In the four regions, I, II, III and IV, the electric fields are described as :

Region I : $E _{ x }= E _0 \sin ( kz -\omega t )$
Region II : $E _{ x }= E _0$
Region III : $E _{ x }= E _0 \sin kz$
Region IV : $E _{ x }= E _0 \cos kz$
The displacement current will exist in the region :

Question

In the four regions, I, II, III and IV, the electric fields are described as :

Region I : $E _{ x }= E _0 \sin ( kz -\omega t )$
Region II : $E _{ x }= E _0$
Region III : $E _{ x }= E _0 \sin kz$
Region IV : $E _{ x }= E _0 \cos kz$
The displacement current will exist in the region :

YESPYQ · Accepted Answer

Correct answer: a) I. Displacement current density, ${J}_{d}={\epsilon }_{0}\frac{dE}{dt}$ Region I : ${E}_{x}={E}_{0}\sin \left(kz-\omega t\right)\ \frac{d{E}_{x}}{dt}=-\omega {E}_{0}\cos (kz-\omega t)\ {J}_{d}=-{\epsilon }_{0}\omega {E}_{0}\cos (kz-\omega t)$ Region II : ${E}_{x}={E}_{0}\Rightarrow \frac{d{E}_{x}}{dt}=0\ {J}_{d}=0$ Region III : ${E}_{x}={E}_{0}\sin (kz)\ \frac{dE}{dt}=0\Rightarrow {J}_{d}=0$ Region IV : ${E}_{x}={E}_{0}\cos (kz)\ \frac{dE}{dt}=0\Rightarrow {J}_{d}=0$

In the four regions, I, II, III and IV, the electric fields are described as : Region I : \(E _{ x }= E _0 \sin ( kz -\o…

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