Laplace Sphere Adfffa

1. **Problem Statement:** Find the potential $u(r, \phi)$ inside a sphere of radius $R=1$ where there are no charges inside, so $\nabla^2 u = 0$, and the potential on the surface of the sphere is constant: $f(\phi) = 1$. Show that the potential outside the sphere behaves like that of a point charge at the origin, i.e., $u(r) \propto \frac{1}{r}$. 2. **Formula and Important Rules:** The Laplace equation in spherical coordinates (assuming azimuthal symmetry, no $\theta$ dependence) is: $$\nabla^2 u = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2 \sin \phi} \frac{\partial}{\partial \phi} \left(\sin \phi \frac{\partial u}{\partial \phi}\right) = 0$$ The general solution inside the sphere can be expressed as a series of Legendre polynomials $P_n(\cos \phi)$: $$u(r, \phi) = \sum_{n=0}^\infty A_n r^n P_n(\cos \phi)$$ The boundary condition on the sphere $r=1$ is: $$u(1, \phi) = f(\phi) = 1$$ Since $f(\phi) = 1$ is constant, it corresponds to $P_0(\cos \phi) = 1$ only. 3. **Solve for the interior potential (Problem #17):** Since $f(\phi) = 1$, only the $n=0$ term is nonzero: $$u(r, \phi) = A_0 r^0 P_0(\cos \phi) = A_0$$ Apply the boundary condition at $r=1$: $$u(1, \phi) = A_0 = 1$$ So, $$\boxed{u(r, \phi) = 1}$$ This means the potential inside the sphere is constant everywhere. 4. **Solve for the exterior potential (Problem #21):** Outside the sphere ($r > 1$), the general solution to Laplace's equation that vanishes at infinity is: $$u(r, \phi) = \sum_{n=0}^\infty B_n r^{-(n+1)} P_n(\cos \phi)$$ Since the boundary condition on the sphere is $u(1, \phi) = 1$, again only the $n=0$ term is nonzero: $$u(r, \phi) = B_0 r^{-1} P_0(\cos \phi) = \frac{B_0}{r}$$ Apply the boundary condition at $r=1$: $$u(1, \phi) = B_0 = 1$$ So, $$\boxed{u(r) = \frac{1}{r}}$$ This is exactly the potential of a point charge at the origin. **Final answers:** - Interior potential: $$u(r, \phi) = 1$$ - Exterior potential: $$u(r) = \frac{1}{r}$$