Question 163: EM waves

Question

The electric field of a plane electromagnetic wave, travelling in an unknown nonmagnetic medium is given by,
$$ E_{\mathrm{y}}=20 \sin \left(3 	imes 10^6 x-4.5 	imes 10^{14} \mathrm{t}ight) \mathrm{V} / \mathrm{m} $$
(where $x, \mathrm{t}$ and other values have S.I. units). The dielectric constant of the medium is $\_\_\_\_$
(speed of light in free space is $3 	imes 10^8 \mathrm{~m} / \mathrm{s}$ )

Accepted Answer

The general standard mathematical form for a plane wave traveling in the positive x -direction is:
$$ \mathrm{y}(\mathrm{x}, \mathrm{t})=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}) $$
Where :
A is the amplitude (maximum displacement).
k is the angular wave number, related to the wavelength $(\lambda)$ by $\mathrm{k}=\frac{2 \pi}{\lambda}$.
$\omega$ is the angular frequency, related to the time period (T) by $\omega=\frac{2 \pi}{T}$.
The speed of propagation of traveling wave is given as,
$$ v=\frac{\omega}{k} $$
From Maxwell's equations, the speed of an electromagnetic wave in a vacuum (c) depends on the fundamental constants of free space:
$$ c=\frac{1}{\sqrt{\mu_0 \epsilon_0}} $$
Where $\mu_0$ is the permeability of free space and $\epsilon_0$ is the permittivity of free space.
When light travels through a material medium, its speed (v) changes depending on the magnetic and electrical properties of that specific medium :
$$ v=\frac{1}{\sqrt{\mu \epsilon}} $$
Where $\mu$ is the absolute permeability and $\epsilon$ is the absolute permittivity of the medium.
Relative to a vacuum :
$\mu=\mu_0 \mu_{\mathrm{r}}$ (where $\mu_{\mathrm{r}}$ is relative permeability)
$\epsilon=\epsilon_0 \epsilon_{\mathrm{r}}$ (where $\epsilon_{\mathrm{r}}$ is relative permittivity, also called the dielectric constant, often denoted as K )
$\mathrm{v}=\frac{1}{\sqrt{\left(\mu_0 \mu_{\mathrm{r}}ight)\left(\epsilon_0 \epsilon_{\mathrm{r}}ight)}}$
$$\Rightarrow $$ $v=\frac{1}{\sqrt{\mu_0 \epsilon_0}} \cdot \frac{1}{\sqrt{\mu_r \epsilon_r}}$
$$\Rightarrow $$ $$ v=\frac{c}{\sqrt{\mu_r \epsilon_r}} $$
As the medium is non-magnetic so its relative permeability is exactly 1 ( $\mu_{\mathrm{r}}=1$ ).
$$ \mathrm{v}=\frac{\mathrm{c}}{\sqrt{1 	imes \epsilon_{\mathrm{r}}}} $$
$$\Rightarrow $$ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\epsilon_{\mathrm{r}}}}$
$$\Rightarrow $$ $\sqrt{\epsilon_{\mathrm{r}}}=\frac{\mathrm{c}}{\mathrm{v}}$
$$\Rightarrow $$ $$ \epsilon_{\mathrm{r}}=\left(\frac{\mathrm{c}}{\mathrm{v}}ight)^2 $$
From the given electric field equation :
$$ E_y=20 \sin \left(3 	imes 10^6 x-4.5 	imes 10^{14} tight) $$
By comparing this to the standard form $\mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \sin (\mathrm{kx}-\omega \mathrm{t})$ :
$\mathrm{k}=3 	imes 10^6 \mathrm{rad} / \mathrm{m}$
$\omega=4.5 	imes 10^{14} \mathrm{rad} / \mathrm{s}$
The speed of the wave in the medium (v) :
$$ \mathrm{v}=\frac{\omega}{\mathrm{k}} $$
$$\Rightarrow $$ $\mathrm{v}=\frac{4.5 	imes 10^{14}}{3 	imes 10^6}$
$$\Rightarrow $$ $$ \mathrm{v}=1.5 	imes 10^8 \mathrm{~m} / \mathrm{s} $$
We are given $c=3 	imes 10^8 \mathrm{~m} / \mathrm{s}$.
$$ \epsilon_{\mathrm{r}}=\left(\frac{\mathrm{c}}{\mathrm{v}}ight)^2 $$
$$\Rightarrow $$ $\epsilon_{\mathrm{r}}=\left(\frac{3 	imes 10^8}{1.5 	imes 10^8}ight)^2$
$$\Rightarrow $$ $$ \epsilon_{\mathrm{r}}=(2)^2=4 $$
Therefore, the dielectric constant of the medium is 4 .
Hence, the correct answer is 4 .

EM waves

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