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A small sphere of charge q1 = 0.840 µC hangs from the end of a spring as in Figure a. When another small sphere of charge q2 = -0.546 µC is held beneath the first sphere as in Figure b, the spring stretches by d = 3.18 cm from its original length and reaches a new equilibrium position with a separation between the charges of r = 4.95 cm. What is the force constant of the spring?

  • physics -

    Use Hooke's Law: |F| = k |d|
    Use Coulomb's Law: |F| = ke |q1 q2| / r^2
    Equate forces and rearrange.
    k = ke |q1 q2|/(|d| r^2)

    Where:
    q1 = 0.840×10^(-6) C
    q2 = -0.546×10^(-6) C
    d = 3.18×10^(-2) m
    r = 4.95×10^(-2) m
    ke = 8.9875517873681764×10^9 N·m^2/C^2 (Coulomb's Constant)

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