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Homework Help: Two spheres hanging, find charge.

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Two identical, small spheres of mass 2.0g are fastened to the ends of a 0.6m long light, flexible, insulating fishline. The fishline is suspended by a hook in the ceiling at it's exact center. The spheres are each given the same electric charge. They are in static equilibrium, creating an angle of 30° between the halves of the string. Calculate the magnitude of the charge on each sphere.

    Here's a sketch of the picture given on the worksheet:
    Red lines are what I drew over the diagram to make a right triangle.

    This being a bonus type question, I was given three possible answers and had to find out which one was the real one:

    q =
    3.7 x 10^-7 C or
    1.2 x 10^3 C or
    1.2 x 10^-3 C.

    2. Relevant equations

    Coulombs Law: [tex]F = k \ \frac{q_1 q_2}{r^2}[/tex]

    3. The attempt at a solution

    First I converted the mass into kilograms: m = 0.002kg

    Then I split the angle into two right triangles to get the distance between the spheres.

    r/2 = Sin15 x 0.3 m
    r = approx. 0.16m

    Then I figured since the spheres are at rest, then a = 0 and Fg = Fe...so I tried coulombs law to solve for q:

    [tex]F_{g} = F_{e}[/tex]
    [tex]mg = k \ \frac{q_1 q_2}{r^2}[/tex]
    [tex]\frac{mgr^2}{k} = q_1 q_2[/tex]

    I ended up with 2.6 x 10^-11...obviously way off any of the true answers.

    Thanks for taking the time to read over all this. =D

    Cheers - Krunklehorn
  2. jcsd
  3. Jan 30, 2012 #2
    You need to look at the forces acting on the balls.
    There is a vertical force = weight of a ball = 2x10^-3x9.81 = 0.0196N
    There is a sideways force, F, due to the repulsion.
    Can you see that F/0.0196 = Tan15
    This will give you F to use in the repulsion equation.
    I got Q = 1.2x10^-7C, on your list there is 1.2x10^-3C
    I use k = 8.98x10^9 in the repulsion equation
  4. Jan 30, 2012 #3
    Oh I see! Fe pushes the spheres to the side while Fg pushes them down to the ground.

    Tan = Opp / Adj so my equation should be:

    [tex]Tan15 Fg = Fe[/tex]

    [tex]\frac{Tan15mgr^2}{2k} = q[/tex]

    Awww too bad it's actually wrong. I am getting 7.03x10^-15 as an answer.
  5. Jan 30, 2012 #4
    Agree with technician. The force on one of the spheres is equal to the x-component of the fishline force(S):







    Last edited: Jan 30, 2012
  6. Jan 30, 2012 #5
    Oh I completely agree with him and appreciate the help immensely....I just don't see why I'm getting such a mangled answer.

    Did I rearrange the equation wrong?
  7. Jan 30, 2012 #6
    The equation


    can(as you know) be rewritten as


    Sum of forces in x-direction equals zero:




    Similarly in the y-direction:




    Now, from simple trigonometry you see that




    This means that you can write Sx as


    And from the equilibrium equation in the x-direction you see that


    Now you have Fe. You then need the distance r between the spheres:


    Filling all this information into the expression for q then gives you

    q=1.187*10-7 C
    Last edited: Jan 30, 2012
  8. Jan 30, 2012 #7
    Thanks for the help! I understand completely now.
  9. Jan 30, 2012 #8
    Great.....well done.....don't always believe published answers
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