GATE EC 2003 Electromagnetics Solution Q 8

GATE EC 2003 Electromagnetics Solution Q 8

Medium 1 has the electrical permittivity \varepsilon_1=1.5\varepsilon_0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity \varepsilon_2=2.5\varepsilon_0\;farad/m and occupies the region to the right of x = 0 plane.

If E1 in medium 1 is E_1=\left(2u_x-3u_y+1u_z\right)volt/m, then E2 in medium 2 is

(a) \left(2.0u_x-7.5u_y+2.5u_z\right)volt/m

(b) \left(2.0u_x-2.0u_y+0.6u_z\right)volt/m

(c) \left(1.2u_x-3.0u_y+1.0u_z\right)volt/m

(d) \left(1.2u_x-2.0u_y+0.6u_z\right)volt/m

Ans : (c)

Explanation

We know that according to the Dielectric-Dielectric Boundary conditions, tangential components of E and normal components od D are continuous. So splitting the given Electric field into tangential (x = 0 plane) and normal components(along z).

{\overrightarrow E}_1=2u_x-3u_y+1u_z {\overrightarrow E}<em>{1t}=-3u_y+u_z={\overrightarrow E}</em>{2t} {\overrightarrow E}_{1n}=2u_x {\overrightarrow D}<em>{1n}={\overrightarrow D}</em>{2n} \therefore \in_1{\overrightarrow E}<em>{1n}=\in_2{\overrightarrow E}</em>{2n} 1.5\in_0.2u_x=2.5\in_0.{\overrightarrow E}_{2n} \therefore {\overrightarrow E}_{2n}=\frac3{2.5}u_x=1.2u_x {\overrightarrow E}<em>2={\overrightarrow E}</em>{2t}+{\overrightarrow E}_{2n} {\overrightarrow E}_2=-3u_y+u_z+1.2u_x

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