1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trouble Finding Charge

  1. Mar 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Two small balls each with a mass of 25.0 g, are attached to silk threads 1.00m long and hung from a common point. When the balls are given equal quantities of negative charge, each thread makes an angle of 20degrees with the vertical. Find the magnitude of the charge on each ball.

    2. Relevant equations
    E = F/q
    F = k*Q1*Q/r^2
    E = k*Q/r^2
    3. The attempt at a solution
    I got the Net Force in the Y-direction to be .6946 N and the distance between the two balls to be .684m, but when I rearrange the equations and solve for Q I get 8.8E-6, but the answer is 2.15E-6
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 16, 2008 #2

    hage567

    User Avatar
    Homework Helper

    Can you show how you got this? I don't think this is right.
     
  4. Mar 16, 2008 #3
    I drew the diagram with the two threads making a 20 degree angles with the vertical. Then since the ball makes a right angle, i got sin 70 degrees to be the Tension force and subtracted the force due to gravity (W = mass * gravity).
     
  5. Mar 16, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Anuja! Welcome to PF! :smile:

    Hint: The tension force is something the question doesn't give you and doesn't ask you for.

    So always try to find an equation which doesn't involve it!

    In this case, can you see a direction for taking components of force which avoids having to know what the tension force is? :smile:
     
  6. Mar 16, 2008 #5
    But wouldn't you use tension to find the y component of the force since Tension in the y-direction and W are going in opposite directions? I can't figure out what way to use the force for the equations.
     
  7. Mar 16, 2008 #6
    The X-component of the tensional force equals the electrostatic force between the two charged balls, but the Y-component of the tensional force would equal a ball's weight.

    I'm not sure why you're trying to calculate the electric field instead of the electric force.
    You have two equations and two unknowns.
     
    Last edited: Mar 16, 2008
  8. Mar 16, 2008 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes, you would, if you needed to know the y component of the force (I take it you mean vertical?).

    But you don't!

    You only need to know the force from the opposite charge.

    In fact, you only need to know its component in any one direction.

    Can you now see a direction which avoids having to know what the tension force is? :smile:
     
  9. Mar 16, 2008 #8
    When I draw the diagram, the 1 m thread serves as the hypotenuse of a right triangle on both sides of the vertical, which serves as a common leg between the two, with angles of 20 and 70 degrees. The two legs come out to .9396m and .342m. Since there are two triangles of the same size, I multiplied .342 by two, and they have .9396 in common.
    I hope that clears up what I was doing. What is the electrostatic equation?
     
  10. Mar 16, 2008 #9
    [tex]F=\frac{KQ_1Q_2}{r^2}[/tex]

    But you know that the x-component of the tension equals the electric force (the equation above), and you know that the y-component of the tension equals the weight of one of the balls.

    Since the charges are equal, the equation above can further be simplified to make [tex]Q_1Q_2=Q^2[/tex]

    You know the distance between the two charges, and you know the angle of the string, so solving for the charge should be trivial.
     
  11. Mar 16, 2008 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Simpler to take components perpendicular to the string (so the tension component is zero) - then electric-force x cos = weight x sin.
     
  12. Mar 16, 2008 #11
    That equation still leaves two variables, F and Q^2
     
  13. Mar 16, 2008 #12
    F is not a variable. F is equal to the x-component of the tensional force and Q is the variable you're trying to find.
     
  14. Mar 16, 2008 #13
    I'm sorry I'm getting so confused,but K = 9E9 and r^2 = (.684m)^2. Q^2 is the variable, but in the equation i'm not getting what F is. .245 is the Weight(y-component) but how do I get the x-component?
     
  15. Mar 16, 2008 #14
    If the y-component of the tension is the weight, ie. Tcos20 = mg, then the x-component of the tension, ie. Tsin20, equals the electric force, kQ^2/r^2

    So then if Tcos20 = mg, what does T equal? Thus, what does Tsin20 equal?
     
  16. Mar 16, 2008 #15
    Thank you Snazzy, tiny-tim, and hage5567 for helping me find the answer. I just don't see how I'm supposed to figure problems like these out, but Thank You again for showing me.
     
  17. Mar 17, 2008 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    … the short answer …

    Hi Anuja! :smile:

    Since you've obviously tried, I'll give you the short answer:

    Consider only the components of forces perpendicular to the string.

    Then:
    T.cos90º + F.cos20º = W.cos70º;

    but cos90º = 0,

    so F = W.tan20º :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trouble Finding Charge
  1. Trouble finding Torque (Replies: 3)

Loading...