Respuesta :
Answer:
a) 0.477 W/m²
b) 13.407 N/C
c) 18.96 N/C
Explanation:
P = Power = 150 W
r = Distance = 5 m
ε₀ = Permittivity of space = 8.854×10⁻¹² F/m
a) Average intensity
[tex]\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2[/tex]
∴ Average intensity is 0.477 W/m²
b) Rms value
[tex]\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C[/tex]
∴ Rms value of the electric field is 13.407 N/C
c) Peak value
[tex]E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C[/tex]
∴ Peak value of the electric field is 18.96 N/C
The average intensity of the light, rms value of the electric field and peak value of electric field are respectively; 0.477 W/m²; 13.407 N/C; 18.96 N/C
What is the value of the electric field?
We are given;
Power; P = 150 W
Distance; r = 5 m
A) Formula for average intensity is;
I = P/A
Where A is area = 4πr² = 4π(5²)
A = 100π
I = 150/(100π)
I = 0.477 W/m²
B) Formula for the rms value of the electric field is;
E_rms = √(I/(ε₀ * c))
where;
ε₀ is permittivity of free space = 8.854 × 10⁻¹² F/m
c is speed of light = 3 * 10⁸ m/s
Thus;
E_rms = √(0.477/(8.854 × 10⁻¹² * 3 * 10⁸))
E_rms = 13.407 N/C
C) The peak value of the electric field is;
E₀ = E_rms * √2
E₀ = 13.407 * √2
E₀ = 18.96 N/C
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