Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    Owokl.gif
    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):

    Sx=S*sin15

    and

    S=Sy/cos15

    Therefore

    q=1.19*10-7

    since

    Fe=m*g*tan15
     
    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

    Fe=(k*q2)/(r2)

    can(as you know) be rewritten as

    q=r*Sqrt[Fe/k]

    Sum of forces in x-direction equals zero:

    Fe-Sx=0

    therefore

    Fe=Sx

    Similarly in the y-direction:

    Sy-w=0

    therefore

    Sy=w=m*g

    Now, from simple trigonometry you see that

    Sy=S*cos15

    and

    Sx=S*sin15

    This means that you can write Sx as

    Sx=(Sy/cos15)*sin15=m*g*tan15

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

    Fe=m*g*tan15

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

    r=2*0.3*sin15

    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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook