Dielectric regions y 0 have relative permittivities 2 and 5. Region 1 has E 1=3 a x+4 a y+2 a z. With no free interfacial charge, the field in region 2 is

$1.2\hat a_x+4\hat a_y+2\hat a_z$

$1.2\hat a_x+4\hat a_y+0.8\hat a_z$

$3\hat a_x+10\hat a_y+0.8\hat a_z$

GATE 2015 EE (EE2) – Question 19

Electromagnetic Fields · Effect of dielectric medium, Capacitance of simple configurations · 1 mark · Multiple choice

Dielectric regions y<0 and y>0 have relative permittivities 2 and 5. Region 1 has $\mathbf E_1=3\hat a_x+4\hat a_y+2\hat a_z$. With no free interfacial charge, the field in region 2 is

$3\hat a_x+1.6\hat a_y+2\hat a_z$
$1.2\hat a_x+4\hat a_y+2\hat a_z$
$1.2\hat a_x+4\hat a_y+0.8\hat a_z$
$3\hat a_x+10\hat a_y+0.8\hat a_z$

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Show answer and explanation

Correct answer: (A) $3\hat a_x+1.6\hat a_y+2\hat a_z$

Explanation

At the plane y=0, x and z components are tangential and continuous: E2x=3 and E2z=2. The normal displacement is continuous without free surface charge.

Thus ε1E1y=ε2E2y, giving E2y=(2/5)4=1.6. The resulting field is **3ax+1.6ay+2az, A**.

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