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;
[tex]E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m[/tex]
The average intensity of the wave is given by;
[tex]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[/tex]
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
