The electric field in an electromagnetic wave is given by $$E = 56.5 \sin\omega\left(\frac{t - x}{c}\right)$$ NC$$^{-1}$$. Find the intensity of the wave if it is propagating along $$x$$-axis in the free space. (Given $$\varepsilon_0 = 8.85 \times 10^{-12}$$ C$$^2$$ N$$^{-1}$$ m$$^{-2}$$)

We are given the electric field $$E = 56.5 \sin\omega\left(\frac{t - x}{c}\right)$$ NC$$^{-1}$$ for a wave propagating along the $$x$$-axis in free space. This expression matches the form $$E = E_0 \sin(\omega t - kx)$$ with $$k = \frac{\omega}{c}$$, so the amplitude is $$E_0 = 56.5 \text{ NC}^{-1}$$.

The intensity (average power per unit area) of an electromagnetic wave in free space is given by the time-averaged Poynting vector as $$I = \frac{1}{2}\varepsilon_0 c E_0^2$$, where the factor $$\frac{1}{2}$$ arises from averaging $$\sin^2$$ over a full cycle.

Squaring the amplitude yields $$E_0^2 = (56.5)^2 = 3192.25 \text{ N}^2\text{C}^{-2}$$, while the product $$\varepsilon_0 c = 8.85 \times 10^{-12} \times 3 \times 10^{8} = 26.55 \times 10^{-4} = 2.655 \times 10^{-3} \text{ C}^2\text{N}^{-1}\text{m}^{-1}\text{s}^{-1}$$.

Substituting these values into the intensity formula gives

$$I = \frac{1}{2} \times 2.655 \times 10^{-3} \times 3192.25$$

$$= \frac{1}{2} \times 8.4754$$

$$= 4.2377 \approx 4.24 \text{ Wm}^{-2}$$.

The correct answer is Option C: $$4.24$$ Wm$$^{-2}$$.