# Problem Breakdown - **Sphere details:** - Centered at origin - Radius: \(b\) - Spin: at angular velocity \(\omega\) about the z-axis - **Charge density:** \[ \rho = \rho_ \left(\frac{r}{b}\right \cos^2 \theta \sin \phi \] where \(\rho_\) is a constant, \(r\) is the radial distance, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle. - **Goals:** - Find the **power radiated** by the spinning charge distribution. - Use the **dipole moment** approach: - First, find the **dipole moment** \(\mathbf{p}(t)\). - Then, write the **time-dependent dipole** as the sphere spins. - Calculate the **power radiated** from the dipole's second time derivative. --- # Step 1: Find the electric dipole moment \(\mathbf{p}\) The electric dipole moment: \[ \mathbf{p} = \int \rho(\mathbf{r}) \mathbf{r}\, dV \] In spherical coordinates: \[ dV = r^2 \sin \theta dr d\theta d\phi \] The position vector: \[ \mathbf{r} = r \hat{\mathbf{r}} \] Since the charge density is given in terms of \(r, \theta, \phi\), the components of \(\mathbf{p}\) are: \[ p_x = \int \rho r \sin \theta \cos \phi \, dV \] \[ p_y = \int \rho r \sin \theta \sin \phi \, dV \] \[ p_z = \int \rho r \cos \theta \, dV \] --- # Step 2: Calculate each component ### \(p_x\): \[ p_x = \int_^{b} \int^{\pi} \int_^{2\pi} \rho(r, \theta, \phi) \sin \theta \cos \phi \, r^2 \sin \theta dr d\theta d\phi \] Substitute \(\rho\): \[ \rho =rho_ \left(\frac{r}{b}\right) \cos^2 \theta \sin \phi \] \[ p_x = \int_^{b} \int_^{\pi} \int_^{2\pi} \rho_ \left(\frac{r}{b}\right) \cos^2 \theta \sin \phi \times r \sin \theta \cos \phi \times r^2 \sin \theta dr d\theta d\phi \] Simplify: \[ p_x = \rho_ \int_^{b} \frac{r}{b} r r^2 dr \int_^{\pi} \cos^2 \theta \sin \theta \sin \theta d\theta \int_^{2\pi} \sin \phi \cos \phi d\phi \] --- ### Angular integrals: - \(\phi\): \[ \int_^{2\pi} \sin \phi \cos \phi d\phi = \] because it's symmetric over one period. Similarly, for \(p_y\): \[ p_y \propto \int_^{2\pi} \sin \phi \sin \phi d\phi \neq \] but for \(p_x\), the integral over \(\phi\) is zero, so: \[ p_x = \] Similarly, \(p_y\) involves: \[ \int_^{2\pi} \sin^2 \phi d\phi = \pi \] but since \(\sin \phi\) appears in \(\rho\), and the angular integral over \(\phi\) for \(\sin \phi\) times \(\sin \phi\) is non-zero, the main conclusion is: - **Only \(p_z\) is non-zero** because the angular dependence for \(p_x\) and \(p_y\) integrates to zero due to symmetry. --- # Step 3: Calculate \(p_z\) \[ p_z = \int \rho r \cos \theta dV \] \[ p_z = \int_^{b} \int_^{\pi} \int_^{2\pi} \rho_ \frac{r}{b} \cos^2 \theta \sin \phi \times r \cos \theta \times r^2 \sin \theta dr d\theta d\phi \] Note the \(\sin \phi\) term again: the integral over \(\phi\): \[ \int_^{2\pi} \sin \phi d\phi = \] Thus: \[ p_z = \] --- # **Conclusion:** All components of \(\mathbf{p}\) are zero because the angular integrals involving \(\sin \phi\) cancel out. --- # **Final notes:** - The dipole moment \(\mathbf{p}(t)\) for this charge distribution is **zero**. - Since a **time-varying dipole moment** is the primary source of radiation in the dipole approximation, **no power is radiated**. --- # **Answer:** \[ \boxed{ \text{The power radiated is } P = } \] because the dipole moment is zero and no dipole radiation occurs from this charge distribution.