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Help electric charge Coulomb's law two charges?

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    http://loncapa.vcu.edu/res/vcu/phys202/zzImages/two-charges-forces.bmp

    Problem#1:
    Two identical conducting balls, A and B, of identical masses m = 10 kg, are suspended in equilibrium by insulating massless strings in length L = 3 m. Both balls make the same angle θ = 30° with the vertical axis. Both masses have equal charge. You can ignore the size of the balls.

    What is the magnitude of the Coulomb force, i.e. electric force, exerted on A from B due to the charges? (From free body diagram analysis)
    What is the amount of charges (do not worry it is positive charge or negative charge) on each ball?



    Problem#2:
    Two identical conducting balls, A and B, of identical masses m = 40 kg, are suspended in equilibrium by insulating massless strings in length L = 2 m. Both balls make the same angle θ with the vertical axis. The angles are very small such that small angle approximation applies (i.e. tan θ = sin θ). Both masses have equal charge Q= 3e-06 C. You can ignore the size of the balls.

    What is the distance r between the two balls?


    2. Relevant equations
    F=q1*q2*k/(r^2)
    where k=9*10^9
    E=F/Q


    3. The attempt at a solution
    the only part i figured out (sadly) is the 1st part to find the Coulomb force. and that was only because the teacher gave us the equation :(
    ((40kg*10)/cos(30))*sin(30)=i got 57.7N for coulomb force
    still don't no why thats the answer, but at least i no how to get it.
    but i don't know what to do from here. i don't know how to find r so i assume it is L given (=3m). so i don't know how to find q (amt of charges).

    for problem#2 i am stumped. i don't understand angle approximation and i can't use the cos/sin method that our teacher gave us. so wat do i do??
     
  2. jcsd
  3. Feb 11, 2013 #2

    tms

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    The problem gave you one small-angle approximation. Another is [itex]\sin\theta \approx \theta[/itex].

    The first step to solve the problem is to draw a diagram showing the forces on each ball. From that calculate the horizontal forces. That equation will include [itex]r[/itex], which you can then solve for.
     
    Last edited by a moderator: Feb 11, 2013
  4. Feb 11, 2013 #3
    T(y)=mg
    F(t)=mg/cos(theta)
    F(e)=mg*tan(theta)

    do i need sin in one of the equations?? wat am i doing wrong. also i dont understand small angle approximation so how to find the angle if i know sin=tan?
     
  5. Feb 11, 2013 #4
    For problem #1, you can calculate r by recognising that the triangle made from the 2 strings and a line connecting the 2 charges is equilateral.
     
  6. Feb 11, 2013 #5

    tms

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    What are these supposed to be?
    For one thing, you are not using conventional orthography; ignoring the conventions hinders communication.
    What don't you understand about the approximations? Look at their series expansions, and think about what happens to the sine and cosine when the angle is very small.
     
  7. Feb 11, 2013 #6
    ohhh i see..yeah so now i kinda understand. cos approaches 1 at small angles, but i still dont know how to use the sin one.
    so the F(t) tension force would be (40*10)/cos(0)=400N
    but that means the total electric force is 0? because i multiply this by sin(0) and get 400*0=0.

    i'm just approaching this like i got the answer to the 1st part on problem#1, coz the two seem very similar. but i have a feeling im doing something wrong.
     
  8. Feb 11, 2013 #7

    tms

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    Again, look at the series expansion of sine (remembering that the expansions use radians, not degrees). I don't know how to make it clearer.
    That is the horizontal component of the tension. Except that the angle is not zero; it is small, but not zero. You will also need the vertical component,.
    Not at all. Again, the angle is small, but not zero.
    You will communicate better if you use conventional orthography and spelling. Things like 'kinda' and 'coz' are okay in casual speech, but writing is more formal, and such spellings and lack of capitals and so forth are at best distractions.
     
  9. Feb 11, 2013 #8
    series expansion?? all i know is that sin is like the inverse of cos, so sin(90) is same as cos(0). but i don't see how that helps me because i planned to use cos/tan like in the last one.

    for the vertical component in my FBD i got mass*gravity=400N. is that right?

    i still don't know how to find the very small angle so if i have to use cos(0.01) instead of cos(0).

     
  10. Feb 11, 2013 #9

    tms

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    The series expansion for sine is
    [tex]
    \begin{align}
    \sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \,x^{2n+1} \\
    &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots.
    \end{align}[/tex]
    The other circular functions have similar expansions. In this case, you want to eliminate all terms in x to a higher power than 1.

    The only vertical force is gravity. The tension has a vertical component, which must offset gravity. Since the angle is small ...

    As for the angle, that is what you are solving for. isn't it?
     
  11. Feb 12, 2013 #10
    i plugged in 1 into the first 3 terms of the sine expansion and got 0.842. this is angle right, not radians?

    for the coulomb force i got mass*gravity=(400N/cos(0.842))*sin(0.842)=5.88N

    5.88=Q^2*k/r^2
    r^2=(3e-06^2*9*10^9)/5.88
    r^2=0.081/5.88
    r=0.117m

    but i'm still gettin the wrong answer. what did i do wrong?
     
  12. Feb 12, 2013 #11

    tms

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    Just use the first term of the expansion. That leads to the approximation [itex]\sin x = x[/itex], as I said above.
     
  13. Feb 12, 2013 #12
    still getting the wrong answer..don't know what i'm doing wrong.

    400N*tan(1)=6.98

    Q=3e-06 C
    K=9x10^-9

    6.98=Q^2*K^2/R^2

    R^2=0.081/6.98
    R=0.108m
     
  14. Feb 12, 2013 #13

    tms

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    First, you can't just assume a particular value for theta. Use the approximations to find a relation between r and theta.
     
  15. Feb 12, 2013 #14
    sin(theta)=tan(theta)=theta=O/H=(r/2)/2m=r/4
    400N*(r/4)=Q^2*K^2/R^2
    400*(R^3)/4=0.081
    R^3=0.00081
    R=0.093

    did i do this right?
     
  16. Feb 12, 2013 #15

    tms

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    What are O and H?
    Why are you squaring k?
     
  17. Feb 12, 2013 #16
    sorry O/H is opposite/hypotenuse. in the right triangle i got r/2 for the opposite (since r is line from A to B) and L=2m for hypotenuse. and no sorry i wasn't squaring k that was a typo.
     
  18. Feb 24, 2013 #17
    can you plz just tell me how to get the answer? i have tried but i just dont understand it o.o
     
  19. Feb 25, 2013 #18

    tms

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    No; it is against the rules here to do that.

    Let's start over from the beginning. Draw the diagram showing all the forces. Each ball is acted on by the tension in the string, gravity, and the electrostatic force. Since the system in in equilibrium, all the forces on each ball will add up to zero. You want to break the forces into their vertical and horizontal components (since the setup is symmetric, you only have to do this for one side.

    So what are the components of the forces on the ball? You'll get two equations in two unknowns: [itex]T[/itex] and [itex]\theta[/itex]. Don't worry about the small-angle approximation yet.

    Use symbols everywhere, not numbers, so you can see more easily what is going on. Let [itex]\theta[/itex] be the angle between the string and the vertically down direction.
     
  20. Feb 26, 2013 #19
    all i know is electrostatic/coulomb force=mass*gravity*tan(θ). and since the problem says tan(θ)=sin(θ), force=mass*gravity*sin(θ)

    for the FBD..don't really know how to draw it. in the negative Y direction i get mass*gravity. don't know how to calculate string tension, i am assuming it is just L given.
     
    Last edited: Feb 26, 2013
  21. Feb 26, 2013 #20

    tms

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    There are three forces on each charged body: gravity, tension, and electrostatic. Gravity is down, electrostatic is horizontal, tension is at an angle [itex]\theta[/itex] to the vertical. Since the bodies are at rest, the forces must cancel. That is, the vertical component of the tension must be equal and opposite to gravity, and the horizontal component of the tension must be equal and opposite to the electrostatic force.
     
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