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Find the Force and Direction of a charge |
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| May9-09, 07:18 AM | #1 |
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Find the Force and Direction of a charge
1. The problem statement, all variables and given/known data
http://www.brainmass.com/homework-he...c-theory/98519 That site has the diagram of the Lab and distances. Rg-r=sqrt 3.7092 + .5332 = 3.75 Fg-r= K(q1)(q2)/Rg-r2= (8.99x109)(1x10-5)(1x10-5)/3.752 = .064 angle 1 = tan-1(.533/3.709) = 8.18 degrees Rb-r=sqrt 2.5532 + 2.7092 = 3.72 Fb-r= K(q1)(q2)/Rg-r2= (8.99x109)(1x10-5)(1x10-5)/3.722 = .065 angle 1 = tan-1(2.533/2.709) = 43 degrees Fxg-r = (.064)Cosine8.18 = .063 Fyg-r = (.064Sine8.18 = .009 Fxb-r = (.065)Cosine43 = .048 Fyb-r = (.065)Sine43 = .044 Fx = .063 + .048 = .111 Fy = .009 + .044 = .053 Force = sqrt of .1112 + .0532 = .123N Angle = tan-1(.64/.65) = 45 degrees + 8.18 = 53 degrees to the + X The lab shows that the total Force = .185N Where did I go wrong?!?! |
| May9-09, 05:21 PM | #2 |
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Recognitions:
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It's kind of hard to tell, but it looks like you calculated the force on the charge at t = 0, that is, as soon as it is released. The problem is asking for the force after 1 second. To figure it out you'd have to write and solve a differential equation for the charge's position as a function of time... seems like a rather unpleasant thing to have to do :-( Are you sure you're supposed to calculate this?
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