Let's break down the problem step by step: ### **Problem Statement (translated):** > With respect to \(x < 0\), \(\vec{P} = 5\sin(\alpha y) \hat{a}_x\), where \(\alpha\) is a constant. Find \(\rho_{pol}\) and \(\rho_{psup}\). Draw a sketch of the \(\vec{P}\) field. --- ### **Step 1: Understanding the Variables** - \(\vec{P}\): Polarization vector - \(\rho_{pol}\): Volume bound charge density - \(\rho_{psup}\): Surface bound charge density --- ### **Step 2: Relevant Formulas** - **Volume bound charge density:** \[ \rho_{pol} = -\nabla \cdot \vec{P} \] - **Surface bound charge density:** \[ \rho_{psup} = \vec{P} \cdot \hat{n} \] where \(\hat{n}\) is the normal vector to the surface. --- ### **Step 3: Compute \(\rho_{pol}\)** Given: \[ \vec{P} = 5\sin(\alpha y)\hat{a}_x \] Calculate the divergence: \[ \nabla \cdot \vec{P} = \frac{\partial}{\partial x} P_x + \frac{\partial}{\partial y} P_y + \frac{\partial}{\partial z} P_z \] Here, - \(P_x = 5\sin(\alpha y)\) - \(P_y = 0\) - \(P_z = 0\) So, \[ \nabla \cdot \vec{P} = \frac{\partial}{\partial x} [5\sin(\alpha y)] + 0 + 0 = 0 \] Therefore, \[ \rho_{pol} = -\nabla \cdot \vec{P} = 0 \] --- ### **Step 4: Compute \(\rho_{psup}\)** At the surface \(x = 0\), the normal vector is \(\hat{n} = \hat{a}_x\) (points outward). \[ \rho_{psup} = \vec{P} \cdot \hat{n} = 5\sin(\alpha y) \] But for \(x < 0\), \(\vec{P} = 5\sin(\alpha y)\hat{a}_x\), and for \(x > 0\), it's assumed that \(\vec{P} = 0\) (since it's not defined there). So at the surface \(x = 0\): \[ \rho_{psup} = \vec{P}(x=0^-) \cdot \hat{a}_x - \vec{P}(x=0^+) \cdot \hat{a}_x = 5\sin(\alpha y) - 0 = 5\sin(\alpha y) \] --- ### **Step 5: Field Sketch** - For \(x < 0\): The vectors point along the \(x\)-axis, and their magnitude varies sinusoidally with \(y\). - For \(x > 0\): \(\vec{P} = 0\). A simple sketch would show arrows along \(x\) (left half-plane), with their lengths varying as \(\sin(\alpha y)\). --- ### **Final Answers** \[ \boxed{ \begin{aligned} &\rho_{pol} = 0 \\ &\rho_{psup} = 5\sin(\alpha y) \end{aligned} } \] **Field Sketch:** - Draw the \(y\)-axis vertically, \(x\)-axis horizontally. - For \(x < 0\), draw arrows along \(x\) whose length varies as \(\sin(\alpha y)\). - For \(x > 0\), no arrows. Let me know if you need the actual sketch!

**Step 1: Calculate the volume bound charge density \(\rho_{pol}\).** Given the polarization vector: \[ \vec{P} = 5 \sin(\alpha y) \hat{a}_x \] which depends only on \(y\), and points along the \(x\)-direction. The bound volume charge density is: \[ \rho_{pol} = - \nabla \cdot \vec{P} \] Since \(\vec{P}\) has only an \(x\)-component that varies with \(y\): \[ \nabla \cdot \vec{P} = \frac{\partial P_x}{\partial x} + \frac{\partial P_y}{\partial y} + \frac{\partial P_z}{\partial z} \] Here: - \(P_x = 5 \sin(\alpha y)\) - \(P_y = 0\) - \(P_z = 0\) Calculating derivatives: \[ \frac{\partial P_x}{\partial x} = 0 \quad \text{(no } x \text{-dependence)} \] \[ \frac{\partial P_y}{\partial y} = 0 \] \[ \frac{\partial P_z}{\partial z} = 0 \] Therefore: \[ \nabla \cdot \vec{P} = 0 \] and so: \[ \rho_{pol} = - 0 = 0 \] --- **Step 2: Calculate the surface bound charge density \(\rho_{psup}\).** The surface charge density at the interface (say, at \(x=0\)) is: \[ \rho_{psup} = \vec{P} \cdot \hat{n} \] where \(\hat{n}\) is the outward normal to the surface. - For the surface at \(x=0\), the normal vector pointing toward \(x>0\) is \(\hat{a}_x\). - The polarization just inside the surface (for \(x<0\)) is: \[ \vec{P} = 5 \sin(\alpha y) \hat{a}_x \] - Outside (for \(x>0\)), \(\vec{P} \approx 0\). Thus: \[ \rho_{psup} = \left.\vec{P}\right|_{x=0^-} \cdot \hat{a}_x = 5 \sin(\alpha y) \] **Final Answer:** \[ \boxed{ \begin{aligned} \rho_{pol} &= 0 \\ \rho_{psup} &= 5 \sin(\alpha y) \end{aligned} } \] **Explanation:** - The volume bound charge density is zero because the divergence of \(\vec{P}\) is zero; \(\vec{P}\) varies only with \(y\), but its \(x\)-component's partial derivative with respect to \(x\) is zero. - The surface bound charge appears where \(\vec{P}\) discontinuously changes at the boundary \(x=0\), resulting in a surface charge density proportional to \(\sin(\alpha y)\).