origin Q 4 0 d and E produced at the origin? We next must find the orientations o 1 Figure from Hallida, Resnick, Walker, field page vectors 583. at the origin. Because q 1 is

19 Question about field from point charges 22-4 THE ELECTRIC FIELD DUE TO A OINT C Consider a proton p and an electron e on an x axis. x S e R p on p and an electron e on on of the electric field due to the electron at (a) point S and ction of What the net iselectric the direction field at (c) of point the net R and electric (d) point fields? at point S and point R? (A) leftward at S, leftward at R Sample roblem (B) leftward at S, rightward at R Net electric field due to three charged particles (C) rightward at S, leftward at R particles (D) with rightward charges at qs, 1 2Q, rightward at R E each a distance d from the origin Q 4 0 d and E produced at the origin? We next must find the orientations o 1 Figure from Hallida, Resnick, Walker, field page vectors 583. at the origin. Because q 1 is

20 C, and object B has a charge electric Electric forces Field on the Question objects? 52F S qba 1 = (d) S q 3 = FAB S µc, FBA q 2 = 2.00 µc, and a = m. The resultant force exerted on q 3 is F net,3 = ( 1.04 i 7.94 j) N. What is the electric field at the location of q 3 due to the other two charges? q 2 a a 2a S F 23 q 3 S F 13 (A) ( 1.04 i 7.94 j) N (B) ( 1.04 i 7.94 j) N/C (C) ( i 1.59 j) MN/C (D) ( 2.08 i 15.9 j) N/C q 1 x Figure (Example 23.2) The Figure from Serwa & S Jewett, pg 696, Ex 2.

21 C, and object B has a charge electric Electric forces Field on the Question objects? 52F S qba 1 = (d) S q 3 = FAB S µc, FBA q 2 = 2.00 µc, and a = m. The resultant force exerted on q 3 is F net,3 = ( 1.04 i 7.94 j) N. What is the electric field at the location of q 3 due to the other two charges? q 2 a a 2a S F 23 q 3 S F 13 (A) ( 1.04 i 7.94 j) N (B) ( 1.04 i 7.94 j) N/C (C) ( i 1.59 j) MN/C (D) ( 2.08 i 15.9 j) N/C q 1 x Figure (Example 23.2) The Figure from Serwa & S Jewett, pg 696, Ex 2.

22 Question about net field (a) (b) (c) Fig Question 1. charge magnit 2 Figure shows two square arras of charged particles. The squares, which are centered on point,are misaligned.the particles are separated b either d or d/2 along the perimeters of the squares. What are the magnitude and direction of the net electric field at? 6q 2q 3q 3q 2q 2q q q q 3q 2q 3q 6q 2q Fig Question 2. 1 Figure from Hallida, Resnick, Walker, page 597, problem 2. 5 Fig axis. ( than a a poin field is their l there a

23 Electric Dipole electric dipole z A pair of charges of equal magnitude q but opposite sign, We are usuall inte separated b a distance, d. that are large compared that z d.at such larg dipole moment: E ( ) proximation, we can neg r () E () p = qd ˆr E q 4 0 z where ˆr is a unit vector z pointing from the negative charge to the positive r ( ) charge. Up here the q The product qd, wh field dominates. dipole, is the magnitude p : of the dipole. (The uni q d Dipole center q p Down here the q p : The direction of is t dipole, as indicated in

24 Electric Field from an Electric Dipole r ( ) r () d z E () z q Dipole center q (a) E ( ) Up here the q field dominates. (b) Fig (a) An electric dipole.the electric field vectors E : and E : ( ) ( ) at point p Down here the q field dominates. After forming a common denominator and multipl E q 2d/z 4 0 z 2 We are usuall interested in the electrical effec that are large compared with the dimensions of the dip that z d.at such large distances,we have d/2z proximation, we can neglect the d/2z term in the deno E 1 qd r 2 0 z. () 2 r( ) 2 3 ( The product qd, which involves the ) two intrins 1 dipole, is the magnitude p of a vector quantit known z 2 p : of the (1 dipole. d/2z) (The 2 1 unit (1 of p : is d/2z) the coulomb-meter.) 2 p : 1 d 2z 2 2 E p z 3 2z We will find an expression for the magnitude of the field along the dipole axis E = E () E ( ) = kq = kq kq (electric di The direction of is taken to be from the negati dipole, as indicated in Fig. 22-8b. We can use the orientation of a dipole. Equation 22-9 shows that, if we measure the el distant points, we can never find q and d separatel; product. The field at distant points would be uncha doubled and d simultaneousl halved. Although E q 2 0 z 3

25 Electric Field from an Electric Dipole r ( ) r () d z E () E ( ) z q Dipole center q Up here the q field dominates. p After forming a common denominator and multipl We are usuall interested in the electrical effec that are large compared with the dimensions of the dip that z d.at such large distances,we have d/2z proximation, we can neglect the d/2z term in the deno E 1 qd r 2 0 z. () 2 r( ) 2 3 ( The product qd, which involves the ) two intrins 1 dipole, is the magnitude p of a vector quantit known z 2 p : of the (1 dipole. d/2z) (The 2 1 unit (1 of p : is d/2z) the coulomb-meter.) 2 The direction of (electric di is taken to be from the negati Down here the q dipole, as indicated in Fig. 22-8b. We can use the field dominates. orientation of a dipole. Equation 22-9 shows that, if we measure the el The effect (a) of the dipole (b) falls off as distant 1/z 3 points, - means we can the never charges find q and d separatel; Fig. largel, 22-8 but (a) An not electric entirel dipole.the cancel each product. other The out. field at distant points would be uncha electric field vectors E : and E : ( ) ( ) at point doubled and d simultaneousl halved. Although E E q 2d/z 4 0 z 2 p : 1 d 2z 2 2 E p z 3 2z We will find an expression for the magnitude of the field along the dipole axis E = E () E ( ) = kq = kq = 2kp z 3 kq where we assumed z >> d q 2 0 z 3

26 r-inverse decas 1 Figure b Neeraj Sood, from rfidjournal.com.

27 Continuous distribution of charge In previous examples, we added up the field from each point charge. But what about the case of a charged object, like a plate or a wire? In just -1 Coulomb of charge, there are more than a quintillion excess electrons!

28 Continuous distribution of charge In previous examples, we added up the field from each point charge. But what about the case of a charged object, like a plate or a wire? In just -1 Coulomb of charge, there are more than a quintillion excess electrons! You do not want to add up the effect of each one b one.

29 Continuous distribution of charge In previous examples, we added up the field from each point charge. But what about the case of a charged object, like a plate or a wire? In just -1 Coulomb of charge, there are more than a quintillion excess electrons! You do not want to add up the effect of each one b one. Solution: treat the charge as a continuous distribution with some charge densit.

30 Charge Densit charge densit The amount of charge in per unit volume of an object. (Here volume could be volume, area, or length) B convention, different smbols can be used in different cases: smbol description units λ charge per unit length C m 1 σ charge per unit area C m 2 ρ charge per unit volume C m 3 For a wire, usuall use charge per length. For a plate, charge per area.

31 Continuous distribution of charge This negativel charged rod is obviousl not a particle. lastic rod of charge Q But we can treat this element as a particle. ds Here is the field the element creates. ds de r x x θ x (a) We need to add up the charge of each little particle ds. Each These x components add. has charge λ ds. These components just Our job is to add all such cancel, so neglect them. components. To be perfectl accurate, we would make the length of ds 0. ds ds ds This is an integral: λ s λ ds (b) (c) de de de de

32 The main trick All this does not mean ou have to be able to do integrals. If ou understand that ou sum up the effect of charges, ou can still figure out what the net field at man points is just b smmetr.

33 dq 1 r for ld g (22-11) Example: Field from a ring of charge -11 as 1 0 ds (z 2 R 2 ). z de de cos θ θ The perpendicular components just cancel but the parallel components add. ds r R ds r 2. θ z (22-12) Vertical components? From each charge λ ds: de = de cos θ ( ) kλ ds = cos θ = = r 2 ( kλ ds (R 2 z 2 ) ) k z λ ds (R 2 z 2 ) 3/2 z R 2 z 2

34 xpress the magnitude of de as for ld g Example: Field from (22-11) a ring of charge dq 1 r as 1 0 ds (z 2 R 2 ). z de de cos θ θ The perpendicular components just cancel but the parallel components add. ds r R ds r 2. θ z (22-12) de = k z λ ds (R 2 z 2 ) 3/2 There are 2πR-worth of little lengths ds. Adding the field for all together: E = k z λ(2πr) (R 2 z 2 ) 3/2 = k q z (R 2 z 2 ) 3/2 since total charge q = 2πRλ b definition.

35 Question s of smmetr, as,replace r 2 with ding component u,but u is identiariable. Replace,around the cir- x x (b) (c) The figure here shows three nonconducting rods, one circular and Fig (a) oint is on an extension of the line of charge. two straight. (b) Each is on a line has of smmetr a uniform of the charge line of charge, of magnitude at perpendicular distance from that line. (c) Same as (b)except that is not on Q along its top half and a line another of smmetr. along its bottom half. For each rod, what is the direction of the net electric field at point? ds, one circular and two de Q along its top half, what is the direction of Q Q Q x x x Q Q Q (a) (b) (c) For (a) it is: (A) up (B) down (C) left (D) right 1 age 590, Hallida, Resnick, Walker.

36 Question s of smmetr, as,replace r 2 with ding component u,but u is identiariable. Replace,around the cir- x x (b) (c) The figure here shows three nonconducting rods, one circular and Fig (a) oint is on an extension of the line of charge. two straight. (b) Each is on a line has of smmetr a uniform of the charge line of charge, of magnitude at perpendicular distance from that line. (c) Same as (b)except that is not on Q along its top half and a line another of smmetr. along its bottom half. For each rod, what is the direction of the net electric field at point? ds, one circular and two de Q along its top half, what is the direction of Q Q Q x x x For (a) it is: (A) up (B) down (C) left (D) right Q Q (a) (b) (c) 1 age 590, Hallida, Resnick, Walker. Q

37 Question s of smmetr, as,replace r 2 with ding component u,but u is identiariable. Replace,around the cir- x x (b) (c) The figure here shows three nonconducting rods, one circular and Fig (a) oint is on an extension of the line of charge. two straight. (b) Each is on a line has of smmetr a uniform of the charge line of charge, of magnitude at perpendicular distance from that line. (c) Same as (b)except that is not on Q along its top half and a line another of smmetr. along its bottom half. For each rod, what is the direction of the net electric field at point? ds, one circular and two de Q along its top half, what is the direction of Q Q Q x x x Q Q Q (a) (b) (c) For (b) it is: (A) up (B) down (C) left (D) right 1 age 590, Hallida, Resnick, Walker.

38 Question s of smmetr, as,replace r 2 with ding component u,but u is identiariable. Replace,around the cir- x x (b) (c) The figure here shows three nonconducting rods, one circular and Fig (a) oint is on an extension of the line of charge. two straight. (b) Each is on a line has of smmetr a uniform of the charge line of charge, of magnitude at perpendicular distance from that line. (c) Same as (b)except that is not on Q along its top half and a line another of smmetr. along its bottom half. For each rod, what is the direction of the net electric field at point? ds, one circular and two de Q along its top half, what is the direction of Q Q Q x x x For (b) it is: (A) up (B) down (C) left (D) right Q Q (a) (b) (c) 1 age 590, Hallida, Resnick, Walker. Q

39 Question s of smmetr, as,replace r 2 with ding component u,but u is identiariable. Replace,around the cir- x x (b) (c) The figure here shows three nonconducting rods, one circular and Fig (a) oint is on an extension of the line of charge. two straight. (b) Each is on a line has of smmetr a uniform of the charge line of charge, of magnitude at perpendicular distance from that line. (c) Same as (b)except that is not on Q along its top half and a line another of smmetr. along its bottom half. For each rod, what is the direction of the net electric field at point? ds, one circular and two de Q along its top half, what is the direction of Q Q Q x x x Q Q Q (a) (b) (c) For (c) it is: (A) up (B) down (C) left (D) right 1 age 590, Hallida, Resnick, Walker.

40 Question s of smmetr, as,replace r 2 with ding component u,but u is identiariable. Replace,around the cir- x x (b) (c) The figure here shows three nonconducting rods, one circular and Fig (a) oint is on an extension of the line of charge. two straight. (b) Each is on a line has of smmetr a uniform of the charge line of charge, of magnitude at perpendicular distance from that line. (c) Same as (b)except that is not on Q along its top half and a line another of smmetr. along its bottom half. For each rod, what is the direction of the net electric field at point? ds, one circular and two de Q along its top half, what is the direction of Q Q Q x x x For (c) it is: (A) up (B) down (C) left (D) right Q Q (a) (b) (c) 1 age 590, Hallida, Resnick, Walker. Q

41 Summar E-field from man charges electric fields of charge distribution Homework E-fields worksheet Hallida, Resnick, Walker: Ch 22, onward from page 597. roblems: 2, 5, 7, 9, 23