Find the Force and Direction of a charge

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SUMMARY

The discussion focuses on calculating the force and direction of a charge in an electromagnetic context. The user applied Coulomb's law using the formula F = K(q1)(q2)/r², with K as Coulomb's constant (8.99 x 10⁹ N m²/C²) and distances calculated using the Pythagorean theorem. The user derived forces of 0.064 N and 0.065 N at specific angles, ultimately calculating a total force of 0.123 N at an angle of 53 degrees. The error identified was in misunderstanding the problem's requirement to calculate the force after one second, necessitating the use of differential equations to determine the charge's position over time.

PREREQUISITES
  • Coulomb's Law for electric forces
  • Pythagorean theorem for distance calculations
  • Basic trigonometry for angle calculations
  • Understanding of differential equations in physics
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  • Study differential equations for modeling charge motion over time
  • Learn advanced applications of Coulomb's Law in dynamic systems
  • Explore vector addition of forces in electromagnetism
  • Investigate numerical methods for solving differential equations in physics
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Students and educators in physics, particularly those focusing on electromagnetism and force calculations, as well as anyone needing to understand the dynamics of charged particles in electric fields.

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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??
 
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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?