VIPSolutionsAssume that the z=0 plane separates two lossless dielectric regions with εr1=2 and εr2=3. If we know that electric field in region 1 is ��̅ = 2��̂�� + 3��̂�� + 5��̂�� , determine ��̅2 =? in region 2 on the boundary.) IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VERY FEW SENTENCES AND MOREOVER I NEED COMPLETE AND CLEAR ANSWER at last explain what we did in each step in just few sentences
Question:
Assume that the z=0 plane separates two lossless dielectric regions with εr1=2 and εr2=3. If we know that electric field in region 1 is ��̅ = 2��̂�� + 3��̂�� + 5��̂�� , determine ��̅2 =? in region 2 on the boundary.) IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VERY FEW SENTENCES AND MOREOVER I NEED COMPLETE AND CLEAR ANSWER at last explain what we did in each step in just few sentences
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Asked by: Test User 100351
Created at: 2025-05-22 22:12:03
Asked by: Test User 100351
Created at: 2025-05-22 22:12:03
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---
**STEP 1: Introduction to the Concept**
When two dielectric media are separated by a boundary (here, the \( z=0 \) plane), the electric field \( \vec{E} \) and electric displacement \( \vec{D} \) satisfy certain boundary conditions:
- The **tangential component** of the electric field (\( \vec{E}_t \)) is **continuous** across the boundary:
\[
\vec{E}_{1t} = \vec{E}_{2t}
\]
- The **normal component** of the electric displacement (\( \vec{D}_n \)) is **continuous** (if no free surface charge is present):
\[
\vec{D}_{1n} = \vec{D}_{2n}
\]
where \( \vec{D} = \varepsilon \vec{E} \), and \( \varepsilon = \varepsilon_0 \varepsilon_r \).
---
**STEP 2: Given Data**
- Relative permittivity in region 1: \( \varepsilon_{r1} = 2 \)
- Relative permittivity in region 2: \( \varepsilon_{r2} = 3 \)
- Electric field in region 1: \( \vec{E}_1 = 2\hat{x} + 3\hat{y} + 5\hat{z} \) (V/m)
---
**STEP 3: Find Tangential and Normal Components**
At the boundary (\( z = 0 \)):
- **Tangential components:** \( E_x \) and \( E_y \)
- **Normal component:** \( E_z \)
So,
- Tangential: \( \vec{E}_{1t} = 2\hat{x} + 3\hat{y} \)
- Normal: \( E_{1n} = 5 \)
---
**STEP 4: Apply Boundary Conditions**
- **Tangential:** \( \vec{E}_{1t} = \vec{E}_{2t} \)
\[
E_{2x} = 2,\quad E_{2y} = 3
\]
- **Normal (Displacement):**
\[
D_{1n} = D_{2n}
\]
\[
\varepsilon_1 E_{1n} = \varepsilon_2 E_{2n}
\]
where \( \varepsilon_1 = \varepsilon_0 \varepsilon_{r1} \), \( \varepsilon_2 = \varepsilon_0 \varepsilon_{r2} \).
Plug in values:
\[
\varepsilon_0 \cdot 2 \cdot 5 = \varepsilon_0 \cdot 3 \cdot E_{2n}
\]
\[
10\varepsilon_0 = 3\varepsilon_0 E_{2n}
\]
\[
E_{2n} = \frac{10}{3}
\]
---
**STEP 5: Write the Final Electric Field in Region 2**
\[
\boxed{
\vec{E}_2 = 2\hat{x} + 3\hat{y} + \frac{10}{3}\hat{z} \ \text{(V/m)}
}
\]
---
**STEP 6: Final Answer in a Few Sentences**
The electric field just on the other side of the boundary (in region 2) is \( \vec{E}_2 = 2\hat{x} + 3\hat{y} + \frac{10}{3}\hat{z} \) V/m. We used boundary conditions to equate tangential electric field components and normal displacement components across the interface.
---
**Step Summary:**
- **Step 1:** Introduced boundary conditions for electric fields at dielectric interfaces.
- **Step 2:** Listed the given parameters.
- **Step 3:** Identified tangential and normal components of the field.
- **Step 4:** Applied boundary conditions to solve for unknowns.
- **Step 5:** Assembled the final answer.
- **Step 6:** Summarized the whole process and provided the concise result.
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