Let's tackle **part (a)** step by step: --- ## Step 1 Find \(\{E}\ \(\vec{D}\), and \(\vec{P}\) in Region \(a \leq r \leq b\) (Dielectric \(\epsilon_1\)) ### Using Gauss's Law for \(\vec{D}\): For a point charge \(Q\) at the center, by symmetry, the displacement field \(\vec{D}\) at radius \(r\) is radial and has the same magnitude at all points on a sphere of radius \(r\): \[ \oint \vec{D} \cdot d\vec{A} = Q_{\text{free, enclosed}} \] \[ D(r) \cdot 4\pi r^2 = Q \] \[ D(r) = \frac{Q}{4\pi r^2} \] **Explanation:** We use Gauss's Law for \(\vec{D}\), which only counts the *free* charge. The symmetry tells us the field points radially. ### Electric Field \(\vec{E}\): \[ \vec{D} = \epsilon_1 \vec{E} \implies \vec{E} = \frac{\vec{D}}{\epsilon_1} \] \[ E(r) = \frac{Q}{4\pi \epsilon_1 r^2} \] **Explanation:** The electric displacement relates to the electric field via the dielectric constant. ### Polarization \(\vec{P}\): \[ \vec{P} = \vec{D} - \epsilon_ \vec{E} \] \[ \vec{P} = \epsilon_1 \vec{E} - \epsilon_ \vec{E} = (\epsilon_1 - \epsilon_) \vec{E} \] \[ \vec{P} = (\epsilon_1 - \epsilon_) \frac{Q}{4\pi \epsilon_1 r^2} \hat{r} \] **Explanation:** \(\vec{P}\) is the difference between total and vacuum permittivity times \(\vec{E}\). --- ## Step 2: Find \(\vec{E}\), \(\vec{D}\), and \(\vec{P}\) in Region \(b \leq r \leq c\) (Dielectric \(\epsilon_2\)) ### Displacement Field \(\vec{D}\): Again, by Gauss's law, for \(r > b\), the free charge enclosed is still \(Q\): \[ D(r) \cdot 4\pi r^2 = Q \implies D(r) = \frac{Q}{4\pi r^2} \] **Explanation:** The *free* charge (point charge at the center) is still the only contributor. ### Electric Field \(\vec{E}\): \[ D(r) = \epsilon_2 E(r) \implies E(r) = \frac{Q}{4\pi \epsilon_2 r^2} \] **Explanation:** Now, the dielectric constant is \(\epsilon_2\). ### Polarization \(\vec{P}\): \[ \vec{P} = (\epsilon_2 - \epsilon_) \vec{E} \] \[ \vec{P} = (\epsilon_2 - \epsilon_) \frac{Q}{4\pi \epsilon_2 r^2} \hat{r} \] **Explanation:** Same reasoning as before, but with \(\epsilon_2\). --- ## **Final Answer (Part a):** ### For \(a \leq r \leq b\) (Dielectric \(\epsilon_1\)): \[ \boxed{ \begin{aligned} \vec{D} &= \frac{Q}{4\pi r^2} \hat{r} \\ \vec{E} &= \frac{Q}{4\pi \epsilon_1 r^2} \hat{r} \\ \vec{P} &= \frac{(\epsilon_1 - \epsilon_) Q}{4\pi \epsilon_1 r^2} \hat{r} \end{aligned} } \] ### For \(b \leq r \leq c\) (Dielectric \(\epsilon_2\)): \[ \boxed{ \begin{aligned} \vec{D} &= \frac{Q}{4\pi r^2} \hat{r} \\ \vec{E} &= \frac{Q}{4\pi \epsilon_2 r^2} \hat{r} \\ \vec{P} &= \frac{(\epsilon_2 - \epsilon_) Q}{4\pi \epsilon_2 r^2} \hat{r} \end{aligned} } \]

## Part (b): Find All Polarization Charges and Show Total Polarization Charge is Zero ### Step 1: Polarization Charge Density The polarization charge density \(\sigma_p\) at the boundary surfaces can be calculated using: \[ \sigma_p = \vec{P} \cdot \hat{n} \] where \(\hat{n}\) is the unit normal vector at the interface. ### Boundaries 1. **At \(r = a\)** (interface between vacuum and dielectric \(\epsilon_1\)): - \(\vec{P} = 0\) (since there is no polarization in vacuum). - Therefore, \(\sigma_{p,a} = 0\). 2. **At \(r = b\)** (interface between dielectric \(\epsilon_1\) and \(\epsilon_2\)): - From the previous calculations: \[ \vec{P}_1 = \frac{(\epsilon_1 - \epsilon_0) Q}{4\pi \epsilon_1 b^2} \hat{r}, \quad \text{(from dielectric \(\epsilon_1\))} \] \[ \vec{P}_2 = \frac{(\epsilon_2 - \epsilon_0) Q}{4\pi \epsilon_2 b^2} \hat{r} \quad \text{(from dielectric \(\epsilon_2\))} \] - The polarization charge will be: \[ \sigma_{p,b} = P_1 - P_2 = \left( \frac{(\epsilon_1 - \epsilon_0) Q}{4\pi \epsilon_1 b^2} - \frac{(\epsilon_2 - \epsilon_0) Q}{4\pi \epsilon_2 b^2} \right) \] **Explanation:** This expression captures the difference in polarization at the boundary between the two dielectrics. ### Step 2: Total Polarization Charge The total polarization charge \(Q_p\) can be found by integrating the polarization charge density over the surface area: \[ Q_p = \sigma_{p,b} \cdot 4\pi b^2 \] Substituting \(\sigma_{p,b}\): \[ Q_p = \left( \frac{(\epsilon_1 - \epsilon_0) Q}{4\pi \epsilon_1 b^2} - \frac{(\epsilon_2 - \epsilon_0) Q}{4\pi \epsilon_2 b^2} \right) \cdot 4\pi b^2 \] \[ Q_p = Q \left( \frac{(\epsilon_1 - \epsilon_0)}{\epsilon_1} - \frac{(\epsilon_2 - \epsilon_0)}{\epsilon_2} \right) \] **Explanation:** The total polarization charge is expressed as a function of the dielectric constants and is a measure of the bound charge at the interface. ### Conclusion: Show Total Polarization Charge is Zero To show that the total polarization charge is zero, we need the condition: \[ \frac{(\epsilon_1 - \epsilon_0)}{\epsilon_1} = \frac{(\epsilon_2 - \epsilon_0)}{\epsilon_2} \] This condition implies that the polarization effects of the two dielectrics cancel each other out, leading to no net polarization charge. --- ## **Final Answer (Part b):** \[ \text{Total Polarization Charge} = Q_p = 0 \quad \text{(if the condition holds)} \]