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Electric Force

  1. Dec 1, 2007 #1
    Two identical small spheres of mass 2 g are fastened to the ends of an insulating thread of length 0.6m. The spheres are suspended by a hook in the ceiling from the centre of the thread. The spheres are given identical electric charges and hang in static equilibrium with an angle of 30 degrees between the string halves. Calculate the magnitude of the charge on each sphere.

    So both the spheres are experiencing the gravitational force, electric force and tension force. And each side to wich each charge is hung is 0.3m. I am guessing we have to use trig since the angle is given. Also since both the spheres are given the same charges, there will be repulsion. We could use F=Kq1q2/r^2. (how do we find r)This is all I can get from this problem. Could someone help me with this?
     
  2. jcsd
  3. Dec 1, 2007 #2

    dst

    User Avatar

    Draw a force triangle. It's the tangent you want.
     
  4. Dec 1, 2007 #3
    umm..not clear..could you explain more please?
     
  5. Dec 1, 2007 #4

    dst

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    Draw a free body diagram of all the forces. You then use tan of a certain angle on there, multiplied by the weight of the sphere (19.6N) to get your force. From there it's manipulation of F=k(q1*q2)/r^2.
     
  6. Dec 1, 2007 #5
    umm..ok..so as i said, the forces acing the gravitational force, tension, and electrical force. So for a sphere, force acing down is mg, in the direction of the thread is the tension. And i hve the angle, SO i use trig to find tension force. Some of these forces is the electrical force? And what abt r? its the distance between the 2 spheres, we dont hav that..
     
  7. Dec 2, 2007 #6
  8. Dec 2, 2007 #7
  9. Dec 2, 2007 #8
    someone help please..
     
  10. Dec 2, 2007 #9
    Work out the 3 forces on one of the spheres vectorially. Since you are saying that they are in an equilibrium, then the sum of the 3 force vectors must be 0.
    Altough you will only to consider the vertical components of the force vectors, so you can only try to consider them, and not worry about the horizontal ones.
     
  11. Dec 2, 2007 #10
    umm..ok...so, Fg+Ft+Ff=0. Since Ff acts in the horizontal direction, Fg+Ft=0? Fg=mg ...I am still lost..
     
  12. Dec 2, 2007 #11
    you need to find the radius between the two spheres to find the force. To do this you should have 2 right triangles with one side 0.3m and angle of 15. Use the the sin ratio to find the opposite of your hypotenuse. Once you have the opposite side, multiply it by 2 (because that's the radius between the two spheres right?). You have your sphere now.

    The charges should both be of the same magnitude for an equal replusion so you may right q^2 when solving. so do your forces in component Fy = 0 so mg = Ft cos 15. You can use this to find Ft and plug it into your Fx = 0 formula which should be. Fe = Ft sin 15.

    Fe is kq^2/r^2.
     
  13. Dec 2, 2007 #12
    um..ok..got it..just one question, when you solved mg=Ftcos15, you used the mass of one one sphere or 2?
     
  14. Dec 2, 2007 #13
    He doesn't need to find the distance between the 2 spheres, it is already stated in the problem that they fastened to the ends of a thread with length L = 0.6 m. So you already know the electric repulsion between the 2 spheres, Fe = kq^2/L^2.
    And this expression: mg = Ft cos 15, is missing out the electric force component, and why do you take the angle to be 15? Isn't the angle that thread forms with the vertical 30 deg? Or am I completly misunderstanding this problem?
    I pictured something like:

    ''''|------- <- Ceiling
    ''''|''/.
    ''''|/
    ''''/
    ./

    Where the 2 spheres are the ".", (just ignore the ''''', it's there so the picture doesn't get ruined), and the 30 deg angle is what the thread (drawed as those "/") form with the vertical (the "|"). Am I having the wrong picture?

    Assuming the above picture is what this is about:
    pinkyjoshi65 start by drawing a sketch of the situation. If you consider the 'upper' sphere, the the vertical forces on it is the electric repulsion in the vertical direction which would be upwards, and downwards you have the vertical component of the tension force and the gravity. In equilibrium, the gravitional force plus the vertical component of the tension force should equal the vertical component of the electric force.
    From here you just need to solve an equation.
     
  15. Dec 2, 2007 #14
    umm..ok..the question states that the spheres are suspended by a hook in the ceiling from the centre of the thread. The length of the entire thread is 0.6m. Now since a hook is attached to the centre of the thread and fixed to the celing, the hypotunuse becomes half of the length of the thread. So now you have an isoceles triangle, with 2 sides equal to 0.3m. We do not know the base of the triangle, which is the distance between the 2 spheres. Hence we divide the triangle in 2 right triangles. Since the triangle is divided into 2, the angle is halfed. So the angle in each rt. triangle becomes 30 degrees.

    Remember that the electric force acts in the horizontal direction.

    So i think shawshank is right..makes sense..
     
  16. Dec 2, 2007 #15
    Here you have that the distance (let's call it r) between the spheres is 0.6 m, so now you 'know' the electric force, Fe = kq^2/r^2.

    What's this angle? Or where is this angle?
    Is this drawing of the situation incorrect?:

    ''''|------- <- Ceiling
    ''''|''/.
    ''''|/
    ''''/
    ./

    Where the dots represent the spheres.
    Isn't it the angle between the hook and the thread that is given?

    The electric force doesn't only act in the horizontal direction, it acts in the direction from sphere 1 to sphere 2.
     
  17. Dec 2, 2007 #16
    as far as I know sphere 1 and 2 are in the horizontal direction..
     
  18. Dec 2, 2007 #17
    also your diagram is wrong.
     
  19. Dec 2, 2007 #18
    "The spheres are suspended by a hook in the ceiling from the centre of the thread"

    So this isn't saying that the thread is hanging by a string attached to it's center?
    Then I don't think I can picture this situation. Do you have a sketch of it in your problem?
     
  20. Dec 2, 2007 #19
    Last edited: Dec 2, 2007
  21. May 11, 2008 #20
    im also stuck on this question..and ^ that diagram is wrong its actually 30degrees between the string halves.

    i came up with

    Fe=mgtan15=kq^2/r^2

    where r^2=0.02412 (derived from cosine law formula)

    i still dont get the right answer...i assumed that the horizontal component of the tension in the string is equal to the electrostatic force as they are at rest..what am i doing wrong?
     
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