Answer:
1341.03 V/m
Explanation:
The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.
[tex]S = \frac{P}{A}[/tex] = cε₀[tex]E^{2} _{rms}[/tex]
⇒ [tex]\frac{P}{A}[/tex] = cε₀[tex]E^{2}_{rms}[/tex]
Where;
P is the power output
A is the area of the beam
c is speed of light
ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m
[tex]E_{rms}[/tex] is the average (rms) value of electric field
Making electricfield [tex]E_{rms}[/tex] the subject of the equation
[tex]E^{2}_{rms}[/tex] = P / Acε₀
[tex]E_{rms}[/tex] = √(P / Acε₀)
But area A = πr²
[tex]E_{rms}[/tex] = √(P / πr²cε₀)
Given:
Output power, P = 15 mW = 0. 015 W
Diameter, d = 2 mm = 0.002 m
⇒ Radius, [tex]r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m[/tex]
Solving for average (rms) value of electric field;
[tex]E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }[/tex]
[tex]E_{rms}[/tex] = 1341.03 V/m
