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?
2 answers
The answer is 62 kV/m.
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.
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.
0 answers