1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring Constant Due to 2 Charges

  1. Aug 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A tiny sphere with a charge of q = +6.1 µC is attached to a spring. Two other tiny charged spheres, each with a charge of -4.0 µC, are placed in the positions shown in the figure, in which b = 3.6 cm. The spring stretches 5.0 cm from its previous equilibrium position toward the two spheres. Calculate the spring constant.

    (See attachment)

    2. Relevant equations

    Coulomb's law:
    F = Kq1q2/r^2

    Spring Constant:
    F = kx

    3. The attempt at a solution

    I determine the net force due to to the individual force caused by F1 and F2:

    F1 = kq1q2/r^2 * < cos(a), -sin(a) >
    F2 = kq1q2/r^2 * < -cos(a), -sin(a) >

    Fnet = F1 + F2

    Where q1 = 4E-6 C, q2 = 6.1E-6 C, r = .0412 m ( by Pythagorean theorem) , a = 61 degrees

    Fnet = < 0, -3.8E-1 >
    ||Fnet|| = 3.8E-1 N ; This is force due to the two charges pulling down on the spring.

    F = kx
    Sub in values:

    3.8E-1 = k*(.036) ===> k = 11

    The spring constant is 11 N/m ?
     

    Attached Files:

  2. jcsd
  3. Aug 25, 2007 #2
    for Fnet = < 0, -3.8E-1 >

    are you sure that this is correct?

    Edit: because each time I do the calculations I get like > 100 N or about that
     
  4. Aug 25, 2007 #3

    learningphysics

    User Avatar
    Homework Helper

    Yeah, I get kq1q2/r^2 for one of the spheres to be 129.48N, then the vertical component is 129.48*sin(61) = 113.2N

    Then 2 of them so 2*113.2N = 226.5N

    226.5N = kx
    226.5N = k(0.05m)

    k = 4530N/m ?
     
    Last edited: Aug 25, 2007
  5. Aug 25, 2007 #4
    Thanks. You're absolutely right. I was making it more complicated than it was with the vectors which should of gotten me the right answer but I goofed up. The net force pulling the spring down is simply a vector pointing straight down. Here's what I was thinking:

    kq1q2/r^2 * [ <cos(a), -sin(a)> + <-cos(a), -sin(a)>]
    = kq1q2/r^2 * <0, -2sin(a)>
    <0, -226.5 >

    Take the magnitude of that vector to find k:

    226.5 = k ( 0.05 m ) ==> k = 4530 N/m

    Thank you again.
     
  6. Aug 26, 2007 #5
    hey,
    what book you are using?
     
  7. Aug 26, 2007 #6
    College Physics

    Giambattista
    Richardson
    Richardson
     
  8. Aug 26, 2007 #7
    thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?