Surface x-0 separates two perfect dielectric with relative permittivities -r1 - 2 and -r2- 8 If E1 - 100 -x-200 -y - 50-z V-m for x -0 Then E2

From boundary nbsp;condition of dieletric dielectric medium. E_t_1=E_t_2 and D_n1 =D_n2 ε _r1E_n1 = ε _r2E_n2 E_n2= ε _r1ε r_2E_n1 = 28× 100 = 25 δ_x E_2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Electric Boundary Condition

Surface x=0 s...

Question

Surface x=0 separates two perfect dielectric with relative permittivities $ϵ_{r 1} = 2$ and $ϵ_{r 2} = 8$
If $_{1} = (100 δ_{x} + 200 δ_{y} - 50 δ_{z}) V / m$ for $x > 0$ Then $_{2}$

100 δ_{x} + 50 δ_{y} - 1.25 δ_{z}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

400 δ_{x} + 200 δ_{y} - 50 δ_{z}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 δ_{x} + 200 δ_{y} - 50 δ_{z}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100 δ_{x} + 800 δ_{y} - 200 δ_{z}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $25 δ_{x} + 200 δ_{y} - 50 δ_{z}$
From boundary condition of dieletric-dielectric medium.
${\to E}_{t_{1}}$ = ${\to E}_{t_{2}}$

and ${\to D}_{n 1} = {\to D}_{n 2}$

$ε_{r 1} {\to E}_{n 1} = ε_{r 2} {\to E}_{n 2}$

${\to E}_{n 2} = \frac{ε_{r 1}}{ε r_{2}} {\to E}_{n 1}$

$= \frac{2}{8} \times 100 = 25 δ_{x}$

$_{2} = 25 δ_{x} + 200 δ_{y} - 50 δ_{z}$

Suggest Corrections

Similar questions

Q. Medium 1 has the electrical permittivity

ε_{1} = 1.5 ε_{0}

farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity

ε_{2} = 2.5 ε_{0}

farad/m and occupies the region to the right of x = 0 plane. If

E_{1}

in medium 1 is

E_{1} = (2 u_{x} - 3 u_{y} + 1 u_{z})

volt/m, then

E_{2}

in meidum 2 is

Q. A current sheet

$\to J = 10 {^u}_{y} A / m$ lies on the dielectric interface x = 0 between two dielectric media with

$ε_{r_{1}} = 1, μ_{r_{1}} = 1$ in Region -1 (x < 0) and $ε_{r_{2}} = 2, μ_{r_{2}} = 2$ in Region-2 (x > 0). If the magneitc field in Region-1 at x = $0^{+}$ is ${\to H}_{1} = 3 {^u}_{x} + 30 {^u}_{y} A / m,$ the magnetic field in Region-2 at $x = 0^{+}$ is

Join BYJU'S Learning Program

Surface x=0 separates two perfect dielectric with relative permittivities ϵr1=2 and ϵr2=8 If −→E1=(100δx+200δy−50δz)V/m for x>0 Then −→E2

The correct option is C 25δx+200δy−50δzFrom boundary condition of dieletric-dielectric medium. →Et1=→Et2 and →Dn1=→Dn2 εr1→En1=εr2→En2 →En2=εr1εr2→En1 =28×100=25δx −→E2=25δx+200δy−50δz

Surface x=0 separates two perfect dielectric with relative permittivities $ϵ_{r 1} = 2$ and $ϵ_{r 2} = 8$
If $_{1} = (100 δ_{x} + 200 δ_{y} - 50 δ_{z}) V / m$ for $x > 0$ Then $_{2}$