Question

Suppose that intensity of a laser is $\left(\frac{315}{\pi }\right)W/m^2.$ The rms electric field, in units of $V/m$ associated with this source is close to the nearest integer is
$\left({ϵ}_0=8.86\times 10^{-12}{C}^{2}N{m}^{-2};c=3\times 10^{8}m{s}^{-1}\right)$

Answer 1

Answer: 194.00

$I={ϵ}_0{E}_{rms}^{2}C$
$E_{rms}^2=\frac{I}{{ϵ}_{0}C}$
$=\frac{315}{\pi {ϵ}_0}\times \frac{1}{C}$
$=\frac{4\times 315}{4\pi {ϵ}_0}\times \frac{1}{3\times 10^8}$
$=\frac{4\times 315\times 9\times 10^9}{3\times 10^8}$
$E_{rms}^2=4\times 315\times 30$
$E_{rms}=2\sqrt{315\times 30}$
$=194.42$