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Finding the electric field magnitude in a parallel plate compacitor. 
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#1
Sep2408, 01:57 PM

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Problem:
The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 9.42 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. What i did: x=.5(Vo + V)t .02m=.5(9.42 x 10^6 m/s + 9.42 x 10^6 m/s)t solve for t=2 x 10^9 then: x=Vot + .5at^2 .0015m=0 + .5(a)(2 x 10^9)^2 solve for a=8.87 x 10^15 then: F=ma F=(9.11 x 10^31)(8.87 x 10^15) F=8.08 x 10^15 then: E=F/q E=8.08 x 10^15/(1.6 x 10^19) E=50524.3 N/C But that is the wrong answer, i was just wondering what i did wrong and how to correct myself. In Addition i was wondering how to do this one?: A long, thin rod (length = 3.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +7.0 C point charge is fixed to one end of the rod, and a 7.0 C point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude 2.0 x 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod. Thank you in advanced!! 


#2
Sep2408, 02:34 PM

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P: 41,325

Also: Don't round off intermediate values (such as your value for t); wait until the last step. 


#3
Sep2408, 02:39 PM

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P: 41,325




#4
Sep2408, 03:26 PM

P: 3

Finding the electric field magnitude in a parallel plate compacitor.
Should i just us the equation: E=kq/r^2 and then use E=F/q ? 


#5
Sep2408, 03:40 PM

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P: 41,325




#6
Sep2408, 04:01 PM

P: 3

ahh ok, i got it right, thank you so much for your help :)



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