# Q5 Solution (Step-wise) **Given:** Potential at the surface of a hollow spherical shell of radius \( R \) is \[ V(R, \theta) = V_ \cos \theta \] where \( V_ \) is a constant. **To Find:** The electric potential \( V(r, \theta) \) for \( r > R \) (outside the sphere). --- ## **Step 1: Use Laplace’s Equation and General Solution** For regions with no charge (\( r > R \)), Laplace's equation applies: \[ \nabla^2 V = \] In spherical coordinates (for azimuthal symmetry, i.e., no \(\phi\) dependence): \[ V(r, \theta) = \sum_{l=}^{\infty} \left[ A_l r^l + \frac{B_l}{r^{l+1}} \right] P_l(\cos\theta) \] - \( P_l \) are Legendre polynomials. - For \( r \to \infty \), the potential should remain finite, so set \( A_l = \) for all \( l \). Thus, outside the sphere: \[ V(r, \theta) = \sum_{l=}^{\infty} \frac{B_l}{r^{l+1}} P_l(\cos\theta) \] **Explanation:** We use Laplace's equation because there are no free charges outside. The general solution for azimuthal symmetry is a sum over Legendre polynomials, but the physically meaningful solution outside the shell only involves terms that decay at infinity. --- ## **Step 2: Apply Boundary Condition at \( r = R \)** Given boundary condition: \[ V(R, \theta) = V_ \cos\theta \] Recall: \[ P_1(\cos\theta) = \cos\theta \] So, only the \( l = 1 \) term is present. \[ V(R, \theta) = \frac{B_1}{R^2} \cos\theta = V_ \cos\theta \] Thus, \[ \frac{B_1}{R^2} = V_ \implies B_1 = V_ R^2 \] So, the potential outside (\( r > R \)) is: \[ V(r, \theta) = \frac{V_ R^2}{r^2} \cos\theta \] **Explanation:** We match the boundary condition at the surface using the form of the Legendre polynomial, finding that only the dipole (\( l=1 \)) term is nonzero. --- ## **Final Answer** \[ \boxed{ V(r, \theta) = V_ R^2 \frac{\cos\theta}{r^2}, \quad \text{for } r > R } \] This is the electric potential outside the hollow spherical shell.