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Electrostatic force problem help with vector parts

  1. Mar 31, 2004 #1
    I've included a picture.

    Two charges Qc and -Qc(Qc = 4 µC) are fixed on the x-axis at x = -7 cm and x = 7 cm, respectively. A third charge Qb = 5 µC is fixed at the origin.
    A particle with charge q = 0.3 µC and mass m = 5 g is placed on the y-axis at y = 14 cm and released. There is no gravity.

    a) Calculate x-component of acceleration of particle.
    b) Calculate y-component of acceleratin of particle.
    c) Magnitude of the net electric force on q at its point of release?
    =====
    Ok, I know the general idea of what I got to do. To get the acceleration, set the electrostatic force equal to mass times acceleration
    F=ma
    f/m=a

    Now I got three charges in a line. For simplicity sake, lets call the charges from left to right in the line 1, 2,3, and charge 4 on the y axis.

    Before I go any further, I realize that this is an iscoceles triangle.

    When I draw the FBD on charge 4, I have a F4_2(force on 4 by 2) going vertical, F4_1 up to the right, and F4_3 down to the right.

    With the use of Coulomb's law, I figured out the electrostatic force from each charge onto charge 4.

    |F4_1|= |(k*q_1*q_4)/(.157m)^2| =>.4408N
    |F4_2|= |(k*q_2*q_4)/(.14m)^2| => .689N
    |F4_3|= |(k*q_3*q_4)/(.157m)^2| => .4408N

    .157m is the sides of the triangles
    .14m is the height of triangle
    q1=4E-6 C
    q2=5E-6 C
    q3=-4E-6 C
    q4=.3E-6 C
    =======
    Ok, I believe I'm doing good so far.

    Now this is where I'm having trouble with. To get the x and y components of acceleration, I have to break up the electrostatic forces...btw, I'm getting these angles by putting the origin on q4 on my FBD.

    x-component F4_1
    .4408N * cos 70
    y-component F4_1
    .4408 sin 70

    x-component F4_2
    0
    y-component F4_2
    .689N*sin 90

    x-component F4_3
    .4408N * cos -70
    y-component F4_3
    .4408 sin -70
    ----------
    Add all the x's together, all the y's together.
    Fx=.302 N
    Fy=.689 N

    a_x=.302N/.005kg=60.4m/sec^2
    a_y=.689N/.005kg= 137.8 m/sec^2

    Surprisingly, my a_y answer is correct, but my a_x answer is not, which makes me think that I screwed up the x-component calcualtions somewhere, but I do not know where...Sorry for long post, but can't some one help me out with my frustration?
     
  2. jcsd
  3. Mar 31, 2004 #2
    Ok, I know whats up...just because its iscoceles triangle, it doesn't mean you have 70 degree angles. Just do inv tan of .14/.07 to get 63.4 degrees, and then the x-components will work out.
     
  4. Mar 31, 2004 #3
    I have no idea what rdn meant by that previous post...

    I believe the x component of your acceleration will be zero sue to symmetry.
    Qc and -Qc are equal and opposite. The force that each one acts on q will be cancelled by the other.

    There will only be a y component of acceleration.
     
  5. Mar 31, 2004 #4

    Janitor

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    Science Advisor

    Paul, the symmetry is not there because the charge on the left is positive while the charge on the right is negative. So the positive test particle (if I may call it that) will be accelerated in a direction that will include a nonzero x component, and more specifically, the x component of acceleration will be in the positive x direction.

    Rdn is correct that the angles involved can be calculated by arctan(2).
     
  6. Apr 1, 2004 #5
    Janitor, your right. Sorry about that, I don't know what I was thinking.

    As for the arctan thing, I see it now. I should have drawn it out before making dumb statements. That is what I get for reading it quick and making snap judgements. Sorry gang.
     
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