Question

Medium 1 has the electrical permittivity \(\epsilon_1=1.5\) farad/m and occupies the region to the left of \(x=0\) plane. Medium 2 has the electrical permittivity \(\epsilon_2=2.5\) farad/m and occupies the region to the right of \(x=0\) plane. If in medium 1 \[ \vec{E}_1=(2u_x-3u_y+1u_z)\ \text{Volt/m} \] then \(\vec{E}_2\) in medium 2 is :

• A. \((2.0u_x-7.5u_y+2.5u_z)\ \text{Volt/m}\)
• B. \((2.0u_x-2.0u_y+0.6u_z)\ \text{Volt/m}\)
• C. \((1.2u_x-3.0u_y+1.0u_z)\ \text{Volt/m}\)
• D. \((1.2u_x-2.0u_y+0.6u_z)\ \text{Volt/m}\)

Hint:

At dielectric boundary: \[ E_{t1}=E_{t2} \] \[ \epsilon_1E_{n1}=\epsilon_2E_{n2} \] Tangential electric field is continuous while normal electric flux density is continuous.

Answer 1

The Correct Option is D

Concept: At dielectric boundary:
• Tangential electric field remains continuous.
• Normal electric flux density remains continuous. Boundary conditions: \[ E_{t1}=E_{t2} \] \[ \epsilon_1E_{n1}=\epsilon_2E_{n2} \]

Step 1: Identify normal and tangential components. Boundary plane: \[ x=0 \] Hence:
• \(x\)-component is normal component
• \(y\)- and \(z\)-components are tangential components Given: \[ \vec{E}_1=(2u_x-3u_y+1u_z) \] Thus: \[ E_{n1}=2 \] Tangential components: \[ E_{t1}=-3u_y+1u_z \]

Step 2: Apply tangential field continuity. Tangential components remain unchanged: \[ E_{t2}=E_{t1} \] Therefore: \[ E_{y2}=-3 \] \[ E_{z2}=1 \]

Step 3: Apply normal flux continuity. Using: \[ \epsilon_1E_{n1}=\epsilon_2E_{n2} \] Substituting: \[ 1.5\times2=2.5\times E_{n2} \] \[ 3=2.5E_{n2} \] \[ E_{n2}=\frac{3}{2.5} \] \[ E_{n2}=1.2 \] Thus: \[ E_{x2}=1.2 \]

Step 4: Write final electric field vector. Therefore: \[ \vec{E}_2=(1.2u_x-3u_y+1u_z)\ \text{Volt/m} \] Hence, the correct option is: \[ \boxed{(C)\ (1.2u_x-3u_y+1u_z)\ \text{Volt/m}} \]