A plane electromagnetic ware travelling in a $$non-magnetic$$ medium is given by $$E=9$$×$$108NC$$−$$1sin$$⁡$$9$$×$$108rads$$−$$1t$$−$$6m$$−$$1x$$ where x is in metre and t is in second. The dielectric constant of the medium is

$$v=$$ω$$k=9$$×$$1086=1.5$$×$$108ms$$−$$1Refractive$$ index $$n=cv=3$$×$$1081.5$$×$$108=2Also$$ $$n=$$μr∈r. For a $$non-magnetic$$ medium μ$$r=1$$.$$Thereforen=$$∈r⇒∈$$r=n2=(2)2=4$$∴ Dielectric constant $$k=$$∈$$r=4$$.

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A plane electromagnetic ware travelling in a non-magnetic medium is given by $E = (9 \times 10^{8} {NC}^{- 1}) \sin [(9 \times 10^{8} {rads}^{- 1}) t - (6 m^{- 1}) x]$ where x is in metre and t is in second. The dielectric constant of the medium is

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Solution

$v = \frac{ω}{k} = \frac{9 \times 10^{8}}{6} = 1 .5 \times 10^{8} {ms}^{- 1}$
Refractive index $n = \frac{c}{v} = \frac{3 \times 10^{8}}{1 .5 \times 10^{8}} = 2$
Also $n = \sqrt{μ_{r} \in_{r}}$ . For a non-magnetic medium $μ_{r} = 1$ .
Therefore
$n = \sqrt{\in_{r}} \Rightarrow \in_{r} = n^{2} = (2)^{2} = 4$
$∴$ Dielectric constant $k = \in_{r} = 4$ .

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Electromagnetic waves and its nature

Remember concepts with our Masterclasses.

A plane electromagnetic ware travelling in a non-magnetic medium is given by E=9×108NC−1sin⁡9×108rads−1t−6m−1x where x is in metre and t is in second. The dielectric constant of the medium is

v=ωk=9×1086=1.5×108ms−1Refractive index n=cv=3×1081.5×108=2Also n=μr∈r. For a non-magnetic medium μr=1.Thereforen=∈r⇒∈r=n2=(2)2=4∴ Dielectric constant k=∈r=4.

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A plane electromagnetic ware travelling in a non-magnetic medium is given by $E = (9 \times 10^{8} {NC}^{- 1}) \sin [(9 \times 10^{8} {rads}^{- 1}) t - (6 m^{- 1}) x]$ where x is in metre and t is in second. The dielectric constant of the medium is