\n– The interface between the two dielectric media D1 and D2 is the x-z plane, which is defined by $y=0$. The normal vector to this interface is along the y-axis, i.e., $\\hat{j}$.
\n– Medium D1 has dielectric constant $\\epsilon_{r1} = 3$, and medium D2 has dielectric constant $\\epsilon_{r2} = 2$. Let’s assume D1 is in the region $y<0$ and D2 is in the region $y>0$.
\n– The electric field in D1 is given as $\\vec{E}_1 = 3\\hat{i} + 2\\hat{j}$.
\n– At the boundary between two dielectric media, two boundary conditions for the electric field and displacement field must be satisfied:
\n 1. The tangential component of the electric field ($\\vec{E}_{tan}$) is continuous across the boundary: $\\vec{E}_{1,tan} = \\vec{E}_{2,tan}$.
\n 2. The normal component of the electric displacement field ($\\vec{D}_{norm}$) is continuous across the boundary (assuming no free charges on the surface): $\\vec{D}_{1,norm} = \\vec{D}_{2,norm}$.
\n– The tangential components of the electric field are those parallel to the x-z plane (along $\\hat{i}$ and $\\hat{k}$). From $\\vec{E}_1 = 3\\hat{i} + 2\\hat{j}$, the tangential component is $\\vec{E}_{1,tan} = 3\\hat{i}$.
\n– Therefore, the tangential component of the electric field in D2 is $\\vec{E}_{2,tan} = 3\\hat{i}$. This means the x-component of $\\vec{E}_2$ is 3 ($E_{2x} = 3$) and the z-component is 0 ($E_{2z} = 0$).
\n– The normal component of the electric field is along the y-axis ($\\hat{j}$). From $\\vec{E}_1$, the normal component is $E_{1y} = 2$.
\n– The electric displacement field is given by $\\vec{D} = \\epsilon_0 \\epsilon_r \\vec{E}$.
\n– The normal component of the displacement field in D1 is $D_{1y} = \\epsilon_0 \\epsilon_{r1} E_{1y} = \\epsilon_0 \\times 3 \\times 2 = 6\\epsilon_0$.
\n– The normal component of the displacement field in D2 is $D_{2y} = \\epsilon_0 \\epsilon_{r2} E_{2y} = \\epsilon_0 \\times 2 \\times E_{2y}$.
\n– Applying the boundary condition $\\vec{D}_{1,norm} = \\vec{D}_{2,norm}$, we have $D_{1y} = D_{2y}$.
\n– $6\\epsilon_0 = 2\\epsilon_0 E_{2y}$.
\n– $6 = 2 E_{2y} \\implies E_{2y} = 3$.
\n– The electric field in D2 is $\\vec{E}_2 = E_{2x} \\hat{i} + E_{2y} \\hat{j} + E_{2z} \\hat{k}$.
\n– Substituting the values we found: $\\vec{E}_2 = 3\\hat{i} + 3\\hat{j} + 0\\hat{k} = 3\\hat{i} + 3\\hat{j}$.
\n<\/section>\n