Method of Images two parallel pipes at potentials of 0 & +V

[Erwin Derek]

Consider the scenario where there are two parallel conducting pipes of radius $R$ separated by a distance $d$, with pipe 1 at a potential of $-V$ and pipe 2 at a potential $+V$. I have seen from many sources that there is a very easy method of images solution to the potential outside the pipes, given that the potential goes to $0$ at $\infty$.

 

Now I thought of an alternate scenario where the only difference is that instead of $-V$ and $+V$ it is now $0$ and $+V$. Can this still be solved with method of images?

I have tried using two line charges of unequal charge and the equipotential surfaces end up being slightly distorted circles, so that doesn't work.

Otherwise, how can we do this with separation of variables?

 

[mfb]

Superposition still applies. You can use the original solution to find the potential distribution if only one pipe is charged, you can find suitable charges for the two pipes and then add two of these solutions together.

 

[Erwin Derek]

Could you please elaborate more, I'm not sure I completely understood that?

Everything after the first sentence flew over my head :P

 

[mfb]

There were just twos sentences! ;)

Find the original symmetric solution, determine the charge density on both pipes. Take one of them, determine the field it produces and find the corresponding potentials of the pipes. Then find a combination of (charge on pipe 1) and (charge on pipe 2) to get the potentials at the pipes right.

 

[Erwin Derek]

I feel like I understand what you're saying but I'm having trouble following this for some reason--

Let's say in the original scenario we have two pipes $-V$ and $+V$, so now we found some charges $L_1=-\lambda$ and $L_2=+\lambda$, so suppose each of these produces a potential on pipes 1 and 2 as $V_{11}$, $V_{12}$, $V_{21}$, and $V_{22}$, so now $V_{11}+V_{21}=-V$ for all points on pipe 1 and $V_{12}+V_{22}=+V$ for all points on pipe 2.

Now I want a scenario with, let's say two pipes $0$ and $+2V$. I need to find a combination of $L_1$ and $L_2$ that will produce these potentials on the surface of the pipes?

Just considering the $0$ pipe alone, how would this be possible? Any combination of the pipes that would produce a $0$ potential on this pipe will also zero out the potential on the other pipe, and if I clone some more charges to try getting pipe 2's to $+2V$, won't I break the 0 potential on pipe 1?

 

[mfb]

Pipe 1 will have a small negative charge, pipe 2 will have a larger positive charge.

You can also interpret this as superposition of "oppositely charged pipes" (with a non-zero field going through the symmetry plane) and "both pipes with the same charge" (with zero orthogonal field at the symmetry plane).

 

[Erwin Derek]

Interesting. If two pipes have opposite charge however, the equipotential surfaces are exactly cylinders, which helps when using parallel pipes. However if they have the same charge, the surfaces are very distorted and do not form circles.

Do you know of a solution to two conducting pipes with potential of $+V$ and $+V$?

Thanks for your help