Two dielectric media D1 and D2 have dielectric constants 3 and 2, respectively. They are separated by x – z plane. A uniform electric field $\vec{E} = 3\hat{i} + 2\hat{j}$ exists in D1. Which one among the following is the correct electric field in D2 at the xz plane ?
$ ec{E} = 2hat{j}$
$ ec{E} = 3hat{i} + 3hat{j}$
$ ec{E} = 3hat{i} - 2hat{j}$
$ ec{E} = 2hat{i} + 3hat{j}$
Answer is Right!
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UPSC Geoscientist – 2024
– 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 $y>0$.
– The electric field in D1 is given as $\vec{E}_1 = 3\hat{i} + 2\hat{j}$.
– At the boundary between two dielectric media, two boundary conditions for the electric field and displacement field must be satisfied:
1. The tangential component of the electric field ($\vec{E}_{tan}$) is continuous across the boundary: $\vec{E}_{1,tan} = \vec{E}_{2,tan}$.
2. The normal component of the electric displacement field ($\vec{D}_{norm}$) is continuous across the boundary (assuming no free charges on the surface): $\vec{D}_{1,norm} = \vec{D}_{2,norm}$.
– 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 $\vec{E}_{1,tan} = 3\hat{i}$.
– 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 ($E_{2z} = 0$).
– The normal component of the electric field is along the y-axis ($\hat{j}$). From $\vec{E}_1$, the normal component is $E_{1y} = 2$.
– The electric displacement field is given by $\vec{D} = \epsilon_0 \epsilon_r \vec{E}$.
– The normal component of the displacement field in D1 is $D_{1y} = \epsilon_0 \epsilon_{r1} E_{1y} = \epsilon_0 \times 3 \times 2 = 6\epsilon_0$.
– The normal component of the displacement field in D2 is $D_{2y} = \epsilon_0 \epsilon_{r2} E_{2y} = \epsilon_0 \times 2 \times E_{2y}$.
– Applying the boundary condition $\vec{D}_{1,norm} = \vec{D}_{2,norm}$, we have $D_{1y} = D_{2y}$.
– $6\epsilon_0 = 2\epsilon_0 E_{2y}$.
– $6 = 2 E_{2y} \implies E_{2y} = 3$.
– The electric field in D2 is $\vec{E}_2 = E_{2x} \hat{i} + E_{2y} \hat{j} + E_{2z} \hat{k}$.
– Substituting the values we found: $\vec{E}_2 = 3\hat{i} + 3\hat{j} + 0\hat{k} = 3\hat{i} + 3\hat{j}$.