Answer:
434.0 V/m
Explanation:
The power output of the laser is:
[tex]P=5 mW = 0.005 W[/tex]
while the radius of the beam is
[tex]r=\frac{5 mm}{2}=2.5 mm = 0.0025 m[/tex]
so the cross-sectional area is
[tex]A=\pi r^2 = \pi (0.0025 m)^2=2.0\cdot 10^{-5} m^2[/tex]
So the intensity of the laser beam is
[tex]I=\frac{P}{A}=\frac{0.005 W}{2.0\cdot 10^{-5} m^2}=250 W/m^2[/tex]
The intensity of a laser beam is related to the magnitude of the electric field by
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
Solving the formula for E, we find
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(250 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=434.0 V/m[/tex]
