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A square piece of foil with area A = 5.0 × 10^-4 m^2 and mass m = 5.0 × 10^-5 kg is suspended along one edge and hangs under its own weight. When light is incident perpendicular to the foil,the foil bends and reaches an equilibrium angle of 2 degrees. If the light is completely reflected by the foil, what is the peak value of the electric field associated with the light?

  • physics - ,

    The answer is 62 kV/m.

  • physics - ,

    I got...

    Torque (by radiation) = F dot R = A*P dot R = A*P cos(theta) R
    P = 0.5 * (electric constant) * E^2

    Torque (by gravity) = -mg sin(theta)

    Torque (by radiation) + Torque (by gravity) = 0

    Then i solved for E. Which gives me a wrong answer. I don't know why.

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