HW+6+solutions

HW+6+solutions - Eq.34 p.i60 I:——-—_e d 0"WW DO...

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Unformatted text preview: Eq.34 p.i60 I:——-—:___e - d 0 "WW. ' DO 7- J VO 1 *thac e R The Tom-l energy dissipa‘red fsfpdt =f o o ()0 dt 2 _. Va RC '70 e _L w— 2. 0 Suppose we have a l micro-Faer capaCH'or‘ charged +0 tOO vol+s. Q 2 CV = 10'” coulombs. One eiecfron charge is 1.6 >4 IO-” coukombs. Ii? to = RC is +he +ime cons+an+, we shall have one eiedron IeFJr ” 1.6xto'” 't — whene/t°=~W—>0F —£-:/BTU|.6XIOIS O This gives t 2 tax 2.3 103.0(6x10’q) = 3‘HLO . 50 1F +he +ime cons+an+ were a: I sec, we'd be down +0 "one efedron“ In a “Hie over halF a minufei Rz . _ ' RI Rin R. + RZ+RR Se‘H'mg Rm " R) . . _ RZR Rm 122% %R we. have R “R, + R1+R Rz—Rfi — m2 =0 R. :JR} +4R1Rz The negarive roo+ is do?“ ordinarily ' 2 realizable. We are leEJr wi+h : R :1/R3 + 4:2,}?1 + R. 2 R: Given V Wha+ is V’? Replace +he ladder +0 +he right OF B J by R. Tha‘r won’+ affed any+hing +0 +he leP+ OF B. We now have: C R' The res'lerance befween 4, R2 R C and A is Juer R, as A ‘_ - we know. Eviden’rlyji-hen, I fl}..— ‘ R 1 I %=(R+RR1) =R+2RZ I¥wewanT-\\—:—=-§: we muer have R: R2 Then 2R2 = UR3+4RIR2+ R. (ZRI‘RJZ : R? + H'RrRz QR: : SRIRZ or R2: 2R: To Jrerminafe The ladder dF+er any secfion, simply conned on a Singie resisfor O‘F value R. 5 1:5" 2| In reer IO Cm Frame 0? pioTes : . A], i S‘ro‘t‘voH' E ._. __._.._._____.___._ y 0 T Z Gm 234,“ .. _ : 0.5 '1+ K \ \€ _ S’rcgvo cm 20cm 03 : o Lin' \I G0 = OQAO '- Number 0F excess elec+rons on nega+ive ploie 2 36° Eo A0 0.5 x i0 x 20 4W8 “ W * 1.66 X iolo eiec‘rrons Frame F, is moving eas+ (11‘ 0.6 C. In i:F ,+he pia+e5 are moving weer a+ 0.6 C. Their eas+-wes+ dimension is 20W = iecm. NS dimension is unchanged, as is +he ver+icoi separa+ion. The number OF elecirons on “the negafive piafe is The same, 1.66 X io'°. Bu+ CT, and E, I are greoi'erJ by +he Facior a” = ' = 1.25 ._ 2 E _ 3/5 _ 0625 5+a’rvoi‘is ” ‘6 I” 0 — ' Crn Frame F2 is moving upward o+ 0.6 C. In F2, +he piaJre dimensions are 20x iOcm, +he ver+icoi separafion is 2Vi~.62 = L6 cm. The number oF eiec+rons is unchanged. 0'2 =0; and E1: E0 = 0.5 si’ai’voiTs/cm. e: .6 {zizs My 2.8 e=+an'*2f= SioLi" E: ZTFO' E,,=Ei zfiwo I WIN)”: _ Eil :Ell : YE; E Z 2 E -- Charge befween A and B is some as charge beTween A’ and 5' (7W : UV? O"=1.ioL+3(T Gauss‘s law . ’ S‘HH holds. cp :2ie~45°)= i2.68° Ean’cos a: l.iO‘i3E The Field of +he pro’f’on (If The posi‘f'ion 01C ‘H‘e plan is e/rz 3 fhe Force on Jrhe pion is Ee. F = 2/r2 = 2.3 x 10"5 dyne. The Fn'eld 0F +he pioh (1+ +he posifion 0F +he profon is (|~,62)e/r2 (Eq. :2 wiTh 8 = O) . The Force on “the profon is 0.64 x 2.3 x 15's dyne) _n_gj equal +0 fhe Porce on The pr‘on. There is ' momen+um I'_n me Held and The Field is changing. Only +he fowl momen+um, proTon momenfum + pion' mOmen+um + Field momenTum, is conserved. This 1's no’r a Two a body Sys+6m ! ...
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HW+6+solutions - Eq.34 p.i60 I:——-—_e d 0"WW DO...

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