The amplitude of the electric field associate | vedclass

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Question

(C) 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} c \epsilon_0

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The intensity $I$ of an electromagnetic wave is related to the amplitude of the electric field $E_0$ by the formula: $I = \frac{1}{2} c \epsilon_0 E_0^2$. Given: $I = \frac{15}{\pi} 	ext{ W m}^{-2}$, $c = 3 	imes 10^8 	ext{ m s}^{-1}$, and $\epsilon_0 = \frac{1}{36\pi} 	imes 10^{-9} 	ext{ F m}^{-1}$. Substituting the values: $\frac{15}{\pi} = \frac{1}{2} 	imes (3 	imes 10^8) 	imes (\frac{1}{36\pi} 	imes 10^{-9}) 	imes E_0^2$. $\frac{15}{\pi} = \frac{3 	imes 10^{-1}}{72\pi} 	imes E_0^2$. $\frac{15}{\pi} = \frac{1}{240\pi} 	imes E_0^2$. $E_0^2 = 15 	imes 240 = 3600$. $E_0 = \sqrt{3600} = 60 	ext{ N C}^{-1}$.

The amplitude of the electric field associated with a light beam of intensity $\frac{15}{\pi} \text{ W m}^{-2}$ is (in $\text{ N C}^{-1}$)

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