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obj-1 - fl ’/7 ’d 21.6 EDENTEFY Apply Coulomb's law...

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Unformatted text preview: fl 6/77/377’? /7/ ’d/ 21.6. EDENTEFY: Apply Coulomb's law and calculate the net charge q on each sphere. SET UP: The magnitude of the charge of an electron is e : l.60><10‘19 C . 1 2 EXECUTE: F x ——4 i2 . This gives M = \/47r60Fr2 = \/47re,(4.57x 10~21 N)(O.200 m)2 =1.43><10“6 C. And ' 72'60 r therefore, the total number of electrons required is n 2 lg] /e =(1.4~3><10'l6 C)/(l.60><10‘19 C/electron) = 890 electrons. EVALUATE: Each sphere has 890 excess electrons and each sphere has a net negative charge. The two like charges repel. 21.1.3. IDENTIFY: Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions. SET UP: Like charges repel and unlike charges attract. lq2q3l and is in the +x direction. Fl must be in EXECUTE: The force E that q2 exerts on (13 has magnitude F2 2 k r2 2 IqIHqu ___k lqzllqal - the —"x direction, so q, must be positive. E 2 F2 gives k r2 r2 1 2 2.00 cm r2 4.00 cm EVALUATE: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 . lqil =lqzl(i) = (300 nC)( ) = 0.750 nC. 21.23. IDENTIFY: Apply Coulomb’s law to calculate the force exerted on one of the charges by each of the other three and then add these forces as vectors. (3) SET UP: The charges are placed as shown in Figure 21.23a. q1=q2=q3=q4=q Figure 21.23a Consider forces on (14. The free-body diagram is given in Figure 21.23b. Take the y—axis to be parallel to the diagonal between q2 and q4 and let + y be in the direction away from qz. Then F2 is in the +y—direction. l q2 EXECUTE: F3 = Fl =———2— 47ko L 2 F =;_47 47m, 2L FIX = 4. sin 45° = —F,/x/§ Fly = +Fl cos 45° = +P;/J§ F” = +17. sin 45° = +F3/x/5 EU = +1”; cos45° = High/5 sz = 0: Fzy = F2 Figure 21.231) RX =PL¥+FVZI +F;X =0 1 q2 1 £12 ‘11 R :1: +F +F = 2N5 ————+————-= “2‘5 y U 2y 3y ( {475%}? 471:502L2 87FEOL2( ) 2 R = q 2 (1+ 2x5). Same for all four charges. 8/:q,L EVALUATE: In general the resultant force on one of the charges is directed away from the opposite corner. The forces are all repulsive since the charges are all the same. By symmetry the net force on one charge can have no component perpendicular to the diagonal of the square. I 21.72. EDENTIFY: Apply F = k for each pair of charges and find the vector sum of the forces that q, and q2 exert on (13. SET UP: Like charges repel and unlike charges attract. The three charges and the forces on 513 are shown in Figure 21.72. Y Figure 21.72 —9 —9 Iq,z213l =(899x109 N.m2/C2)(5.00x1/0 C)(6.002><10 C) r1 (0.0500 In) 6=36.9°. F“=+1§cos49=8.63><10‘5 N. FU =+F,sina=6.48x10'5 N. -9 —9 F2 =k|q2_:b|=(8,99x109 Macaw r2 (0.0300 In) FZX =0, Fly =—F2 =——1.20x10“‘ N. P; =1?” +F21 =8.63x10'5 N. Fy =1?xy +1?2y =6.48x10‘5 N+(—1.20x10’4 N)=—5.52x10‘5 N. EXECUTE: (a) Fl =k =1.079x10*‘ C. =1.20><10“t C. r S s z y F (b) F=./Ff+Ff =1.02><10“‘ N. tan¢= F = 0.640. (D = 32.6° , below the +x axis. x EVALUATE: The individual forces on q3 are computed from Coulomb’s law and then added as vectors, using components. Mani 21.74. IDENTIFY: Apply ZFX =0and ZFy =0 to one of the spheres. SET UP: The free-body diagram is sketched in Figure 21.74. Fe is the repulsive Coulomb force between the spheres. For small 0, sine z tan-0. . __ _ _ mgsin0_ _E: z. =1 EXECUTE: Zlirszmg—Fc -—0and 2Fy —Tcosl9-—mg—0.So cosy —Fe— d2 .But tang sm9 2L , 2 2L 2L m so d3: kq and d=[—q——) . mg 27teomg EVALUATE: d increases when q increases. T Fe TsinO Figure 21.74 ...
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