Solving Laplace equations, electrostatics

[Brian-san]

Homework Statement

a) Consider a conducting sphere of radius R whose surface is maintained at a potential $\Phi(R)=\Phi_0cos\theta$. Assuming that there are no charges present (inside or outside), what is the potential inside and outside the sphere?

b) Consider a cylindrical conducting can of radius R and height h. The side and the bottom walls of the can are grounded while the top face is maintained at $\Phi_0(r,\phi)$. Find the electrostatic potential inside the can.

Homework Equations

Laplace Equation
$$\nabla^2\Phi=0$$

The Attempt at a Solution

For part a, we already derived the general solution for spherical coordinates in class. The boundary condition seems to imply symmetry in the azimuthal angle. So the general solution is given by
$$\Phi(r,\theta)=\sum_{l=0}^{\infty}\left(A_lr^l+\frac{B_l}{r^{l+1}}\right)P_l(cos\theta)$$

If we assume that the potential goes to zero at infinity, then all $A_l=0$. Furthermore, $B_l=0, l\geq2$ as, we need only a cosine term at r=R. The solution is then found by plugging in the boundary condition, so
$$\Phi_0cos\theta=\frac{B_0}{R}+\frac{B_1}{R^2}cos\theta$$

From this, $B_0=0, B_1=\Phi_0R^2$ and the specific solution is given by
$$\Phi(r,\theta)=\frac{\Phi_0R^2}{r^2}cos\theta$$

I put this expression back into the Laplace equation and got zero, and it satisfies the boundary condition, so I believe must be the correct solution. The only piece I'm not sure of is using the fact that the potential is zero at infinity.

For part b, I located the general solution in cylindrical coordinates, which is
$$\Phi(r,\phi,z)=\sum_{n=0}^{\infty}\left(A_nJ_n(kr)+B_nN_n(kr)\right)\left(C_nsin(n\phi)+D_ncos(n\phi)\right)\left(E_nsinh(kz)+F_ncosh(kz)\right)$$

Mathematically, the boundary conditions are represented as
$$\Phi(r\leq R,\phi,z=0)=0, \Phi(r=R,\phi,0<z<h)=0, \Phi(r,\phi,z=h)=\Phi_0(r,\phi)$$

I'm not exactly sure how to apply the boundary conditions in this problem, as I've never encountered the Bessel functions before, which appear in the radial part of the solution. I think the boundary condition at z=0 implies $C_n=0, n\geq 0, D_n=0, n\geq 1$ since the potential is zero for all angles, so then
$$\Phi(r,\phi,z)=\left(A_1J_1(kr)+B_1N_1(kr)\right)\left(D_1F_1cosh(kz)\right)$$

I'm not sure if this step is right, so how would I proceed from here?

 

[gabbagabbahey]

$$\Phi(r,\theta)=\frac{\Phi_0R^2}{r^2}cos\theta$$

I put this expression back into the Laplace equation and got zero, and it satisfies the boundary condition, so I believe must be the correct solution. The only piece I'm not sure of is using the fact that the potential is zero at infinity.

This is fine for the potential outside the sphere.What about the potential inside?

For part b, I located the general solution in cylindrical coordinates, which is
$$\Phi(r,\phi,z)=\sum_{n=0}^{\infty}\left(A_nJ_n(kr)+B_nN_n(kr)\right)\left(C_nsin(n\phi)+D_ncos(n\phi)\right)\left(E_nsinh(kz)+F_ncosh(kz)\right)$$

Mathematically, the boundary conditions are represented as
$$\Phi(r\leq R,\phi,z=0)=0, \Phi(r=R,\phi,0<z<h)=0, \Phi(r,\phi,z=h)=\Phi_0(r,\phi)$$

You also require that $\Phi$ be finite everywhere, including at $r=0$...what does that tell you about $B_n$?

I'm not exactly sure how to apply the boundary conditions in this problem, as I've never encountered the Bessel functions before, which appear in the radial part of the solution. I think the boundary condition at z=0 implies $C_n=0, n\geq 0, D_n=0, n\geq 1$ since the potential is zero for all angles, so then
$$\Phi(r,\phi,z)=\left(A_1J_1(kr)+B_1N_1(kr)\right)\left(D_1F_1cosh(kz)\right)$$

I'm not sure why you think this...plugging $z=0$ into your expression does not give $\Phi=0$ for all $0\leq r\leq R$ except for the trivial case $A_1=B_1=0$.

Instead, plug $z=0$ into your general solution and you should realize that $F_n=0$ is the only way the potential will be zero for all $\phi$ and all $0\leq r\leq R$.