## Potential at the end of a rod

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.

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 Recognitions: Gold Member 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?
 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).

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## Potential at the end of a rod

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

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