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Unformatted text preview: -e 7 ' i 1 g 56 of the text, note that tne Pythagorean theorem tells us that the 4.00 {JG 7 F the charge of the l' ht ' ‘ ‘ exerts on the 3.00 #0 Change of the fight gigatélrégZde. The magnitude of the force that the 4.00 ,uC charge _ 1 lQl 9 « FH_.__ I. 2 -0 1 wig; (5.00 ><10—2 m )2 143.2 N. rE‘he force is repulsive since both charges have the same sign. If we denote by 4') the angle hetween the m-axis and the line connecting the 4.00 M) charge with the origin, we can express the force F1 in component form as E 2 43.2 N (— cos qii— sin 95;) = 43.2 N (—0.800? — 0.6003) = -—34.5 N’i— 25.9 N}. - The force of the —3.00 n0 charge on the charge of the light brigade has magnitude 1 long! 2 (9.00 x109 N-m2/02 )(3.00 x 10-6 execs x 10-5 C) F = —- z T 2 4m r2 (4.00x10-9m)2 5051‘- This force is attractive since the two charges have opposite signs. Write the force on the charge of the light brigade in Cartesian form as t —p 1. F2 = 50.6 Ni. The total force is the vector sum: from 2 F. + E = —34.6 Ni— 25.9 Nj+ 50.6 Ni : 16,0 Ni— 25.9 N}. b) The magnitude of the total force is Ftoga} 3 N)2 4* Z c) The acute angle 9 that the total force makes with the x-direction is found from Fy total _ —25.9 N Fm total _ 16.0N : "‘1'62 => 9 x —58.3 . tané‘ z Therefore, the total force points along a direction 58.3“ below the x—axis. 16.27 a) Find the electric field produced by each charge at point P from the general expression E 2: "L912: are) 7‘2 Where f‘ is 3 unit vector pointing radially away from the charge. In general, the magnitude of E is E: _1_i¢il_ Limo 7‘2 Therefore: the magnitude of the electric fieid produced at P by the 2.00 ,uC charge is = (9.00 x :09 N-rnZ/CE )(200 x 10-6 C) 2.1 13N .' (4.00111? 13x0 /C E1 (Q7750 {7/ The charge called Billy produces a field Where lql is located of magnitude E (9.00 x 109 N-m2/02 )} — 2.00 x 10-3 C 1 I l_ 5 Since Bill}r is negative, this field is directed toward Billy, parallel to (Hi). Therefore, 733: —(4.50 X 105 N/C)i. Using the Pythagorean theorem to find the distance of Milly from the point Where a" = «(1.00 m)2 + (2.00 m}2 x 2.24 m. Thus, the charge called Milly produces a field where I 19} is located, we have ql is located of magnitude 9 _2 2 M3 E2 : (9.00x10 Ng2/4Cm))l:.00x10 C] 21.79x106N/C. E2 =1.79 x 106 N/o(cos9i+ 51119;) 2.00113 - —1.00rn a : . 1 6\T ' ' 179x 0 i/C[(2.24711)1+(2.24m l3] 2 (1.60 x 106 N/C)§— (0.799 x 10“ N/C)j. I The total electric field at the place Where M is to be piawd is the vectm sum 0f the fie} Milly and Billy at that location, ds produced by —a E=E1+E2 2 (4.50 x 106 N/C)l+ {1.60 X 196 N/ofi— (0.799 x106 = (“2.90 x 106 N/Cfi— (0.799 x 106 N/Cfi. gative. The force on this N/CJE C) Your charge is He charge is F 2 9E : (—3.00 x 1073 C)((—2.90 ><106 N/efi— (0.799 x 105 N/C)j] = (8.70 x 103 N)l+ (2.40 x 103 IN}. 598 CHAPTER 16. ELECTRICAL CHARGES, ELECTRICAL FORCES, AND .. . d) The magnitude of the force is F 2 (8.70 x103 N)? + (—2.40 x 103 N)2 = 9.03 x103 N. 16.32 a) The charges shown in Figure 13.32 on page 758 of the text are point charges, so the magnitude of the electric fieid each creates is 1 lQi _ three 4"? ' E Referring to Figure 13.32, the charge on the :r-axis produces a field at the origin of magnitude 9W. 2 2 ~ MS .. zwwzaiw 108 5/0. E1 (0.500 m)? Since the charge producing this eiectric field is negative, the field is directed toward the charge creating it, parailel to i, so the field vector is E; 2 (5.40 x 105 N/C)i. The charge on the y—axis produces a field at the origin of magnitude (9.00 X 109 N-m2/02)|1(}.0 X 10—6 CI = . 5N1 . (0.2mm? 244x10 i/C E2: Since the charge producing this electric field is positive, the field is directed away from the charge creating it, parallel to ~j, so the fieid vector is a. m (44.4 x 105 51/0); The total electric field at the origin is the vector sum of the fields produced by both charges at that location, 1?: = E1 + a m (5.40 x 105 won — (14.4 x 105 N/C )3. b) The magnitude of the field at the origin is E = (5.40 x105 N/o)2 4 (—14.4 x 105 we 2 = 15.4 x 105 N/C. c) If a proton is placed at the origin, the force on it is f = qfi : (1.602 x 10’1g C) [(5.40 X105 N/C)i m (24.4 x 105 N/C )3] = (8.65 x 10-"14 Nfi- (23.1 ><10"14 N); d) The magnitude of the force on the proton is F a (8.65 x 10-14 N)2 + (—23.1 x 10-14 N)2 = 24.7 x 10"” N. The acceleration of the proton is found from Newton’s second law. Using the magnitudes of the vectors, we have F 24.7 x 10w14 N Fzmoz—fi: a:— _ "~————— = 14 2‘ m T 1.57 x 10—27 kg 1.48 x 10 mg 16.40. 601 16.40 Since the torque on an electric dipole is 1'" 2 if x E, to find the torque on the water molecule, we need to know its dipole moment ff and the electric fieid I3”) in which it finds itseif. Referring to Figure P40 on page 759 of the text, we see that the magnitude of the torque on the dipole is .— = {13‘ x E: = pEsinl? a (6.0 x 10—3“ c-m)(250 N/C) sin 120° 21.3 x 10-2? N-m. The vector product right-hand rule indicates that the direction of the torque vector is perpendicular to and out of the page. This problem illustrates the principle behind the microwave oven. Water makes up a high percentage of most food. When micrOWaves pass through the food from one direction, they exert a torque on the water molecules which causes them to rotate until 6 3 0° and the torque they experience vanishes. The microwaves then reflect off the inside of the oven and come though the food from a different direction, causing the water molecules to rotate so they align with the new direction. As this process continues, the water molecules rotate quite a bit and this collective motion contributes to the thermodynamic internal energy. An advantage of microwave heating is that it doesn’t need to start at the surface and work its way in, like other forms of heating, such as conduction. Rather, microwave heating occurs uniformly throughout all of the food. Although some people refer to microwaving their food as “nuking” it, as you see, there are no nuciear processes involved. 16.50. 605 16.50 In Example 16.17 on page 738 of the text, we saw that two oppositely charged parallel plates produce a. a; electric field between them of magnitude E = i7- 50 ' 47FEQE X 3.06 m5 2 Z 2 r M = 2.7 x 10 C m . “1* U EOE 471' 4mm) x 109 NEH/c?) / h) Since 103 mm = 1 m, we have (103)2 mm2 2 106 rim-12 : 1 m2. Hence the number of fundamental units of charge per square millimeter is 2.7 x 104 cm? :17 X 108 mm—z (106 Hung/r112) (1.602 x10“19 C) ' 16.58 a) The force on the particle of mass m and charge 9‘ is the electric force, Whose magnitude is Feiectric : To find the particle’s acceleration, apply Newton’s second law using vector magnitudes, f q E Felectric : mo ::> quE : me :> o = The acceleration of a charged particle in a uniforrn E field is constant, so we can use the kinematic equations for motion with a constant acceleration, letting i be in the direction of motion and choosing the origin at the initial position of the proton: ' 39(t) = $0 + vmgt + ::> tfi 2—ma 293m Vex IQFE‘ For a proton to travel 1.00 In, the time is 20.00 m)(1.6? x 10“27 kg) M : ms gm“ ‘ (1.502 X 10-19 oxsoo N/C) 8'34 X10 3' “Mawmsmmmmwmmmmemm b) For an electron, the time is t _ 2(1.00 m)(9.11 x 18—31 kg) _ 1 9, 10M? elem“ (1.502 x 1049 C)(300 N/C) * ‘ D X S‘ a) The ratio of the two times is: tproton __ X 10%; S telectron X 10"“? s :: 42.8. Now use the equation for t at the end of part a) to evaluate the above ratio Without any numbers: 2 x mproton lfproton __ / mproton 42 8 tElectron 2 1: melectron melectron qE So the ratio of times depends only upon the ratio of the masses (as long as both particles have the same chare‘el. -\A.-....-i -u VJ\\J\JK)A1/\_l" 16.66 Referring to Figure P. 56 on page 7'61, choose a coordinate system with ipointing right, pointing up, and origin at the launch point of the electron. With these choices, we make the important observation that, since We are projecting an electron into the given uniform electric field, the motion of this electron is equivalent to the motion of a massive projectile in a gravitational field. In other words, we can treat this problem exactly like we treated projectile motion problems in Chapter 4, with the only difference being that (my = —eE/m rather than —g. The magnitude of the electrical force on the electron is F = fglE : 6E. Apply Newton’s second law to the electron to find the magnitude of its acceleration. Using vector magni— tudes, we have Plate! BE Rota} : ma => 3 z m m In the chosen coordinate system the acceleration is down, and ye =0m _ . $0 — 0 m oyo : o0 s1ni9 Umo : B0 6059 BE 2 ay = —m— cm 2 0 m/s . m Therefore, the equations for the velocity and position components are . 3E Dy“) _ b0 8mg — -v$(t) 2 U9 c086 eE t9 t = 9 t 310?) = (1:0 sin l9)t — 3 ml ) 00(008 ) Impact occurs Where y 2 0 in, so E t2 0 In 3 (nosinflhf ~— (L) 3-. m .4 Use the quadratic formula to solve this for t, and thus, to find the time the electron is in the air. The two solutions are t : l} s, which corresponds to the launch time (i.e., the time at which 3; = 0 m l» and 2 Zmo tze—Eflsintlm 0 try eE sin 6. Finally, use this time in the equation for :c to find the distance the electron travels: x = (no cos 6% = (no Cost?) (2*mZOEsmg) _ 2mo§ eE : sin 501]" cos 600° (1.602 x 10-19 C}{250 N/G) = 0.492 In . sin 9 cos 8 ...
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This note was uploaded on 10/01/2009 for the course PHYS 102 taught by Professor Henry during the Spring '09 term at Johns Hopkins.

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hw1 - -e 7 ' i 1 g 56 of the text, note that tne...

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