The electric field in a plane electromagnetic | vedclass

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Question

(B) The amplitude of the electric field is $E_{0} = 200 	ext{ V/m}$.The intensity $I$ of the electromagnetic wave is given by $I = \frac{1}{2} \varep

Vedclass · Accepted Answer

$354$

Vedclass · Answer

(B) The amplitude of the electric field is $E_{0} = 200 	ext{ V/m}$.The intensity $I$ of the electromagnetic wave is given by $I = \frac{1}{2} \varepsilon_{0} E_{0}^{2} c$.For a perfectly reflecting surface,the radiation pressure $P$ is given by $P = \frac{2I}{c}$.Substituting the expression for $I$ into the pressure formula: $P = \frac{2}{c} \left( \frac{1}{2} \varepsilon_{0} E_{0}^{2} c ight) = \varepsilon_{0} E_{0}^{2}$.Using $\varepsilon_{0} = 8.85 	imes 10^{-12} 	ext{ F/m}$,we calculate $P = 8.85 	imes 10^{-12} 	imes (200)^{2}$.$P = 8.85 	imes 10^{-12} 	imes 40000 = 8.85 	imes 4 	imes 10^{-8} = 35.4 	imes 10^{-8} = \frac{354}{10^{9}} 	ext{ N/m}^{2}$.Comparing this with $\frac{x}{10^{9}} 	ext{ N/m}^{2}$,we get $x = 354$.

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