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(C) The intensity $I$ of an electromagnetic wave is related to the $rms$ electric field $E_{rms}$ by the formula: $I = \epsilon_0 E_{rms}^2 c$.Rearran

Vedclass · Accepted Answer

$194$

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(C) The intensity $I$ of an electromagnetic wave is related to the $rms$ electric field $E_{rms}$ by the formula: $I = \epsilon_0 E_{rms}^2 c$.Rearranging for $E_{rms}^2$,we get: $E_{rms}^2 = \frac{I}{\epsilon_0 c}$.Given $I = \frac{315}{\pi} \ W/m^2$,$\epsilon_0 = 8.86 	imes 10^{-12} \ C^2 N^{-1} m^{-2}$,and $c = 3 	imes 10^8 \ m/s$.We know that $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \ N m^2 C^{-2}$,so $\frac{1}{\epsilon_0} = 4\pi 	imes 9 	imes 10^9 = 36\pi 	imes 10^9$.Substituting these values into the equation:$E_{rms}^2 = \frac{315}{\pi} 	imes (36\pi 	imes 10^9) 	imes \frac{1}{3 	imes 10^8}$$E_{rms}^2 = 315 	imes 36 	imes 10^9 	imes \frac{1}{3 	imes 10^8}$$E_{rms}^2 = 315 	imes 12 	imes 10 = 37800$.Taking the square root: $E_{rms} = \sqrt{37800} \approx 194.42 \ V/m$.The nearest integer is $194$.

Suppose that the 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. Given: $\epsilon_0 = 8.86 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ and $c = 3 \times 10^8 \ m/s$.

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