week17physics - Could I see a solution to this? 28) F=qE...

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Week 17 Homework Chapter 23 Sean Gaudette 6) E=kq/r^2=(9x10^9)(2x10^-7)/(.075)^2=320,000 N/C ***640,000 N/C*** 10) |E2|=|E1| kq/x^2=kq/(x-d)^2 x-d/x=square root(q2/|q1|)=square root(2/5) x=d/1-square root(2/5)=***2.7d*** 12) |Enet|=k*2q/r^2 r=a/square root(2) ***|Enet|=kq/a^2*** 16) |Enet|=2E*sin theta=2[kq/(d/2)^2+r^2)]*[(d/2)/square root((d/2)^2+r^2)] =kqd/[(d/2)^2+r^2]^(3/2)=kqd/r^3=***-kp/r^3*** 24) ??? I really do not know what to do here.
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Unformatted text preview: Could I see a solution to this? 28) F=qE ma=qE E=ma/q=9.11x10^-31/(-1.6x10^-19)x1.80x10^9i=***-1.02 x 10^-2i N/C*** 38) The minimum difference between measured values in the table is 1.6 x 10^-19, therefore this is the charge of the electron. 46) initial U=-p*E=-pE*cos theta Final U=-p*E=-pEcos(pi-theta)=pEcos theta Work=Ufinal-Uinitial=PE*cos theta-(-PE*cos theta)=2PE*cos theta...
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This note was uploaded on 05/09/2010 for the course MGMT e33 taught by Professor Allansmith during the Spring '10 term at Aarhus Universitet, Aarhus.

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