Answer
(a) $I = 2.2\times 10^{-6}~W/m^2$
(b) $E_0 = 0.041~V/m$
Work Step by Step
(a) We can find the intensity of the light wave:
$I = \frac{P}{A}$
$I = \frac{P}{4\pi r^2}$
$I = \frac{2.5\times 10^4~W}{(\pi) (3.0\times 10^4~m)^2}$
$I = 2.21\times 10^{-6}~W/m^2$
$I = 2.2\times 10^{-6}~W/m^2$
(b) We can find the electric field amplitude:
$I = \frac{1}{2}~E_0^2~\epsilon_0~c$
$E_0^2 = \frac{2~I}{\epsilon_0~c}$
$E_0 = \sqrt{\frac{2~I}{\epsilon_0~c}}$
$E_0 = \sqrt{\frac{(2)(2.21\times 10^{-6}~W/m^2)}{(8.854\times 10^{-12}~F/m)~(3.0\times 10^8~m/s)}}$
$E_0 = 0.041~V/m$
