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(B) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula: $I = \frac{1}{2} \epsilon_0 c

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$2\sqrt{3} 	imes 10^2 \, N/C$

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(B) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula: $I = \frac{1}{2} \epsilon_0 c E_0^2$.Rearranging for $E_0$,we get: $E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$.Given: $I = \frac{500}{\pi} \, W/m^2$,$\epsilon_0 = \frac{1}{36\pi 	imes 10^9} \, F/m$,and $c = 3 	imes 10^8 \, m/s$.Substituting the values:$E_0 = \sqrt{\frac{2 	imes (500/\pi)}{(1 / (36\pi 	imes 10^9)) 	imes 3 	imes 10^8}}$$E_0 = \sqrt{\frac{1000}{\pi} 	imes \frac{36\pi 	imes 10^9}{3 	imes 10^8}}$$E_0 = \sqrt{1000 	imes 12 	imes 10} = \sqrt{120000} = \sqrt{12 	imes 10^4} = 2\sqrt{3} 	imes 10^2 \, N/C$.

The intensity of light from a source is $\left( \frac{500}{\pi} \right) \, W/m^2$. Find the amplitude of the electric field in this wave.

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