Answer
(a) $1.0mW/m^2$
(b) $0.71KN/C$
Work Step by Step
(a) We know that
$I_{\circ}=\frac{P}{A}$
We plug in the known values to obtain:
$I_{\circ}=\frac{1.5\times 10^{-3}}{\pi\times (6\times 10^{-4})^2}$
$I_{\circ}=1.3274\times 10^3$
Now $I=I_{\circ}e^{-\mu d}$
We plug in the known values to obtain:
$I=(1.3274\times 10^3)e^{-(470)(0.03)}$
$\implies I=9.98\times 10^{-4}W/m^2=1.0mW/m^2$
(b) We know that
$E=\sqrt{\frac{2I}{C\epsilon_{\circ}}}$
We plug in the known values to obtain:
$E=\sqrt{\frac{2\times 9.98\times 10^{-4}}{3\times 10^8\times 8.85\times 10^{-12}}}$
$E=0.71KN/C$
