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Electric Energy and Spring Constant Problem

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Two spheres are mounted on identical horizontal springs and rest on a frcionless table, as in the drawing. When the spheres are uncharged, the spacing between them is 0.0500 m, and the springs are unrestrained. When each sphere has a charge of +1.60 microC, the spacing doubles. Assuming that the spheres have a negligible diameter, determine the spring constant of the springs.

    2. Relevant equations

    E=KE_t+KE_r+U_g+PE_sp+PE_e

    Energy is equal to the sum of all energies, translational kinetic, rotational kinetic, gravitational potentional, potential spring, and potential electric.

    Potential gravitational and Rotaional Kinetic are not applicable

    KE_t=(1/2mv^2)
    PE_sp=(1/2)kx^2
    PE_e=[k(q_1)(q_2)]/r^2

    3. The attempt at a solution

    E=KE_t+PE_sp+PE_e
    Ef=[(1/2)m_1f V_1f^2 +(1/2)M_2f V_2f^2]+[1/2kx_f^2]+[(k_e)(q^2)]/r_f^2]

    Ei=0+[1/2kx_i^2]+[(k_e)(q^2)]/r_i^2]

    Ef=Ei
    [(1/2)m_1f V_1f^2 +(1/2)M_2f V_2f^2]+[1/2kx_f^2]+[(k_e)(q^2)]/r-f^2]=[1/2kx_i^2]+[(k_e)(q^2)]/ri^2]

    (1/2)[m_1f V_1f^2 +M_2f V_2f^2]+[kx_f^2]+[(2k_e)(q^2)]/rf^2]= (1/2)[Kx_i^2]+[(2k_e)(q^2)]/r_i^2]

    (m1=m2)

    [mf Vf^2]+[kx_f^2]+[(2k_e)(q^2)]/rf^2]=[(2k_e)(q^2)]/r_i^2]

    Mv^2+x_f^2+(1/r_f^2)=(1/r_i^2)

    So then....

    (1/2) [m_1v_1f^2+m_2v_2f^2]+Kx_f^2+[(k_e)(q^2)]/r^2]=0

    K=-[[k_e(q^2)/r^2]-[mv_f^2]]/x^2

    K=-[(8.99e9)(2.56e6)]/(.01)]-m(1.73^2)

    k=-29.929(x+7.689)

    and i'm stuck....I don't know what I did wrong really...I don't know how to find the mass though....

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 4, 2007 #2
    Consider only one of the spheres. When the electric force pushes on it it does a certain amount of work, right? You could figure that out through integration (or if you dont know any calculus, I'm sure there is some formula you could use). Set that equal to the increase in potential energy in one of the springs and youve got your answer.
     
  4. Feb 4, 2007 #3
    That sounds like a fantastic idea. However, with my lack of knowledge in calculus and the limitations of algebra, I cannot find the EPE. It is not given in the problem. In algebra we only know EPE to be EPE=qV....the voltage is not given in the problem.

    In addition, this chapter of which the homework is assigned, does not cover EPE. Thus, there must be an alternative solution....any ideas?

    Thanks for your help though, I do appreciate it :)
     
  5. Feb 5, 2007 #4
    My friend from school enlightened me on this matter and it is properly performed with algebraic physics as follows.

    F=(kq_1q_2)/r^2

    K=(1/4piE_o) when E_0=8.85e-12

    F=[(K(1.6e-6)^20/(0.1^2)]
    F=2.302

    F=k_spx^2
    K=(f/x^2)
    K=(2.302/0.0025)
    K=920.8

    yay
     
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