Results 1 to 2 of 2

Thread: Need some Vector Analysis help, probably easy

  1. #1
    Join Date
    Feb 2006
    Location
    Canada
    Posts
    2,254
    Mentioned
    21 Post(s)
    Quoted
    238 Post(s)

    Default Need some Vector Analysis help, probably easy

    Hey guys, just working on some Phys 112. I'm definitely rusty at it! My question is this:

    and so far I have:

    I can't seem to figure out how to solve for the net force...

  2. #2
    Join Date
    Apr 2012
    Posts
    56
    Mentioned
    0 Post(s)
    Quoted
    0 Post(s)

    Default

    this may help.similiar problem= two point charges Q1 and Q2 are held in place 4.50cm apart.

    Another point charge -2.50μC of mass 5.5g is initially located 3cm from each of these charges and released from rest.

    You observe that the initial acceleration of -2.50μC is 344m/s² upward, parallel to the line connecting the two point charges.

    Find Q1 and Q2.

    solution=

    In order for Q to move upward it must be attracted to q1 and repelled by q2. q1 and q2 must also have the same charge, but opposite signs as just mentioned, so that the perpendicular force will vanish.

    Just a reminder, Coulomb's Law, will be needed:
    F = kQ1*Q2/r^2
    k = 9.0x10^9 Nm^2/C^2

    Upward force on Q is (first convert mass from g to kg):
    F = ma = (0.0055)(344) = 1.892 N

    If Fv is the vertical component of force from q1 then the force from q2 is also Fv. So the total force is 2Fv and it must equal F just calculated.

    2Fv = 1.892
    Fv = 0.946 N
    This is related to the total force between q1 and q by:
    cos(A) = Fv/(total force)
    A = the angle between an upward vector at Q to the line from Q to q1.
    cos(A) = 2.25/3 = 0.75
    total force = Fv/cos(A) = 0.946/0.75 = 1.26133 N

    Using Coulombs law (stated at the start) and first converting charge from microC to C and distance in cm to m:

    1.26133 = (9.0x10^9)[2.5x10^(-6)][q1]/(.03)^2
    1.26133 = [(9.0)(2.5)x10^3][q1]/(.03)^2
    1.26133 = [(9.0)(2.5)x10^3][q1]/[9x(10^-4)]
    1.26133 = [(2.5)x10^7][q1]
    q1 = 1.26133 / [(2.5)x10^7]
    q1 = 0.5045 x 10^(-7) C

    We can the write:
    q1 = +0.05045 microC
    q2 = -0.05045 microC

    Check by seeing if i get the right force between Q and q1:
    F = (9.0x10^9)[2.5x10^(-6)][0.05045x10^(-6)]…
    F = (9.0x10^9)[2.5x10^(-6)][0.05045x10^(-6)]…
    F = [2.5x10][0.05045]
    F = 1.26125
    And this agrees with the total force calculated above

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •