Why Is the Potential Infinite at the End of a Charged Rod?

[schaefera]

Hi all!

So I've been doing some work with potential calculations, and I seem to be running into a bit of trouble with figuring out how to think about the potential due to a rod at the very edge of it.

Imagine an insulating rod with charge Q is placed along the positive x-axis, with its left end at the origin. I can calculate the potential anywhere on the y-axis, and anywhere on the negative x-axis quite easily. But when I try to take the limiting case of a point at the left end of the rod, something strange happens: I can let y approach 0, or x approach 0 in either of the expressions, and I find that the potential at the end of the rod is infinite. This is because the potential contains the natural log.

I contend that this doesn't make physical sense! You see, what if I had instead charged the rod to a charge Q+q. If the potential when charge Q is already on the rod is infinite at the end, how can I place the extra bit q on the rod? But I know that Q can have any value, which means I should be able to make Q large enough that it includes that extra bit, had I wanted to from the start.

 

[Drakkith]

I don't understand. Why are you trying to find the potential between 2 points on the rod if the rod is charged with Q? Wouldn't there be 0 potential?

 

[schaefera]

I think not- for one, it's not a conductor, and for two this is relative to a 0 potential at infinity, the way that I'm thinking about it (I'm integrating dV over the entire rod to get V, but this all presupposes that V=0 at infinite distances).

 

[Drakkith]

I think not- for one, it's not a conductor, and for two this is relative to a 0 potential at infinity, the way that I'm thinking about it (I'm integrating dV over the entire rod to get V, but this all presupposes that V=0 at infinite distances).

Do you have a charge on the end of the rod, or is the whole rod charged?

 

[haruspex]

Imagine an insulating rod with charge Q is placed along the positive x-axis, with its left end at the origin. I can calculate the potential anywhere on the y-axis, and anywhere on the negative x-axis quite easily. But when I try to take the limiting case of a point at the left end of the rod, something strange happens: I can let y approach 0, or x approach 0 in either of the expressions, and I find that the potential at the end of the rod is infinite. This is because the potential contains the natural log.
I contend that this doesn't make physical sense!

The defiance of physical sense reflects the unreality of the model. A uniform charge over a one-dimensional manifold will always produce infinite potential on the manifold itself.
To avoid this you need to make it at least 2 dimensional, but might as well make it 3. 5.8.7 in http://www.astro.uvic.ca/~tatum/celmechs/celm5.pdf derives the (gravitational) potential due to a uniform solid cylinder.