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Coulomb's law problem

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Two balls, each with a mass of 500 mg, are attached to the same point in the ceiling by strings with the length of 40 cm. The balls have been given equal and opposite charges, which caused them to separate in a way that the strings formed a 60-degree angle. Find the charges and the tension force in the string. Make a sketch.

    The answers ought to be
    q=~0,2*10-6 C
    Ft=~0,0044 N

    2. Relevant equations

    Fg=mg
    FC=kq1q2/d2

    3. The attempt at a solution

    http://desmond.imageshack.us/Himg513/scaled.php?server=513&filename=fdafdafa.png&res=medium [Broken]

    An equilateral triangle is formed with the strings, so the distance between the balls is also
    d=40cm=0,4m

    Gravitational force on a ball
    Fg=mg=~5*10-3 N (g=9,8 m/s2)

    In order for equillibrium to be reached, the component vector of gravity, which is at a right angle with the string, must be equal to the component vector of Coulomb's force, which is also at a 90o with the string.
    That component vector of gravity is
    Fg1=Fg*cos60o=2,5*10-3

    So if FC=kq1q2/d2, then

    Fg1=FC*cos30o, from which we get that
    FC=5*sqrt3*10-3/3 N
    and that
    q1=q2=~0,2*10-6 C

    So far it seemed to me that everything went correctly, but I dont get the tension force to be 0,0044 N. Using sines and cosines on Fg and FC, or Pythagoran theorem for that matter, I got

    Ft=Ft1+Ft2
    Ft1=cos30o*Fg=5*sqrt3*10-3/2
    Ft2=cos60o*FC=2,5*sqrt3*10-3/3
    Ft=~0,0058 N

    Now I've seen many mistakes in answers in this textbook, but Im not so sure it's the textbook which has a fault in it this time. So please tell me, did I go wrong and where, or is the answer in the textbook once again wrong?

    (I didnt mark up all the angles, but they should be pretty easy to figure out. If you dont understand some of my calculations or think they are wrong, be sure to shout out or ask!)

    Thanks in advance
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 26, 2012 #2

    ehild

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    You made calculation errors. m=500 mg=0.5 kg. g=9.8. Force of gravity is 4.9 N.
    Do not round off the numerical results too early.

    ehild
     
  4. Mar 26, 2012 #3
    500 milligrams is equal to 0.0005 kg. So I cant have made a mistake there.
    The reason I rounded them up so early (I'm usually pretty precise) is because the answer given in the back is always calculated that way. I was just trying to get a close answer.

    Thanks in advance
     
  5. Mar 26, 2012 #4
    I would really appreciate if someone would solve the problem independently to see what answers they get. If they match with the book's, I must have made a theoretical error somewhere (I dont think its a calculation error since I've checked the numbers twice). If they match with my answers, it must be the book what contains the error.
    I hope it's not too much to ask.

    What confuses me though, is that the charge I get matches with the book's answer.

    Thanks in advance
     
  6. Mar 26, 2012 #5

    ehild

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    Sorry, I made the mistake!:redface:Your calculation is correct (althought I do not like that early rounding)

    It is an other way for the calculation using the right triangle made of the forces of gravity, Coulomb force and tension.


    ehild
     
  7. Mar 26, 2012 #6
    What you mean is I've made a mistake on calculating the magnitude of the tension? Could you specify please? Because I seem to miss it.

    fawk3s
     
  8. Mar 26, 2012 #7

    ehild

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    No, your calculation is right. But there is an easier way to get it:

    T=mg/cos(30°), (and tan(30°)=Fc/mg; )

    ehild
     
  9. Mar 26, 2012 #8
    Oh, alrighty. But does this mean you got the same answer, and I have to assume the textbook is wrong? Sorry, just have to ask again for confirmation.
     
  10. Mar 26, 2012 #9

    ehild

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    I have got the same answer:Tension=4.9x10-3/cos(30°)=5.7x10-3 N.

    ehild
     
  11. Mar 26, 2012 #10
    Thank you for your help ehild, I really appreciate it. :smile:
     
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