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Electrostatics: 3 balls on a string, calculate net forces.

  1. Feb 17, 2007 #1
    1. The problem statement, all variables and given/known data
    3 balls are on a ?string?/line?. Each are equally displaced, 72cm. Ball A has a charge of -50e5c, Ball B a charge of +25e2, Ball C -10e10.
    1. What is the net force applied on the middle ball (Ball B)?
    2. Assuming a force to the right(ungiven name of Ball D?) (of Ball C) is positive, find the net force applied on Ball C..


    2. Relevant equations
    F=kq1q2/d^2
    k=9x10^9 n*m^2/c^2

    3. The attempt at a solution

    These are all theoretical values.. I know the basic structure of the problem, but totally guesstimating the values. With that said.. To try and get net force of Ball B..
    I did
    F(ab)=kq(a)q(b)/d^2
    F(bc)=kq(b)q(c0/d^2
    ...
    F(ab)+F(bc)=net force applied on Ball B.. I think that part was easy..
    net force=-4.34x10^24, work on screenshot
    [​IMG]
    ===
    Kind of lost for Part 2.. How do I find out the force for D? Or the charge of D? Or the distance (baby blue)?

    On the test, what I think I did was add charges A+B, set equal to C+D, or A+B-C=D.. Then use that value of charge for D, assume 72cm (I wanna know why I should assume), calculate force.. I don't think that's how it works though..
    ===
    I thought up an interesting theory just now.. All charges (balls) have same net forces, but vary +/-..?
    Here's another pic
    [​IMG]
    I forgot to draw it, but there's supposed to be a "blue vector equal to magenta vector".
     
    Last edited: Feb 17, 2007
  2. jcsd
  3. Feb 17, 2007 #2

    Doc Al

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    Staff: Mentor

    Did the problem actually state that there was a fourth ball? Or did you just assume that's what was meant? I'm guessing that part 2 was just to find the net force on ball C. (The stuff about to the right being positive--that's just so you use the correct sign convention.)
     
  4. Feb 17, 2007 #3
    Problem did not say 4th ball (I assumed so because there was a positive "force" to right of Ball C)

    PS, my 1st question, answered, love this forums.
    Loving to see that more people are willing to spend their time to help others. I may be inspired to be a good physics teacher :D

    So Doc Al, I added a 3rd edit.. thoughts?
     
    Last edited by a moderator: Feb 28, 2007
  5. Feb 17, 2007 #4

    Doc Al

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    Staff: Mentor

    My thoughts? Get rid of ball D!

    The force on ball C is just F(A on C) + F(B on C). Figure them out and add them up! (Since forces are vectors, directions--signs--matter.)
     
  6. Feb 17, 2007 #5
    Umm..
    Would that mean for net force of A, add Force(charge of A+C)+Force(charge a+b)?

    With Ball B, I supposed that I used the one on the left and the one on the right..

    Or am I not looking at it from the right perspective?

    How would this problem be different if there was no force on the right?
     
  7. Feb 18, 2007 #6

    Doc Al

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    Staff: Mentor

    There are three balls. They exert electrostatic forces on each other. To find the net force on any one (say ball A) add up the individual forces on that ball (A) from the other two (B & C). That means calculating the force that B exerts on A and the force that C exerts on A. (To find the individual forces between any two balls, use Coulomb's law.) Follow the same logic to find the net force on any of the balls.

    If I understand the problem correctly, there is no extra "force on the right"--that's just your misinterpretation of the instructions to use positive numbers to represent a force to the right (and negative numbers for forces to the left).
     
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