Question

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 . At the window, the electric field of the wave has an rms value 0.0600 V/m .
How much energy does this wave carry through the window during a 30.0-s commercial? Express your answer with the appropriate units.

Answer 1

Answer:

The energy of the wave is 1.435 x 10⁻⁴ J

Explanation:

Given;

area of the window, A = 0.5 m²

the rms value of the field, E = 0.06 V/m

The peak value of electric field is given by;

$E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m$

The average intensity of the wave is given by;

$I_{avg} = \frac{c \epsilon_o E_o^2 }{2}\\\\I_{avg} = \frac{(3*10^8)( 8.85*10^{-12}) (0.0849)^2 }{2}\\\\I_{avg} = 9.569*10^{-6} \ W/m^2$

The average power of the wave is given by;

P = I x A

P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)

P = 4.784 x 10⁻⁶ W

The energy of the wave is given by;

E = P x t

E = (4.784 x 10⁻⁶ W)(30 s)

E = 1.435 x 10⁻⁴ J

Therefore, the energy of the wave is 1.435 x 10⁻⁴ J