Electric Field Boundary Analysis
We need to find the electric field ($\vec{E}_{\text{right}}$) on the right side of the $yz$-plane boundary, given the field on the left ($\vec{E}_{\text{left}}$) and the ratio of permittivities.
Boundary Conditions Application
The interface is the $yz$-plane, meaning the normal vector is $\hat{n} = \hat{i}$. The boundary conditions for the electric field at a dielectric interface are:
- Tangential Component Continuity: The tangential component of $\vec{E}$ is continuous across the boundary. $\vec{E}_{\text{left}, \text{tan}} = \vec{E}_{\text{right}, \text{tan}}$
- Normal Component Relation: The normal component of the electric displacement field ($\vec{D}$) is continuous for a chargeless boundary. $\vec{D}_{\text{left}, \text{norm}} = \vec{D}_{\text{right}, \text{norm}}$, which implies $\epsilon_{\text{left}} E_{\text{left}, \text{norm}} = \epsilon_{\text{right}} E_{\text{right}, \text{norm}}$.
Decomposing the Left Electric Field
The electric field on the left is given as $\vec{E}_{\text{left}} = c(\hat{i} + \hat{j} + \hat{k})$.
- Normal Component: The component normal to the $yz$-plane (along $\hat{i}$) is $E_{\text{left}, \text{norm}} = \vec{E}_{\text{left}} \cdot \hat{i} = c(\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} = c$.
- Tangential Components: The components parallel to the $yz$-plane (along $\hat{j}$ and $\hat{k}$) are $\vec{E}_{\text{left}, \text{tan}} = \vec{E}_{\text{left}} - E_{\text{left}, \text{norm}}\hat{i} = c(\hat{i} + \hat{j} + \hat{k}) - c\hat{i} = c(\hat{j} + \hat{k})$.
Calculating the Right Electric Field Components
Using the boundary conditions and the given permittivity ratio ($\epsilon_{\text{left}} : \epsilon_{\text{right}} = 1 : 2$, so $\epsilon_{\text{right}} = 2\epsilon_{\text{left}}$):
- Tangential Field: From continuity, $\vec{E}_{\text{right}, \text{tan}} = \vec{E}_{\text{left}, \text{tan}} = c(\hat{j} + \hat{k})$. This means $E_{\text{right}, y} = c$ and $E_{\text{right}, z} = c$.
- Normal Field: Using $\epsilon_{\text{left}} E_{\text{left}, \text{norm}} = \epsilon_{\text{right}} E_{\text{right}, \text{norm}}$, we get $\epsilon_{\text{left}}(c) = (2\epsilon_{\text{left}}) E_{\text{right}, \text{norm}}$. Solving for $E_{\text{right}, \text{norm}}$ gives $E_{\text{right}, \text{norm}} = \frac{\epsilon_{\text{left}} c}{2\epsilon_{\text{left}}} = \frac{c}{2}$. This is the component along $\hat{i}$, so $E_{\text{right}, x} = \frac{c}{2}$.
Final Electric Field Vector
Combine the normal and tangential components to find $\vec{E}_{\text{right}}$:
$ \vec{E}_{\text{right}} = E_{\text{right}, \text{norm}}\hat{i} + \vec{E}_{\text{right}, \text{tan}} $
$ \vec{E}_{\text{right}} = \frac{c}{2}\hat{i} + c(\hat{j} + \hat{k}) $
$ \vec{E}_{\text{right}} = c\left(\frac{1}{2}\hat{i} + \hat{j} + \hat{k}\right) $
This matches Option C.
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