Electric Field of Point Charge at y=r and an Infinitely Long Cylinder

[MrBlank]

Let point charge q be at y=r. Let there be an infinite conducting plane along the x-axis and z-axis that is neutrally charged. In this case, the method of mirror charges can be used. The plane is replaced by a point charge -q at y=-r. The electric field for y > 0 is the same in both cases. This is a well known case.

Let point charge q be at y=r. Let there be an infinitely long conducting wire along the x-axis that is neutrally charged. In this case, the method of image charges can not be used. What is the force on q due to the separation of charge that it causes in the wire?

It occurs to me that treating the wire as an infinely long cylinder would help, but I’m not sure how to do the calculations.

 

[kuruman]

It occurs to me that treating the wire as an infinely long cylinder would help, but I’m not sure how to do the calculations.

How would that help? In the case of the wire you have a line that is an equipotential and a linear charge distribution $\lambda(x)$ to deal with. In the case of a cylinder you would have a two dimensional surface charge density $\sigma(x,\theta)$. I am not sure how to do the calculation either at this point, but it seems to me that the one-dimensional case would be less complicated.

 

[Motocross9]

I believe the solution is to replace the problem by a cylinder of radius R (I'm using R instead of r for clarity) where the top half ($\phi$ going from 0 to $\pi$) is at potential $+V_{0}$ and the bottom half ($\phi$ going from 0 to $-\pi$) at $-V_{0}$. Once you have done this, the solution is (I can attach pictures of how I solved this if needed)

$V_{in}(s,\phi)=(2V_{0}/\pi)[\arctan(s\sin(\phi)/(R+s\cos(\phi)))+\arctan(s\sin(\phi)/(R-s\cos(\phi)))]$

$V_{out}(s,\phi)=(2V_{0}/\pi)[\arctan(R\sin(\phi)/(s+R\cos(\phi)))+\arctan(R\sin(\phi)/(s-R\cos(\phi)))]$

If you plug in s=0 corresponding to the placement of the neutral wire in the problem, you get a potential of zero. The solution when s is infinite is zero. Hope this helps.

Note: while the solution to the problem I gave is correct, I am not 100% confident that this gives the solution to the problem in question.

 

[kuruman]

I suppose $s$ is a length along the cylinder and $\phi$ is the polar angle in cylindrical coordinates. What do $V_{in}(s,\phi)$ and $V_{out}(s,\phi)$ represent? The original problem is about a long conducting wire with a charge placed near it at distance $r$. You want to replace this with a conducting cylinder. Fine. However, if $V_{in}$ is the potential inside this cylinder, then it must be constant and your expression does not reflect this.

Unless I don't understand what you are trying to do, I think that demanding that half the cylinder be at $+V_0$ and the other half at $-V_0$ contradicts the premises of the original question and is not an appropriate approach. Please explain why you think it is.

Also, what does $R$ represent in your expressions? It seems to me that, if you are considering a cylinder, the expressions must have in them the distance $r$ of the charge from the cylinder axis and the radius $R$ of the cylinder.

On edit:

Note: while the solution to the problem I gave is correct, I am not 100% confident that this gives the solution to the problem in question.

I don't think it does.

 

[Motocross9]

I suppose $s$ is a length along the cylinder and $\phi$ is the polar angle in cylindrical coordinates. What do $V_{in}(s,\phi)$ and $V_{out}(s,\phi)$ represent? The original problem is about a long conducting wire with a charge placed near it at distance $r$. You want to replace this with a conducting cylinder. Fine. However, if $V_{in}$ is the potential inside this cylinder, then it must be constant and your expression does not reflect this.

Unless I don't understand what you are trying to do, I think that demanding that half the cylinder be at $+V_0$ and the other half at $-V_0$ contradicts the premises of the original question and is not an appropriate approach. Please explain why you think it is.

Also, what does $R$ represent in your expressions? It seems to me that, if you are considering a cylinder, the expressions must have in them the distance $r$ of the charge from the cylinder axis and the radius $R$ of the cylinder.

When using the method of images, it is my understanding that you look for a situation that satisfies the same boundary conditions that the original problem does, but is easier to solve. In what I presented, I did exactly that. Where I am not confident is the fact that I never actually used any image charges, I just replaced the problem with another one that meets the same B.C.'s. That's my understanding of it and, for example, Griffiths book explains the method of images in a similar way--I am not sure if there is added complexity where you must form a set of image charges to solve the problem or not.

$V_{in}$ represents the potential inside the cylinder ($s<R$), and $V_{out}$ represents the potential outside the cylinder ($s>R$).

You would be right that the potential inside would have to be zero if the cylinder was connected at +-$\pi$, but the problem represents two conductors pasted together, separated by an insulator or something, so the potential inside the cylinder will not be zero.

I used R instead of r so there is no confusion between s and r (some people use s for $\sqrt(x^2+y^2)$).

 

[kuruman]

When using the method of images, it is my understanding that you look for a situation that satisfies the same boundary conditions that the original problem does, but is easier to solve. In what I presented, I did exactly that.

Your understanding is correct, but in this case you did not find a situation that satisfies the same boundary conditions. In the original problem, the cylindrical conductor is an equipotential. That means that when $s=R$, the potential is the same (not necessarily zero) and independent of $\phi$.

The potentials that you have presented show that, when $s=R$, $V(R,\pi+\phi)=-V(R,\phi).$ What you propose as a substitute solution has a boundary condition that depends on $\phi$ and, therefore, does not match the boundary conditions of the original problem.

 

[Motocross9]

Your understanding is correct, but in this case you did not find a situation that satisfies the same boundary conditions. In the original problem, the cylindrical conductor is an equipotential. That means that when $s=R$, the potential is the same (not necessarily zero) and independent of $\phi$.

The potentials that you have presented show that, when $s=R$, $V(R,\pi+\phi)=-V(R,\phi).$ What you propose as a substitute solution has a boundary condition that depends on $\phi$ and, therefore, does not match the boundary conditions of the original problem.

I understand what you are saying; however, the cylinder does not represent the neutral conducting rod in the original problem. The problem I proposed only has a cylindrical shell with different potentials on the top and bottom half, nothing else.

This is similar to a point charge above an infinite conducting plane. An entirely new problem is proposed where there is a point charge at distance d above the xy-plane and another point of equal and opposite charge a distance d below the xy-plane. What I did was attempt to create a new problem that satisfies the boundary conditions in the original problem. The cylinder problem I gave yields a potential of zero when s=0, where the rod was in the original problem and it goes to zero at infinity. It likely needs tweaking so that the original problem is modeled exactly (from the solution I gave, I don't believe the charge of the point charge in the original problem is there). Moreover, any solution to Laplace's equation in cylindrical coordinates would satisfy the same boundary conditions, since they all would yield a potential of zero at the z-axis and go to zero at infinity. So what I did isn't correct. There is obviously more to the method of images than just meeting the first-glance, obvious boundary conditions.

So the solution isn't correct, but I believe a similar method of solution will yield an answer with information about the charge of the point charge being there.

Edit: after looking at this again, if someone is to solve this using Laplace's equation in some manner, the trial solution should be Z-dependent, in addition I realize this is actually a case of solving Poisson's equation. Disregard my original attempt, but the idea might help you solve it in a similar manner.

 

[Meir Achuz]

None of the above work, and I won't be of too much help either. If the wire is along the z axis, the first easy step is finding the potential of the point charge along the z axis. Then remove the wire and find a charge distribution along the z axis that cancels the original potential. This involves an integral equation I can't solve.

 

[Motocross9]

None of the above work, and I won't be of too much help either. If the wire is along the z axis, the first easy step is finding the potential of the point charge along the z axis. Then remove the wire and find a charge distribution along the z axis that cancels the original potential. This involves an integral equation I can't solve.

That's an interesting way to solve. Could you show what you did? But if someone was familiar enough with Poisson's equation, you could solve it that way with the point charge being a delta function in terms of charge density. The uniqueness theorem guarantees a solution as long as the charge density and the potential at the boundaries are specified .

 

[Meir Achuz]

$-\frac{q}{\sqrt{d^2+x^2}}=\int_{-\infty}^{+\infty}\frac{\lambda(x')dx'}{x'-x}$ is the integral equation to solve for $\lambda(x)$.