JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 18)
Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by
E = 20cos(2 $$\times$$ 1010 t $$-$$ 200x) V/m. The dielectric constant of the medium is equal to : (Take $$\mu$$r = 1)
E = 20cos(2 $$\times$$ 1010 t $$-$$ 200x) V/m. The dielectric constant of the medium is equal to : (Take $$\mu$$r = 1)
9
2
$${1 \over 3}$$
3
Explanation
Given, electric field,
E = 20 cos(2 $$\times$$ 1010t $$-$$ 200 x) V/m
Comparing with the standard equation,
E = E0 cos($$\omega$$t $$-$$ kx) V/m, we get
Wave constant, k = 200
Angular frequency, $$\omega$$ = 2 $$\times$$ 1010 rad/s
Speed of the wave, $$v = {\omega \over k} = {{2 \times {{10}^{10}}} \over {200}} = {10^8}$$ m/s
Refractive index, $$\mu = {c \over v} = {{3 \times {{10}^8}} \over {{{10}^8}}} = 3$$
As we know the relation between the refractive index and dielectric constant,
$$\mu = \sqrt {{\varepsilon _r}{\mu _r}} $$
Substituting the value in the above equations, we get
$$3 = \sqrt {{\varepsilon _r}(1)} $$
$${\varepsilon _r} = 9$$
Thus, the dielectric constant of the medium is 9.
E = 20 cos(2 $$\times$$ 1010t $$-$$ 200 x) V/m
Comparing with the standard equation,
E = E0 cos($$\omega$$t $$-$$ kx) V/m, we get
Wave constant, k = 200
Angular frequency, $$\omega$$ = 2 $$\times$$ 1010 rad/s
Speed of the wave, $$v = {\omega \over k} = {{2 \times {{10}^{10}}} \over {200}} = {10^8}$$ m/s
Refractive index, $$\mu = {c \over v} = {{3 \times {{10}^8}} \over {{{10}^8}}} = 3$$
As we know the relation between the refractive index and dielectric constant,
$$\mu = \sqrt {{\varepsilon _r}{\mu _r}} $$
Substituting the value in the above equations, we get
$$3 = \sqrt {{\varepsilon _r}(1)} $$
$${\varepsilon _r} = 9$$
Thus, the dielectric constant of the medium is 9.
Comments (0)
