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HW3T_SoL_4_F11 - a'Piiyszm ’ ‘ HOMEWORK SET 3...

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Unformatted text preview: a 'Piiyszm ’ ‘ HOMEWORK SET 3» Instructor: Professor Bandy “ ’“i 5"» ‘3 Finite Cl} Calculate the electric potential energy of the group of charges in Fig. I . 63 A water molecule perpendicular to an electric field has 1.5 x I 0'21 f more potential energy than a water molecule aligned with the field. its dipole moment is 6.2 x 1’ 0’30 Cm. Calculate . A proton with an initial speed of 750,000 m/s is brought to rest by an electric field. 3. Did the proton move into a region of higher potential or lower potential? Explain. b. Calculate the potential difference that stopped the proton. a. in Fig. 2, which of the capacitor plates is positive? Explain. b. Calculate inside the capacitor. e. Calculate the potential energy of a proton at the midpoint of the capacitor. Calculate the electric potential at a point centered in the charge distribution of Fig. I . A ~40. 0 11C point charge and +20.0 nC point charge are 15 cm apart on the x-axis. 3. Calculate the electric potential at the point on the x—axis where the electric field is zero. 1). Find the magnitude and direction of the electric field at the point on the x-axis Where the electric potential is zero. The annulus shown in Fig. 3 carries a uniform surface charge density 6. 2. Find an expression for the potential at an arbitrary point P on its axis. b. If a —-> 0 show that this expression. reduces to the potential on the axis of a. imiformly charged disk. A dipole with charges of i q and a separation of 2a is located a distance x from a point charge +Q. The dipole moment vector is perpendicular to the x axis. See Fig. 4. 3. Find expressions for the magnitude of the net torque and net force on the dipole in the limit of x >> a. bi What is the direction of the net force? What is the final position of dipole? The electric potential at points in the xy plane (in m) is given by V = 2.0x“9 — 3.0322 in V. In unit-vector notation, find the electric field E at the point (3.0 m, 2.0 m). 10. Two electrons are fixed 2.0 cm apart. Another electron is shot horn infinity and stops midway between the two. What is its initial speed? 0 V 3.0 cm 300 V +2 nC E ’l 1' 3 l I /.\ a i I \ l I 3 one i’ ‘l 3 cm i 3 / \\ l l \ i l 1/] \ g : CQ ______ _ ~O C s I ~] . ~I " 3 m n 100 V 200 V Figure 1 Figure 2 +61 Figure 3 +Q 2a ____._._____ «e—-————x-————> Figure 4 —q floHawo/zk SET 3 (I! z! 5' ® ’4? $2 CL. A f? » wi’fikmv‘fi " “ l . I ‘- J“;s" , r 1,, ‘ ’“ =r A a/ =— k/ A 4” + J :33 0,3,, 2... r {meagfifé’ J“ i “ FL; Y 9 1 WV; W413}: 33.x n a: (MHZ) « «I 2x 4) ’2 -‘7 V «ix/{ng + < D{ ) + (#2:.— ij/xlb (0.03m) (0.03m («03:41) M9 , ' LAW :zwfwx/a NM W123; f... @ 6 —+ / S’)<102/—« U; 7”: Pi'gpgwlw/M if: 1 WM (41.“ “ (/fl ’ J U] 5 7:5 'fA/flfiiéfL '73 5‘ 'l A I:3 ,4. 50 C4 :2 " 4* L; F w : ’nt 2 X10 "4 szpoL‘E P " / P I | 7;! ..._ . 2. U” 5 “FEQHO AND Urn—pg: wfia <1 .m- S 2! 2; Li r: L/H-f [’5‘ >416) f [,5’xm J” M_ .. VI 0‘ = ~f>E army/0223‘“ Q), at 5‘ '75,}; (390 mp; > :0 IF AVM =5 v7; > Va +3, Slows douw ) Wile/EM/zmea nN Mov’fd Fwam LanM?éNWAL 7; rWé” FW%“VWAL’ b 71 «in AV 3 cmsagukmou 5F gugeg‘f (4 4}ng a (/2 4‘ Kg \_ 2 3 7V? ” We "2 ’mw’: cw Aw M 132.9%» mm 2? .A , I e, 5 A. a 4 3-5;“ "‘3‘ {5'41"an ~‘.{.o><1:*(~é.o§7; ax 3T 3% E: ..U___, A A N ' A A \I/ ((38%) w 12% +l257z- M (Max Mafiwfig W m -—-——————-——“__J ...
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