The relation between electric field vector ${\bf{E}}$, the displacement vector ${\bf{D}}$ and the polarization vector ${\bf{P}}$ for a dielectric placed in electric field ${\bf{E}}$ is given by

${\bf{D}} = {{\rm{\varepsilon }}_0}{\bf{E}} + {\bf{P}}$

${\bf{P}} = {{\rm{\varepsilon }}_0}{\bf{E}} + {\bf{D}}$

${\bf{P}} = \;{\bf{D}} + {\bf{E}}$

${\bf{E}}\; = \;{\bf{D}} + {\bf{P}}$

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NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

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The relation between electric field vector

\mathbf{E}

, the displacement vector

\mathbf{D}

and the polarization vector

\mathbf{P}

for a dielectric placed in electric field

\mathbf{E}

is given by

View answer and explanation

Correct option: C

\mathbf{D} = {\mathrm{\varepsilon }_0}\mathbf{E} + \mathbf{P}

Option C (

\mathbf{D} = {\mathrm{\varepsilon }_0}\mathbf{E} + \mathbf{P}

) is correct because it represents the fundamental relationship between the electric displacement vector, electric field vector, and polarization vector in a dielectric material.

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What is order of energy of X-rays (

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{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.

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c

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\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:-