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Coulombs law with statics

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Two small spheres of 15 g each are suspended from a common point by threads of length 35 cm. Each thread makes an angle with the vertical of 20 degrees. Each sphere carries the same charge. Find the magnitude of this charge.


    2. Relevant equations
    F=(kq1q2)/r^2


    3. The attempt at a solution
    Well I basically found out the force from each ball from F = mg, than I found out the distance between them with trigonometry which was 23.94 cm.
    I then plugged this into the equation to get:
    0.147 = kq^2/(23.94^2).
    I understand why this is wrong as this force is not the force between the two... I am a bit stuck on how you work backwards with this question so you get the force between the two first.
    Thanks :)
     
    Last edited: Nov 15, 2009
  2. jcsd
  3. Nov 15, 2009 #2

    rl.bhat

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    Homework Helper

    Hi physmania, welcome to PF.
    In the equilibrium position, three forces are acting on each sphere. Can you identify them?
    Once you do that, identify one of the forces which is neither horizontal nor the vertical. Resolve this force in to vertical and horizontal components. Find ΣFx and ΣFy. And proceed.
     
  4. Nov 15, 2009 #3
    Thaks for the reply.
    Is the equilibrium position the position it is in the question? As this is where the forces are keeping each ball.
    Are the forces magnetic, electric, and the work force?
    I did (mg x d) to get the work force, and then split it up into x and y. The x components cancelled out and the y component doubled.. I then used this force in the equation
    0.098 = k(q^2)/(23.9^2). I got it wrong.
    I got no idea really.
     
  5. Nov 15, 2009 #4

    ideasrule

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    There's no such thing as the "work force"; mgd is potential energy. mg is force.

    There are three forces: electrostatic repulsion, tension (in the strings), and gravity. Try drawing a free-body diagram on one of the balls and writing out Newton's second law for both the x and y directions.
     
  6. Nov 15, 2009 #5
    My bad.
    Alright i got it out... It still doesn't make complete sense to me though.
    I was first getting confused by finding the x and y components than finding the resultant of that for the tenision and then using the distance between the two spheres. You have to do them seperately and then find the resultant of the charges in the x and y direction.
    However, I don't understand how the forces from weight and the acceleration on one sphere are related to the electrostatic repulsion between the two? While working it out you only use one sphere, I guess this is because they are equal, but they are not added togethor or anything. What if one of the spheres weighed 50 grams and was at an angle of 40 degrees.
    Sorry im struggling to get my head around it haha.
    Thanks for the help!
     
  7. Nov 16, 2009 #6

    ideasrule

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    You do them separately, but you don't find the "resultant of the charges". You set Fnet=0 for both the x and y directions because the balls aren't moving.

    Electrostatic repulsion is F=kq1q2/r^2. Notice that it involves both charges.

    Are you familiar with free-body diagrams? You isolate each object and separately analyze the forces on it. For this question, you only need to analyze one ball because the two balls are identical. However, there's no harm in analyzing both to prove to yourself that it doesn't change the answer.
     
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