Question

The plane 𝑧=0 separates two linear dielectric media with relative permittivities 𝜀_𝑟1=4 and 𝜀_𝑟2=3, respectively. There is no free charge at the interface. If the electric field in the medium 1 is 𝐸⃗ ₁=3𝑥̂+2𝑦̂ +4𝑧̂, then the displacement vector \(\overrightarrow{D_2}\) in medium 2 is: (𝜀₀ is the permittivity of free space)

• A. (3𝑥̂ + 4𝑦̂ + 6𝑧̂)𝜀₀
• B. (3𝑥̂ + 6𝑦̂ + 8𝑧̂)𝜀₀
• C. (9𝑥̂ + 6𝑦̂ + 16𝑧̂)𝜀₀
• D. (4𝑥̂ + 2𝑦̂ + 3𝑧̂)𝜀₀

Answer 1

The Correct Option is C

Step 1: General conditions at the interface 

At the boundary between two dielectric media, the following conditions hold:

• The tangential components of the electric field (\( \vec{E_t} \)) are continuous:
• The normal components of the displacement field (\( \vec{D_n} \)) are continuous:

Step 2: Separate tangential and normal components of \( \vec{E_1} \)

The electric field in medium 1 is: \[ \vec{E_1} = 3\hat{x} + 2\hat{y} + 4\hat{z}. \] - Tangential components (\( \vec{E_{1t}} \)): These are the components parallel to the interface (\( z = 0 \)): \[ \vec{E_{1t}} = 3\hat{x} + 2\hat{y}. \] - Normal component (\( \vec{E_{1n}} \)): This is the component perpendicular to the interface (\( z \)-direction): \[ \vec{E_{1n}} = 4\hat{z}. \]

Step 3: Determine \( \vec{E_2} \) in medium 2

Using the continuity conditions:

• Tangential components remain unchanged:
• Normal components scale with the ratio of relative permittivities:

Step 4: Calculate \( \vec{D_2} \) in medium 2

The displacement vector \( \vec{D} \) is related to \( \vec{E} \) by: \[ \vec{D} = \epsilon_0 \epsilon_r \vec{E}. \] For medium 2 (\( \epsilon_r = 3 \)): \[ \vec{D_2} = \epsilon_0 \cdot 3 \cdot (3\hat{x} + 2\hat{y} + \frac{16}{3}\hat{z}). \] Distribute \( \epsilon_0 \cdot 3 \): \[ \vec{D_2} = \epsilon_0 (9\hat{x} + 6\hat{y} + 16\hat{z}). \]

Final Answer:

The displacement vector in medium 2 is: \[ \vec{D_2} = (9\hat{x} + 6\hat{y} + 16\hat{z})\epsilon_0. \]