In a medium of dielectric constant $K$, the electric field is ${\bf{E}}$. If ${\varepsilon _0}$ is permittivity of the free space, the electric displacement vector is

$\frac{{K{\bf{E}}}}{{{\varepsilon _0}}}$

$\frac{{\bf{E}}}{{K{\varepsilon _{0\;}}}}$

$\frac{{{\varepsilon _0}{{\bf{E}}_\;}}}{K}$

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NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

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In a medium of dielectric constant

K

, the electric field is

\mathbf{E}

. If

{\varepsilon _0}

is permittivity of the free space, the electric displacement vector is

View answer and explanation

Correct option: D

K{\varepsilon _0}\mathbf{E}

Option D (

K{\varepsilon _0}\mathbf{E}

) is correct because the electric displacement vector in a medium is given by the product of the dielectric constant, permittivity of free space, and the electric field.

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What is order of energy of X-rays (

{E_X}

), radio waves (

{E_R}

) and microwave (

{E_M}

)? The amplitude of the sinusoidally oscillating electric field of plane wave is

60\,\,\frac{V}{m}

. Then the amplitude of the magnetic field is An EM wave from air enters a medium. The electric fields are

{E_1} = {E_{01}}\hat x\cos \left[ {2\pi v\left( {\frac{z}{c} - t} \right)} \right]

in air and

{E_2} = {E_{02}}\hat x\cos \left[ {k\left( {2z - ct} \right)} \right]

in medium, where the wave number k and frequency ν refer to their values in air. The medium is non-magnetic.
If

{\varepsilon _{{r_1}}}

and

{\varepsilon _{{r_2}}}

refer to relative permittivities of air and medium respectively, which of the following options is correct? Electric field in a plane electromagnetic wave is given by

E = 50\sin \,(500x - 10 \times {10^{10}}t)\,v/m.

The velocity of electromagnetic wave in this medium is
(Given,

c

= speed of light in vacuum) If an electromagnetic wave is propagation in a medium with permeability 

\mathrm{\varepsilon }

and permittivity

\mathrm{\mu }

, then

\sqrt {\frac{\mathrm{\mu }}{\mathrm{\varepsilon }}}

is the In an E.M. wave the average energy density is associated with:-