Recent content by Cmon101

  1. C

    Two fixed charges and one free to move problem.

    I forgot to put a negative sign in front of distance since it was to the left. -.58 m is correct. Thanks so much for sticking through this with me and helping me out cupid!
  2. C

    Two fixed charges and one free to move problem.

    But if we assume 3 must be on the left of one done we have to change x-.301 to x+.301.
  3. C

    Two fixed charges and one free to move problem.

    I got .58 meters and still incorrect. I do not have the actual answer.
  4. C

    Two fixed charges and one free to move problem.

    Then from (.301+X)/X = SQROOT[(Q3Q2)/(Q3Q1)] We divide the left side of the equation to get: .301/X+1=SQROOT[(Q3Q2)/(Q3Q1)] Move the 1: .301/X=SQROOT[(Q3Q2)/(Q3Q1)]-1 Move the X: .301=[SQROOT[(Q3Q2)/(Q3Q1)]-1]X Get X alone: (.301)/[SQROOT[(Q3Q2)/(Q3Q1)]-1]=X And that's the...
  5. C

    Two fixed charges and one free to move problem.

    So K(Q3Q1)/ X2 + K(Q3Q2) /(.301+X)2 is the same as: K(Q3Q1)/ X2 = K(Q3Q2) /(.301+X)2 (If we use the absolute value for the charges) And Then we can turn this into (.301+X)2 / (X)2 = [K(Q3Q2)]/[K(Q3Q1)] Then the Ks cancel on the right side leaving you with (.301+X)2 / (X)2 =...
  6. C

    Two fixed charges and one free to move problem.

    Wait I guess it could look like this Q1(1.39uC)-----.301m-----Q2(-3.22 uC)-----Q3(3.33) X being the distance from 1 to 3. Actually does this make sense? I think Q3 on the left is actually the right setup cause in this one wouldn't it be impossible for the forces to equilibrate no matter how...
  7. C

    Two fixed charges and one free to move problem.

    Damn. Still didn't work. So if we assume the setup is like this Q3(3.33 uC)------X----Q1(1.39uC)-----.301m-----Q2(-3.22 uC) The equation would turn into this correct: K(Q3Q1)/ X2 + K(Q3Q2) /(.301+X)2 = 0 Or more plainly K(Q3Q1)/ X2 =...
  8. C

    Two fixed charges and one free to move problem.

    OMG good call. I kept thinking it had to either be Q1------Q2----Q3 or Q1---Q3---Q2 and I kept thinking that just doesn't make sense logically cause of the charges but Q3------Q1----Q2 does make sense! Alright altering my equation to take the new positioning into account. Will tell you the result.
  9. C

    Two fixed charges and one free to move problem.

    How do I use the force of Q1 on Q2 after I find it? I thought since one and two are stationary that force did not matter. Sorry I'm kind of slow at understanding this stuff.
  10. C

    Two fixed charges and one free to move problem.

    Homework Statement Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q1= 1.39x10-6 C and Q2= -3.22 x10-6 C are fixed at their positions, distance 0.301 m apart, and the charge Q3= 3.33 x10-6 C is moved along the straight line. For...
Back
Top