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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

    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
    y-component F4_2
    .689N*sin 90

    x-component F4_3
    .4408N * cos -70
    y-component F4_3
    .4408 sin -70

    Attached Files:

  2. jcsd
  3. Mar 31, 2004 #2
    Yes your calculation of a_y is correct.

    and a_x by my calculation is 395.4605714 m/sec^2

    Fx = F1x + F3x
    = F1*cos(x) + F3*cos(x)
    where x is angle between x-axis and F1, which is equal to that between x-axis and F3.
    and cos(x) = 0.157/0.070 because it's right triangle.

    if you calculate above
    Fx = 2*F1*cos(x) ----------- because |F1| = |F3|
    = 2*F1*(0.157/0.070)
    = 1.977302857 N
    and a_x = 1.977302857/0.005 = 395.4605717 m/sec^2
  4. Mar 31, 2004 #3
    And by the way, don't forget to tell me whether my calculation is right.

    Good luck. :biggrin:
  5. Mar 31, 2004 #4


    User Avatar
    Staff Emeritus
    Science Advisor

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

    Where did you get "70"? Since your right triangles have legs of length 7 and 14, the angle at the base (at 1 and 3) are arctan(14/7)= 63.4 degrees. The opposite angle is 90- 63.4= 26.6 degrees (and the angle at the top of the isosceles triangle is twice that: 53.2 degrees.
    The x-component of F4_1 is 0.4408* cos (63.4) and the y-component is
    0.4408* sin(64.4)
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