Let's solve the problem step by step as requested. --- ### **Given:** A nonconducting spherical shell of radius \( R \) has a surface charge density: \[ \sigma(\theta) = \sigma_0 \cos^2(\theta) \] --- ## **(a) Find the exterior multipole moments \( q_{lm} \).** The exterior multipole moments for a surface charge density are given by: \[ q_{lm} = \int Y_{lm}^*(\theta, \phi) \sigma(\theta) R^2 \sin\theta \, d\theta \, d\phi \] Plug in \( \sigma(\theta) \): \[ q_{lm} = \sigma_0 R^2 \int_0^{2\pi} d\phi \int_0^\pi Y_{lm}^*(\theta, \phi) \cos^2\theta \sin\theta \, d\theta \] Because \( \sigma(\theta) \) is independent of \( \phi \), only \( m = 0 \) terms survive: \[ Y_{l0}(\theta, \phi) = \sqrt{\frac{2l+1}{4\pi}} P_l(\cos\theta) \] So \[ q_{l0} = \sigma_0 R^2 \int_0^{2\pi} d\phi \int_0^\pi \sqrt{\frac{2l+1}{4\pi}} P_l(\cos\theta) \cos^2\theta \sin\theta \, d\theta \] \[ = \sigma_0 R^2 2\pi \sqrt{\frac{2l+1}{4\pi}} \int_0^\pi P_l(\cos\theta) \cos^2\theta \sin\theta \, d\theta \] Let \( x = \cos\theta \), so \( dx = -\sin\theta d\theta \): Limits: \( \theta = 0 \Rightarrow x = 1 \), \( \theta = \pi \Rightarrow x = -1 \) \[ q_{l0} = \sigma_0 R^2 2\pi \sqrt{\frac{2l+1}{4\pi}} \int_{1}^{-1} P_l(x) x^2 (-dx) \] \[ = \sigma_0 R^2 2\pi \sqrt{\frac{2l+1}{4\pi}} \int_{-1}^{1} P_l(x) x^2 dx \] Now, expand \( x^2 \) in terms of Legendre polynomials: \[ x^2 = \frac{2}{3}P_2(x) + \frac{1}{3}P_0(x) \] So, \[ \int_{-1}^1 P_l(x) x^2 dx = \frac{2}{3} \int_{-1}^1 P_l(x) P_2(x) dx + \frac{1}{3} \int_{-1}^1 P_l(x) P_0(x) dx \] Orthogonality: \[ \int_{-1}^1 P_l(x)P_{l'}(x)dx = \frac{2}{2l+1}\delta_{ll'} \] So, only \( l=0 \) and \( l=2 \) survive: - For \( l=0 \): \( \frac{1}{3} \cdot \frac{2}{1} = \frac{2}{3} \) - For \( l=2 \): \( \frac{2}{3} \cdot \frac{2}{5} = \frac{4}{15} \) So, \[ q_{00} = \sigma_0 R^2 2\pi \sqrt{\frac{1}{4\pi}} \cdot \frac{2}{3} \] \[ = \sigma_0 R^2 2\pi \cdot \frac{1}{2\sqrt{\pi}} \cdot \frac{2}{3} = \sigma_0 R^2 \frac{2\pi}{2\sqrt{\pi}} \cdot \frac{2}{3} = \sigma_0 R^2 \frac{2\pi}{3\sqrt{\pi}} \] \[ q_{20} = \sigma_0 R^2 2\pi \sqrt{\frac{5}{4\pi}} \cdot \frac{4}{15} \] \[ = \sigma_0 R^2 2\pi \cdot \frac{\sqrt{5}}{2\sqrt{\pi}} \cdot \frac{4}{15} = \sigma_0 R^2 \frac{2\pi}{2\sqrt{\pi}} \sqrt{5} \cdot \frac{4}{15} = \sigma_0 R^2 \frac{\pi \sqrt{5}}{15\sqrt{\pi}} \cdot 4 = \sigma_0 R^2 \frac{4\pi \sqrt{5}}{15\sqrt{\pi}} = \sigma_0 R^2 \frac{4\sqrt{5\pi}}{15} \] All other moments vanish. --- ## **(b) Find the interior multipole moments \( \tilde{q}_{lm} \).** Interior multipole moments: \[ \tilde{q}_{lm} = \int Y_{lm}^*(\theta, \phi) \sigma(\theta) R^{-(l+1)} R^2 \sin\theta d\theta d\phi = R^{-(l+1)} q_{lm} \] So, \[ \tilde{q}_{lm} = R^{-(l+1)} q_{lm} \] --- ## **(c) Find the potential inside and outside the sphere.** The general solution for the potential is: - **Outside (\( r > R \))**: \[ V_{\text{out}}(r, \theta) = \frac{1}{\epsilon_0} \sum_{l,m} \frac{q_{lm}}{r^{l+1}} Y_{lm}(\theta, \phi) \] - **Inside (\( r < R \))**: \[ V_{\text{in}}(r, \theta) = \frac{1}{\epsilon_0} \sum_{l,m} \tilde{q}_{lm} r^{l} Y_{lm}(\theta, \phi) \] But only \( m = 0 \) terms survive. So: \[ V_{\text{out}}(r, \theta) = \frac{1}{\epsilon_0} \left[ \frac{q_{00}}{r} Y_{00}(\theta, \phi) + \frac{q_{20}}{r^3} Y_{20}(\theta, \phi) \right] \] \[ V_{\text{in}}(r, \theta) = \frac{1}{\epsilon_0} \left[ \tilde{q}_{00} r^0 Y_{00}(\theta, \phi) + \tilde{q}_{20} r^{2} Y_{20}(\theta, \phi) \right] \] With: \[ Y_{00} = \sqrt{\frac{1}{4\pi}} \] \[ Y_{20} = \sqrt{\frac{5}{4\pi}} P_2(\cos\theta) = \sqrt{\frac{5}{4\pi}} \left(\frac{3\cos^2\theta - 1}{2}\right) \] Plug in the moments calculated above. --- ## **Summary of Final Answers:** ### **Multipole moments:** \[ q_{00} = \sigma_0 R^2 \frac{2\pi}{3\sqrt{\pi}} \] \[ q_{20} = \sigma_0 R^2 \frac{4\sqrt{5\pi}}{15} \] All others vanish. \[ \tilde{q}_{00} = \frac{q_{00}}{R} \] \[ \tilde{q}_{20} = \frac{q_{20}}{R^3} \] --- ### **Potential:** **Outside (\( r > R \)):** \[ V_{\text{out}}(r, \theta) = \frac{1}{\epsilon_0} \left[ \frac{q_{00}}{r} Y_{00} + \frac{q_{20}}{r^3} Y_{20} \right] \] **Inside (\( r < R \)):** \[ V_{\text{in}}(r, \theta) = \frac{1}{\epsilon_0} \left[ \tilde{q}_{00} Y_{00} + \tilde{q}_{20} r^2 Y_{20} \right] \] --- **If you want the potentials explicitly in terms of \( \sigma_0, R, \theta \), substitute the values above for \( q_{00}, q_{20}, Y_{00}, Y_{20} \).** --- **If you need the explicit expanded forms for the potential, let me know!**

Certainly! Let's proceed step-by-step, providing detailed calculations and explanations for each part. --- ## **Problem Restatement** A nonconducting spherical shell of radius \( R \) has a surface charge density: \[ \sigma(\theta) = \sigma_0 \cos^2 \theta \] Find: a) The **exterior multipole moments** \( q_{lm} \). b) The **interior multipole moments** \( \tilde{q}_{lm} \). c) The **electric potential** both **inside** and **outside** the sphere. --- ## **Part (a): Compute the exterior multipole moments \( q_{lm} \)** ### **Step 1: General expression for multipole moments** The multipole moments \( q_{lm} \) are given by: \[ q_{lm} = \int \sigma(\theta) R^2 \sin \theta \, d\theta \, d\phi \, Y_{lm}^*(\theta, \phi) \] Since \( \sigma(\theta) \) is independent of \( \phi \), **only \( m=0 \) terms** will be non-zero because: \[ Y_{lm}(\theta, \phi) = \text{function of } \theta \times e^{i m \phi} \] and integrals over \( \phi \) vanish unless \( m=0 \). --- ### **Step 2: Simplify for \( m=0 \)** \[ q_{l0} = R^2 \int_0^{2\pi} d\phi \int_0^{\pi} d\theta \, \sigma(\theta) \sin \theta \, Y_{l0}^*(\theta, \phi) \] \[ Y_{l0}(\theta, \phi) = \sqrt{\frac{2l+1}{4\pi}} P_l(\cos \theta) \] and \( Y_{l0}^* = Y_{l0} \) because these are real. So, \[ q_{l0} = R^2 \times 2\pi \times \sqrt{\frac{2l+1}{4\pi}} \int_0^{\pi} \sigma(\theta) P_l(\cos \theta) \sin \theta \, d\theta \] --- ### **Step 3: Insert \( \sigma(\theta) \)** \[ \sigma(\theta) = \sigma_0 \cos^2 \theta \] Thus, \[ q_{l0} = R^2 \times 2\pi \times \sqrt{\frac{2l+1}{4\pi}} \times \sigma_0 \int_0^{\pi} \cos^2 \theta P_l(\cos \theta) \sin \theta \, d\theta \] ### **Step 4: Change variable** Let \( x = \cos \theta \), so: \[ dx = -\sin \theta d\theta \Rightarrow \sin \theta d\theta = - dx \] When \( \theta = 0 \), \( x=1 \); when \( \theta = \pi \), \( x=-1 \). Rearranged integral: \[ q_{l0} = R^2 \times 2\pi \times \sqrt{\frac{2l+1}{4\pi}} \times \sigma_0 \int_{x=1}^{-1} x^2 P_l(x) (- dx) = R^2 \times 2\pi \times \sqrt{\frac{2l+1}{4\pi}} \times \sigma_0 \int_{-1}^{1} x^2 P_l(x) dx \] --- ### **Step 5: Expand \( x^2 \) in Legendre polynomials** Using the Legendre polynomial expansion: \[ x^2 = \frac{2}{3} P_2(x) + \frac{1}{3} P_0(x) \] which is a standard expansion. ### **Step 6: Compute the integral** \[ I_l = \int_{-1}^1 x^2 P_l(x) dx = \frac{2}{3} \int_{-1}^1 P_2(x) P_l(x) dx + \frac{1}{3} \int_{-1}^1 P_0(x) P_l(x) dx \] Using orthogonality of Legendre polynomials: \[ \int_{-1}^1 P_m(x) P_n(x) dx = \frac{2}{2n + 1} \delta_{mn} \] Thus, - For \( l=0 \): \[ I_0 = \frac{2}{3} \times \frac{2}{5} \delta_{2,0} + \frac{1}{3} \times 2 \times \delta_{0,0} = 0 + \frac{1}{3} \times 2 = \frac{2}{3} \] - For \( l=2 \): \[ I_2 = \frac{2}{3} \times 2 \times \delta_{2,2} + \frac{1}{3} \times 2 \times \delta_{0,2} = \frac{2}{3} \times \frac{2}{5} + 0 = \frac{4}{15} \] - For other \( l \neq 0,2 \), the integral is zero because of orthogonality. --- ### **Step 7: Final expression for \( q_{l0} \)** \[ q_{l0} = R^2 \times 2\pi \times \sqrt{\frac{2l+1}{4\pi}} \times \sigma_0 \times I_l \] which simplifies to: \[ q_{l0} = R^2 \times 2\pi \times \sqrt{\frac{2l+1}{4\pi}} \times \sigma_0 \times \left\{ \begin{array}{ll} \displaystyle \frac{2}{3} & l=0 \\ \displaystyle \frac{4}{15} & l=2 \\ 0 & \text{others} \end{array} \right. \] --- ### **Step 8: Explicit calculations** **For \( l=0 \):** \[ q_{00} = R^2 \times 2\pi \times \sqrt{\frac{1}{4\pi}} \times \sigma_0 \times \frac{2}{3} \] Calculate: \[ \sqrt{\frac{1}{4\pi}} = \frac{1}{2 \sqrt{\pi}} \] Therefore, \[ q_{00} = R^2 \times 2\pi \times \frac{1}{2 \sqrt{\pi}} \times \sigma_0 \times \frac{2}{3} \] Simplify numerator: \[ 2\pi \times \frac{1}{2 \sqrt{\pi}} = \frac{\pi}{\sqrt{\pi}} = \sqrt{\pi} \] Thus, \[ q_{00} = R^2 \times \sqrt{\pi} \times \sigma_0 \times \frac{2}{3} = \frac{2}{3} R^2 \sigma_0 \sqrt{\pi} \] --- **For \( l=2 \):** \[ q_{20} = R^2 \times 2\pi \times \sqrt{\frac{5}{4\pi}} \times \sigma_0 \times \frac{4}{15} \] Calculate: \[ \sqrt{\frac{5}{4\pi}} = \frac{\sqrt{5}}{2 \sqrt{\pi}} \] So, \[ q_{20} = R^2 \times 2\pi \times \frac{\sqrt{5}}{2 \sqrt{\pi}} \times \sigma_0 \times \frac{4}{15} \] Simplify: \[ 2\pi \times \frac{\sqrt{5}}{2 \sqrt{\pi}} = \pi \times \frac{\sqrt{5}}{\sqrt{\pi}} = \sqrt{\pi} \times \sqrt{5} \] Therefore, \[ q_{20} = R^2 \times \sqrt{\pi} \times \sqrt{5} \times \sigma_0 \times \frac{4}{15} = \frac{4}{15} R^2 \sigma_0 \sqrt{5\pi} \] --- ## **Summary of \( q_{lm} \):** \[ \boxed{ \begin{aligned} q_{00} &= \frac{2}{3} R^2 \sigma_0 \sqrt{\pi} \\ q_{20} &= \frac{4}{15} R^2 \sigma_0 \sqrt{5 \pi} \\ \text{All other } q_{lm} &= 0 \end{aligned} } \] --- ## **Part (b): Interior multipole moments \( \tilde{q}_{lm} \)** The interior moments relate to the exterior ones via: \[ \boxed{ \tilde{q}_{lm} = \frac{q_{lm}}{R^{l+1}} } \] Thus, \[ \begin{aligned} \tilde{q}_{00} &= \frac{q_{00}}{R} = \frac{2}{3} R \sigma_0 \sqrt{\pi} \\ \tilde{q}_{20} &= \frac{q_{20}}{R^{3}} = \frac{4}{15} \frac{\sigma_0 \sqrt{5 \pi}}{R} \end{aligned} \] --- ## **Part (c): Potential inside and outside** ### **Outside (\( r > R \))** \[ V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_{00}}{r} Y_{00} + \frac{q_{20}}{r^3} Y_{20} \right] \] where: \[ Y_{00} = \sqrt{\frac{1}{4\pi}} = \frac{1}{2 \sqrt{\pi}} \] \[ Y_{20} = \sqrt{\frac{5}{4\pi}} \times \frac{3\cos^2 \theta - 1}{2} \] Inserting \( q_{00} \) and \( q_{20} \): \[ V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{\frac{2}{3} R^2 \sigma_0 \sqrt{\pi}}{r} \times \frac{1}{2 \sqrt{\pi}} + \frac{\frac{4}{15} R^2 \sigma_0 \sqrt{5 \pi}}{r^3} \times \sqrt{\frac{5}{4\pi}} \frac{3 \cos^2 \theta - 1}{2} \right] \] Simplify: - First term: \[ \frac{\frac{2}{3} R^2 \sigma_0 \sqrt{\pi}}{r} \times \frac{1}{2 \sqrt{\pi}} = \frac{\frac{2}{3} R^2 \sigma_0}{r} \times \frac{1}{2} = \frac{1}{3} \frac{R^2 \sigma_0}{r} \] - Second term: Calculate \( q_{20} Y_{20} \) coefficient: \[ \frac{4}{15} R^2 \sigma_0 \sqrt{5 \pi} \times \sqrt{\frac{5}{4\pi}} \times \frac{3 \cos^2 \theta - 1}{2} \] Note: \[ \sqrt{5\pi} \times \sqrt{\frac{5}{4 \pi}} = \sqrt{5\pi} \times \frac{\sqrt{5}}{2 \sqrt{\pi}} = \frac{5}{2} \] Thus, \[ q_{20} Y_{20} = \frac{4}{15} R^2 \sigma_0 \times \frac{5}{2} \times \frac{3 \cos^2 \theta - 1}{2} = \frac{4}{15} \times \frac{5}{2} \times \frac{1}{2} R^2 \sigma_0 (3 \cos^2 \theta - 1) \] Calculate numerator: \[ \frac{4 \times 5}{15 \times 2} = \frac{20}{30} = \frac{2}{3} \] So, \[ q_{20} Y_{20} = \frac{2}{3} R^2 \sigma_0 (3 \cos^2 \theta - 1) \] Finally, potential outside: \[ V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{3} \frac{R^2 \sigma_0}{r} + \frac{2}{3} R^2 \sigma_0 \frac{3 \cos^2 \theta - 1}{r^3} \right] \] --- ### **Inside (\( r < R \))** \[ V_{\text{in}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \til

Certainly! Continuing from where we left off: --- ## **Potential inside the sphere (\( r < R \))** \[ V_{\text{in}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \tilde{q}_{00} Y_{00} + \tilde{q}_{20} r^2 Y_{20} \right] \] Using the expressions for \( \tilde{q}_{00} \) and \( \tilde{q}_{20} \): \[ \tilde{q}_{00} = \frac{2}{3} R \sigma_0 \sqrt{\pi} \] \[ \tilde{q}_{20} = \frac{4}{15} \frac{\sigma_0 \sqrt{5 \pi}}{R} \] Now, substitute these into the potential: \[ V_{\text{in}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{2}{3} R \sigma_0 \sqrt{\pi} \times \frac{1}{2 \sqrt{\pi}} + \frac{4}{15} \frac{\sigma_0 \sqrt{5 \pi}}{R} r^2 \times \sqrt{\frac{5}{4\pi}} \times \frac{3 \cos^2 \theta - 1}{2} \right] \] Simplify each term: - **First term:** \[ \frac{2}{3} R \sigma_0 \sqrt{\pi} \times \frac{1}{2 \sqrt{\pi}} = \frac{2}{3} R \sigma_0 \times \frac{1}{2} = \frac{1}{3} R \sigma_0 \] - **Second term:** Calculate \( \tilde{q}_{20} r^2 Y_{20} \): \[ \frac{4}{15} \frac{\sigma_0 \sqrt{5 \pi}}{R} r^2 \times \sqrt{\frac{5}{4\pi}} \times \frac{3 \cos^2 \theta - 1}{2} \] Note: \[ \sqrt{5 \pi} \times \sqrt{\frac{5}{4 \pi}} = \frac{5}{2} \] Thus, \[ \tilde{q}_{20} r^2 Y_{20} = \frac{4}{15} \times \frac{\sigma_0}{R} \times \frac{5}{2} \times r^2 \times \frac{3 \cos^2 \theta - 1}{2} \] Simplify numerator: \[ \frac{4 \times 5}{15 \times 2} = \frac{20}{30} = \frac{2}{3} \] Therefore, \[ \tilde{q}_{20} r^2 Y_{20} = \frac{2}{3} \frac{\sigma_0}{R} r^2 \times \frac{3 \cos^2 \theta - 1}{2} = \frac{1}{3} \frac{\sigma_0}{R} r^2 (3 \cos^2 \theta - 1) \] --- ## **Final expressions for potentials:** ### **Inside (\( r<R \))** \[ V_{\text{in}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{3} R \sigma_0 + \frac{1}{3} \frac{\sigma_0}{R} r^2 (3 \cos^2 \theta - 1) \right] \] ### **Outside (\( r > R \))** \[ V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{3} \frac{R^2 \sigma_0}{r} + \frac{2}{3} R^2 \sigma_0 \frac{3 \cos^2 \theta - 1}{r^3} \right] \] --- ## **Summary:** | **Quantity** | **Expression / Notes** | |---|---| | **Exterior multipole moments** | \(\displaystyle q_{00} = \frac{2}{3} R^2 \sigma_0 \sqrt{\pi}\), \(\displaystyle q_{20} = \frac{4}{15} R^2 \sigma_0 \sqrt{5 \pi}\) | | **Interior multipole moments** | \(\displaystyle \tilde{q}_{00} = \frac{q_{00}}{R}\), \(\displaystyle \tilde{q}_{20} = \frac{q_{20}}{R^3}\) | | **Potential outside** | \(\displaystyle V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_{00}}{r} Y_{00} + \frac{q_{20}}{r^3} Y_{20} \right]\) | | **Potential inside** | \(\displaystyle V_{\text{in}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \tilde{q}_{00} Y_{00} + \tilde{q}_{20} r^2 Y_{20} \right]\) | --- If you'd like, I can also provide the explicit forms of the potentials in terms of \(\theta\). Let me know!

Certainly! Here are the final, consolidated answers for each part: --- ### **(a) Exterior multipole moments \( q_{lm} \):** \[ \boxed{ \begin{aligned} q_{00} &= \frac{2}{3} R^2 \sigma_0 \sqrt{\pi} \\ q_{20} &= \frac{4}{15} R^2 \sigma_0 \sqrt{5 \pi} \\ \text{All other } q_{lm} &= 0 \end{aligned} } \] --- ### **(b) Interior multipole moments \( \tilde{q}_{lm} \):** \[ \boxed{ \begin{aligned} \tilde{q}_{00} &= \frac{q_{00}}{R} = \frac{2}{3} R \sigma_0 \sqrt{\pi} \\ \tilde{q}_{20} &= \frac{q_{20}}{R^3} = \frac{4}{15} \frac{\sigma_0 \sqrt{5 \pi}}{R} \\ \text{All other } \tilde{q}_{lm} &= 0 \end{aligned} } \] --- ### **(c) Electric potential:** **Outside (\( r > R \)):** \[ \boxed{ V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_{00}}{r} Y_{00} + \frac{q_{20}}{r^3} Y_{20} \right] } \] which explicitly becomes: \[ V_{\text{out}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{3} \frac{R^2 \sigma_0}{r} + \frac{2}{3} R^2 \sigma_0 \frac{3 \cos^2 \theta - 1}{r^3} \right] \] --- **Inside (\( r < R \)):** \[ \boxed{ V_{\text{in}}(r, \theta) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{3} R \sigma_0 + \frac{1}{3} \frac{\sigma_0}{R} r^2 (3 \cos^2 \theta - 1) \right] \] --- These are the complete final results for the multipole moments and potentials.