Question

Consider two media 1 and 2 having permittivities \(\epsilon_0\) and \(\epsilon_2 (= 2\epsilon_0)\), respectively. The interface between the two media aligns with the x-y plane. An electric field \(\mathbf{E}_1 = 4\hat{i} - 5\hat{j} - \hat{k}\) exists in medium 1. The magnitude of the displacement vector \(\mathbf{D}_2\) in medium 2 is \rule{1cm{0.15mm} \(\epsilon_0\). (up to two decimal places)}

Hint:

Remember the mnemonic "E-tangential is continuous, D-normal is continuous" for dielectric boundaries without free charges. First, identify the normal and tangential directions based on the interface plane. Then, apply the conditions component-wise to find the fields in the second medium.

Answer 1

Correct Answer: 12.85

Step 1: Understanding the Concept:
This problem requires the application of boundary conditions for electromagnetic fields at the interface between two linear dielectric media. The key principle is that the tangential component of the electric field (\(\mathbf{E}\)) and the normal component of the electric displacement (\(\mathbf{D}\)) are continuous across a boundary with no free surface charge.
Step 2: Key Formula or Approach:
The boundary conditions at the interface (\(z=0\)) are: 1. Tangential component of \(\mathbf{E}\) is continuous: \(\mathbf{E}_{1t} = \mathbf{E}_{2t}\) 2. Normal component of \(\mathbf{D}\) is continuous: \(D_{1n} = D_{2n}\) The relation between \(\mathbf{D}\) and \(\mathbf{E}\) is \(\mathbf{D} = \epsilon \mathbf{E}\).
The interface is the x-y plane, so the normal vector is \(\hat{k}\) and tangential vectors lie in the x-y plane.
Step 3: Detailed Explanation:
Given: Permittivity of medium 1: \(\epsilon_1 = \epsilon_0\) Permittivity of medium 2: \(\epsilon_2 = 2\epsilon_0\) Electric field in medium 1: \(\mathbf{E}_1 = 4\hat{i} - 5\hat{j} - \hat{k}\) First, we separate \(\mathbf{E}_1\) into its tangential and normal components: Tangential component (parallel to x-y plane): \(\mathbf{E}_{1t} = 4\hat{i} - 5\hat{j}\) Normal component (perpendicular to x-y plane): \(\mathbf{E}_{1n} = -1\hat{k}\) Apply the boundary condition for the tangential \(\mathbf{E}\) field: \[ \mathbf{E}_{2t} = \mathbf{E}_{1t} = 4\hat{i} - 5\hat{j} \] Now, we find the electric displacement vector \(\mathbf{D}_1\) in medium 1: \[ \mathbf{D}_1 = \epsilon_1 \mathbf{E}_1 = \epsilon_0 (4\hat{i} - 5\hat{j} - \hat{k}) \] Separate \(\mathbf{D}_1\) into its normal component: Normal component: \(D_{1n} = -\epsilon_0\) (the coefficient of \(\hat{k}\)) Apply the boundary condition for the normal \(\mathbf{D}\) field: \[ D_{2n} = D_{1n} = -\epsilon_0 \] We now have the components to construct the vector \(\mathbf{D}_2\). The vector \(\mathbf{D}_2\) is composed of its tangential part \(\mathbf{D}_{2t}\) and its normal part \(D_{2n}\hat{k}\). \[ \mathbf{D}_{2t} = \epsilon_2 \mathbf{E}_{2t} = (2\epsilon_0) (4\hat{i} - 5\hat{j}) = 8\epsilon_0\hat{i} - 10\epsilon_0\hat{j} \] The normal part is \(D_{2n}\hat{k} = -\epsilon_0\hat{k}\). So, the full displacement vector in medium 2 is: \[ \mathbf{D}_2 = \mathbf{D}_{2t} + D_{2n}\hat{k} = (8\epsilon_0\hat{i} - 10\epsilon_0\hat{j}) - \epsilon_0\hat{k} = \epsilon_0 (8\hat{i} - 10\hat{j} - \hat{k}) \] The question asks for the magnitude of \(\mathbf{D}_2\): \[ |\mathbf{D}_2| = |\epsilon_0 (8\hat{i} - 10\hat{j} - \hat{k})| = \epsilon_0 \sqrt{8^2 + (-10)^2 + (-1)^2} \] \[ |\mathbf{D}_2| = \epsilon_0 \sqrt{64 + 100 + 1} = \epsilon_0 \sqrt{165} \] Calculating the numerical value: \[ \sqrt{165} \approx 12.8452... \] The magnitude of \(\mathbf{D}_2\) in units of \(\epsilon_0\), up to two decimal places, is 12.85. Step 4: Final Answer:
The magnitude of the displacement vector \(\mathbf{D}_2\) is 12.85\(\epsilon_0\).