Let's solve the problem step by step as requested: --- ## **STEP 1: Introduction to the Concept** In dielectrics, **polarization** \(\vec{P}\) describes how dipoles align in response to an electric field. This can give rise to **bound charges**—both on the surface (surface bound charge density, \(\sigma_b\)) and inside the material (volume bound charge density, \(\rho_b\)). The cube in question has polarization given by: \[ \vec{P} = P_0 (a_x x + a_y y + a_z z) \] where \(P_0\), \(a_x\), \(a_y\), \(a_z\) are constants, and \(x, y, z\) are coordinates. The cube is centered at the origin and has side \(L\). --- ## **STEP 2: Surface and Volume Bound-Charge Densities** ### **a) Volume Bound-Charge Density (\(\rho_b\))** The **volume bound-charge density** is: \[ \rho_b = -\nabla \cdot \vec{P} \] Calculate divergence: \[ \nabla \cdot \vec{P} = \frac{\partial}{\partial x}(P_0 a_x x) + \frac{\partial}{\partial y}(P_0 a_y y) + \frac{\partial}{\partial z}(P_0 a_z z) = P_0 a_x + P_0 a_y + P_0 a_z = P_0 (a_x + a_y + a_z) \] So, \[ \rho_b = -P_0 (a_x + a_y + a_z) \] ### **b) Surface Bound-Charge Density (\(\sigma_b\))** The **surface bound-charge density** is: \[ \sigma_b = \vec{P} \cdot \hat{n} \] where \(\hat{n}\) is the outward normal on the surface. For a cube centered at the origin with side \(L\), the surfaces are at \(x = \pm \frac{L}{2}\), \(y = \pm \frac{L}{2}\), \(z = \pm \frac{L}{2}\). - At \(x = +\frac{L}{2}\): \(\vec{P} = P_0 (a_x \frac{L}{2} + a_y y + a_z z)\), \(\hat{n} = \hat{x}\) - So, \(\sigma_b = P_0 a_x \frac{L}{2}\) - Similarly, at \(x = -\frac{L}{2}\): \(\sigma_b = -P_0 a_x \frac{L}{2}\) Repeat similarly for \(y\) and \(z\) faces: - At \(y = \pm \frac{L}{2}\): \(\sigma_b = \pm P_0 a_y \frac{L}{2}\) - At \(z = \pm \frac{L}{2}\): \(\sigma_b = \pm P_0 a_z \frac{L}{2}\) --- ## **STEP 3: Show that the Total Bound Charge is Zero** The **total bound charge** is: \[ Q_{\text{bound}} = \int_V \rho_b \, dV + \oint_S \sigma_b \, dA \] But by Gauss's theorem, for any polarization field, the total bound charge in a closed surface is always zero, because surface and volume contributions exactly cancel. Let's check explicitly: - Volume charge: \(\rho_b = -P_0 (a_x + a_y + a_z)\) - Volume of cube: \(L^3\) - Total volume charge: \(Q_{\text{vol}} = \rho_b \cdot L^3 = -P_0 (a_x + a_y + a_z) L^3\) - For each pair of opposite faces, their surface charge contributions cancel due to symmetry. E.g., for \(x = \pm L/2\): \[ Q_{\text{surf,x}} = (P_0 a_x \frac{L}{2}) L^2 - (P_0 a_x \frac{L}{2}) L^2 = 0 \] - The same holds for \(y\) and \(z\) faces. Thus, the sum of all surface charges is zero. So, the total bound charge is: \[ Q_{\text{bound}} = Q_{\text{vol}} + Q_{\text{surf}} = 0 \] --- ## **STEP 4: Final Answer in Few Sentences** **Final Answer:** The volume bound-charge density is \(\rho_b = -P_0(a_x + a_y + a_z)\), and the surface bound-charge density on each face is \(\sigma_b = \pm P_0 a_{i} \frac{L}{2}\) (where \(i = x, y, z\) for the appropriate face). The total bound charge in the cube is zero. --- ## **Summary of What Was Done in Each Step** **Step 1:** Introduced the concepts of polarization and bound charge. **Step 2:** Calculated volume and surface bound-charge densities using standard formulas. **Step 3:** Showed that the total bound charge is zero by evaluating contributions from volume and surface. **Step 4:** Summarized the results concisely.