The equation of the electric field of an electromagnetic wave propagating through free space is given
by: E=$$\sqrt{377}$$ $$\sin(6.27\times10^{3}t-2.09\times 10^{-5}x)N/C$$
The average power of the electromagnetic wave is $$\left(\frac{1}{\alpha}\right)W/m^{2}$$. The value of $$\alpha$$ is______
$$\left(Take \sqrt{\frac{\mu_{\circ}}{\epsilon_{\circ}}}=377 in SI units \right)$$

We are given the electric field $$E = \sqrt{377} \sin(6.27 \times 10^3 t - 2.09 \times 10^{-5} x)$$ N/C and must determine $$\alpha$$ such that the time‐averaged power density equals $$\frac{1}{\alpha}$$ W/m$$^2$$.

For a plane electromagnetic wave, the instantaneous power per unit area is described by the Poynting vector, and its time‐averaged value (or intensity) is

$$ \langle S \rangle = \frac{E_0^2}{2Z_0} $$

In this expression, $$E_0$$ denotes the amplitude of the electric field, while $$Z_0 = \sqrt{\mu_0/\epsilon_0}$$ represents the intrinsic impedance of free space. The factor of $$\tfrac{1}{2}$$ arises because $$\langle \sin^2(\omega t)\rangle = \tfrac{1}{2}$$ over a full cycle.

From the given wave, one reads off the amplitude $$E_0 = \sqrt{377}$$ N/C, and since $$Z_0 = \sqrt{\mu_0/\epsilon_0}$$ takes the value $$377\ \Omega$$ in SI units, substitution yields

$$ \langle S \rangle = \frac{E_0^2}{2Z_0} = \frac{(\sqrt{377})^2}{2 \times 377} = \frac{377}{2 \times 377} = \frac{1}{2} \text{ W/m}^2 $$

Because this result must equal $$\frac{1}{\alpha}$$, we set $$\frac{1}{\alpha} = \frac{1}{2}$$, which leads directly to $$\alpha = 2$$.

The answer is 2.