Question

The amplitude of the electric field of a parallel beam of electromagnetic wave of intensity 8 wm2 is

Answer 1

so, here is the answer of your question mate. I hope it will help you.

Answer 2

The amplitude of the electric field $E_{0}= 77.6\dfrac{N}{C} or NC^{-1}$

Explanation:

We have,

The intensity of a parallel beam of electromagnetic wave, $I=8\dfrac{W}{m^{2}}$

To find, the amplitude of the electric field $E_{0}=$

We know that,

The intensity of the electric field of a parallel beam, $I=(\dfrac{2I}{\phi_{0} C} )^{\dfrac{1}{2}}$

where, C = 3 × $10^{8} \dfrac{m}{s}$

$=(\dfrac{2\times 8}{8.85\times 10^{-12}\times  3\times 10^{8}})^{\dfrac{1}{2}}$

$=(\dfrac{18}{8.85\times 10^{-4}\times  3})^{\dfrac{1}{2}}$

= 77.6$\dfrac{N}{C}$ or $NC^{-1}$

Hence, the amplitude of the electric field $E_{0}= 77.6\dfrac{N}{C} or NC^{-1}$