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Find the Force and Direction of a charge

  • Thread starter topM
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Homework Statement


http://www.brainmass.com/homework-help/physics/electromagnetic-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?!?!
 

Answers and Replies

  • #2
diazona
Homework Helper
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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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