Pri asked in Science & MathematicsPhysics · 1 decade ago

By how much does the spring stretch in cm?

Two 34*10^-6 C charges are attached to the opposite ends of a spring of spring constant 150 N/m and equilibrium length 50 cm .

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  • 1 decade ago
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    Original length = 50 cm = 0.5 m

    Let the spring stretch by distance x.

    Then new length = 0.5 + x

    Each charge Q = 34 * 10^-6 C

    Spring constant k = 150 N/m

    Electrostatic force = 9*10^9*Q^2/(0.5 + x)^2

    Tension in the spring = (1/2)k*x^2

    For equilibrium,

    9*10^9*Q^2/(0.5 + x)^2 = (1/2)k*x^2

    9*10^9* (34*10^-6)^2/(0.5 + x)^2 = (1/2)*150*x^2

    10.404/(0.5+x)^2 = 75 * x^2

    10.404/75 = x^2 * (0.5+x)^2

    Taking square root on both sides,

    sqrt(10.404/75) = x * (0.5 + x)

    0.372 = x * (0.5 + x)

    0.372 = 0.5 x + x^2

    x^2 + 0.5 x - 0.372 = 0

    x = [-0.5 +- sqrt(0.5^2 + 4*0.372)]/2

    x cannot be negative.

    x = [-0.5 + sqrt(1.738)]/2

    x = (-0.5 + 1.32)/2

    x = 0.82/2

    x = 0.41 m

    x = 41 cm

    Ans: 41 cm

  • 6 years ago

    check your formula it is F=kx

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