Boundary Conditions for infinite grounded cylinder (Laplace Equation)

[mccragre]

Homework Statement

Find the potential outside of a long grounded conducting cylindrical rod of radius
R placed perpendicular to a uniform electric field E0.

Homework Equations

V(s,$$\phi$$) = $$a_{0}$$+$$b_0{}$$ln(s) + $$\sum$$($$A_n{}$$cos(n$$\phi$$)+$$B_n{}$$sin(n$$\phi$$))*($$C_n{}$$$$s^n{}$$+$$D_n{}$$$$s^{-n}$$)

The sum being from n=1 to infinity

The problem is independent of Z (on which the axis of rod lies)

The Attempt at a Solution

I know how to solve these types of problems, but I need the boundary conditions first so that I can begin solving for the coefficients. I know that the inner boundary condition is V(R,$$\phi$$)=0 (since it is grounded), but I'm stuck on other boundary conditions. I also know that the potential on the entire inside is zero. But I don't think the boundary condition V(0,$$\phi$$) is relevant in this case since we are talking about the outside potential.

I also know we can't set potential at infinity equal to zero since it was defined as a "long rod".

I'm also a little confused on what effect the E field has on the boundary conditions. I know that the charge will rearrange on the cylinder so that it creates an opposing E field on the inside to cancel out the external E field.

Any hints on what other boundary conditions there are would be much appreciated!

 

[gabbagabbahey]

You can't say that $V$ is zero at infinity, but certainly it must be finite everywhere, including at $s\to\infty$. What does that tell you about the coefficients $b_0$ and $C_n$?

What would you expect the net charge on the conducting cylinder to be? What does that tell you the highest order non-zero multipole moment due only to the cylinder (and the charge density induced on it) would be? Would you expect the potential due to the cylinder to be small or large compared to the potential due to just the uniform external field? What does that tell you about the total potential for $s\gg R$?Note: You may as well choose your coordinate axes so that the external field points along the positive z-axis

 

[mccragre]

Well this would mean that $$C_n{}$$ and $$B_n{}$$ would have to be zero if at infinity the potential can only be finite. That certainly makes solving for $$A_n{}$$ and $$B_n{}$$ easier since $$D_n{}$$ can just collapse into them. For those two the boundary condition at V(R,phi)=0 would probably suffice.

As for the multipole terms I know there will have to be a dipole term since the positive charge and negative charge will separate to two sides of the cylinder (since external E is orthogonal top and bottom). I'm unsure of the highest order multipole term though. I would expect the potential due to the cylinder to be smaller compared to the one created by the uniform field. So I guess the total potential for s>>R would be approximately just due to the uniform field.

 

[gabbagabbahey]

As for the multipole terms I know there will have to be a dipole term since the positive charge and negative charge will separate to two sides of the cylinder (since external E is orthogonal top and bottom). I'm unsure of the highest order multipole term though.

By highest order, I mean highest order in $r$ (the distance from the center of the cylinder/origin)...since the monopole moment must be zero, that would be the dipole term which falls off as $r^{-2}$.

I would expect the potential due to the cylinder to be smaller compared to the one created by the uniform field. So I guess the total potential for s>>R would be approximately just due to the uniform field.

Right, the potential due to just the induced charge density on the cylinder will fall off proportional to $r^{-2}$ (or faster), so as long as the external potential falls off slower than that with distance (it does), it will be significantly larger for $s\gg R$.

So, what is the potential corresponding to just the external field? Doesn't that give you another boundary condition?

 

[paranormal]

The boundary conditions for this problem are as follows:

1. At the inner boundary, V(R,φ) = 0, since the rod is grounded.

2. At the outer boundary (infinity), V(∞,φ) = 0, since the rod is infinitely long and there is no potential at infinity.

3. The potential on the entire inside of the cylinder is zero, V(s,φ) = 0, since the charge on the cylinder rearranges to create an opposing electric field that cancels out the external electric field.

4. The potential must be continuous across the boundaries, so V(R,φ) = V(R,0) = V(R,2π).

Using these boundary conditions, you can solve for the coefficients in the potential function and determine the potential outside of the cylinder. The external electric field will affect the boundary conditions by influencing the charge distribution on the cylinder, as mentioned above.