A plane electromagnetic wave in a non-magnetic dielectric medium is given by $$\vec{E} = \vec{E_0}(4 \times 10^{-7}x - 50t)$$ with distance being in meter and time in seconds. The dielectric constant of the medium is :

We are given the electric field of a plane electromagnetic wave in a non-magnetic dielectric medium: $$\vec{E} = \vec{E_0} \sin(4 \times 10^{-7}x - 50t)$$ where distance is in meters and time in seconds. The medium is non-magnetic, meaning its relative permeability is 1.

First, we recognize the general form of a plane wave: $$\vec{E} = \vec{E_0} \sin(kx - \omega t)$$ where $$k$$ is the wave number and $$\omega$$ is the angular frequency. Comparing this with the given wave, we identify:

• Wave number, $$k = 4 \times 10^{-7}$$ rad/m
• Angular frequency, $$\omega = 50$$ rad/s

The speed of the wave in the medium, denoted by $$v$$, is related to $$k$$ and $$\omega$$ by the formula: $$v = \frac{\omega}{k}$$ Substituting the values: $$v = \frac{50}{4 \times 10^{-7}} = \frac{50}{4} \times 10^{7} = 12.5 \times 10^{7} = 1.25 \times 10^{8} \text{ m/s}$$

We know the speed of light in vacuum, $$c = 3 \times 10^{8}$$ m/s. The refractive index $$n$$ of the medium is given by the ratio of the speed of light in vacuum to the speed in the medium: $$n = \frac{c}{v}$$ Substituting the values: $$n = \frac{3 \times 10^{8}}{1.25 \times 10^{8}} = \frac{3}{1.25} = 2.4$$

For a non-magnetic dielectric medium, the relative permeability $$\mu_r = 1$$. The refractive index $$n$$ is related to the dielectric constant $$\kappa$$ (relative permittivity) by: $$n = \sqrt{\kappa \mu_r} = \sqrt{\kappa \cdot 1} = \sqrt{\kappa}$$ Therefore, solving for $$\kappa$$: $$\kappa = n^{2}$$ Substituting $$n = 2.4$$: $$\kappa = (2.4)^{2} = 5.76$$

The calculated dielectric constant is 5.76. Comparing with the given options:

• A. 2.4
• B. 5.8
• C. 8.2
• D. 4.8

We see that 5.76 is closest to 5.8. Therefore, the dielectric constant of the medium is approximately 5.8.

Hence, the correct answer is Option B.