Electric Potential of a Charged Disk

  • #1
525
16

Main Question or Discussion Point

I was working on E&M I homework with my friend, and the final question was to find the electric potential at any point on the positive x axis of a charged disk (where the x axis is perpendicular to the centre of the disk) We solved this easily enough, starting with a point charge and integrating the electric potential in steps. We got to the end and had a formula for the electric potential at point x (it was the correct answer). Thing is, we know that potential is always in reference to some 'zero' point (infinity in the case of a point charge), so we decided to figure out if that point was infinity for a disk as well, and if not, what it would be. I worked on that for a while and came out with something... but neither of us is really sure about how to interpret it. I'm still in first year so I'm very inexperienced with coming up with this sort of thing on my own and I apologize for any "bad" methods I used. I fully expect to have messed something up, so please tell me what I've done wrong. If I managed to do everything right, my questions are at the end of the pdf that I've attached (I was having trouble with latex, so I just scanned some notes that I made about this)

Help is greatly appreciated. Sorry if anything is unclear.
 

Attachments

Answers and Replies

  • #2
478
32
I always get a bit uneasy when dealing with infinities. But I see 2 things in your analysis:

1. The very first thing you assume in your analysis (in taking the electric potential due to a point charge) is that the potential at infinity is zero.

2. As x -> infinity, x >> r and so Sqrt[x^2+r^2] is approximately Sqrt[x^2] or just x. So in your equation Vx=q/(2*Pi*r^2)*(Sqrt[x^2+r^2] -x), the (Sqrt[x^2+r^2] -x) could be said to approach zero. Maybe that's not technically mathematically correct, but physically that's what has to happen.
 
  • #3
rcgldr
Homework Helper
8,682
520
One thing to note is that as x becomes >> r, the disk approximates a point charge with respect to the distance x. The limit with a finite r and as x -> ∞ should be the same as a point charge.
 
  • #4
525
16
I always get a bit uneasy when dealing with infinities. But I see 2 things in your analysis:

1. The very first thing you assume in your analysis (in taking the electric potential due to a point charge) is that the potential at infinity is zero.

2. As x -> infinity, x >> r and so Sqrt[x^2+r^2] is approximately Sqrt[x^2] or just x. So in your equation Vx=q/(2*Pi*r^2)*(Sqrt[x^2+r^2] -x), the (Sqrt[x^2+r^2] -x) could be said to approach zero. Maybe that's not technically mathematically correct, but physically that's what has to happen.
Ah... so I just failed at limits. I guess I was looking for some sort of "constant divided by infinity" term rather than actually taking the limit properly. If having the reference point at infinity works, then I guess the rest of what I did is kind of pointless. I found that other x_0 but it's kind of a useless one because it's dependent on x. Well, thanks for showing me my error. I can't believe that neither my friend or me saw that.
 

Related Threads on Electric Potential of a Charged Disk

Replies
5
Views
864
  • Last Post
Replies
4
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
9
Views
896
Replies
5
Views
1K
Replies
3
Views
3K
Replies
1
Views
866
Replies
13
Views
3K
Top