The equation of the electric field of an elec | vedclass

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Question

(B) The given electric field equation is $E = E_0 \sin(\omega t - kx)$,where $E_0 = \sqrt{377} 	ext{ V/m}$.The intensity (average power per unit area

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The given electric field equation is $E = E_0 \sin(\omega t - kx)$,where $E_0 = \sqrt{377} 	ext{ V/m}$.The intensity (average power per unit area) of an electromagnetic wave is given by $I = \frac{1}{2} c \varepsilon_0 E_0^2$.Since $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we can write $I = \frac{1}{2} \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \varepsilon_0 E_0^2 = \frac{1}{2} \sqrt{\frac{\varepsilon_0}{\mu_0}} E_0^2$.Given $\sqrt{\frac{\mu_0}{\varepsilon_0}} = 377$,then $\sqrt{\frac{\varepsilon_0}{\mu_0}} = \frac{1}{377}$.Substituting the values: $I = \frac{1}{2} 	imes \frac{1}{377} 	imes (\sqrt{377})^2 = \frac{1}{2} 	imes \frac{1}{377} 	imes 377 = \frac{1}{2} 	ext{ W/m}^2$.Comparing this with $\frac{1}{\alpha} 	ext{ W/m}^2$,we get $\alpha = 2$.

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