Potential at the origin due to an infinite set of point charges

  • #1
SherlockHolmie
14
1
Moved from a technical forum.
Summary: Potential at origin of an infinite set of point charges with charge (4^n)q and distance (3^n)a along x-axis where n starts at 1.

From V=q/r, we find Vtotal=sum from 1 to infinity of (4/3)^n(q/a), which diverges. There cannot be infinite potential because there is a finite electric field at the point.

The only way I could think of solving this is by adding together all of the 1/v, getting the sum from 1 to infinity of (3/4)^n(a/q), but I'm not sure if 1/v is still considered linear and thus obeying the superposition principle.

The only other method I could think of to find V is to find the general electric field, but this will be very complicated as every point charge's electric field will have different values and directions.
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
15,369
4,356
every point charge's electric field will have different values and directions
I understand the former, but can you explain the latter ?
 
Last edited:
  • #3
jbriggs444
Science Advisor
Homework Helper
11,690
6,372
There cannot be infinite potential because there is a finite electric field at the point.
The potential field can diverge without its gradient being infinite anywhere.

Easy example is an infinite uniformly charged plane. The field gradient is constant and the potential increases uniformly.
 
  • Like
Likes Delta2 and Nugatory
  • #4
SherlockHolmie
14
1
I understand the former, but can you explain the latter ?
Each charge has a different magnitude, and they're at different locations, so other than on the horizontal axis, each point charge will have a different direction making a very complicated electric field.
 
  • #5
BvU
Science Advisor
Homework Helper
15,369
4,356
Isn't it so that at the origin all E are pointing in the same direction ?
 
  • #6
SherlockHolmie
14
1
The potential field can diverge without its gradient being infinite anywhere.

Easy example is an infinite uniformly charged plane. The field gradient is constant and the potential increases uniformly.

We were told there was a finite solution to the potential by the professor. The electric field is non-uniform and decreasing as it moves away from the origin in the negative x-direction, so if I do work between two points, there should be a potential difference.

What method would I use to find this potential difference?
 
  • #7
SherlockHolmie
14
1
Isn't it so that at the origin all E are pointing in the same direction ?
Yes, but in order to find the potential from the electric field, we need a function. The only way I know of how to get the electric field at the origin gives a constant that can't be integrated from -infinity to 0
 
  • #8
BvU
Science Advisor
Homework Helper
15,369
4,356
Agreed. Don't know how to tackle this one...
I agree V diverges, so I don't expect to get a finite answer by integrating ##E## from ##-\infty ## to ##0##.

Is this the entire exercise ? an exercise from a book ?
 
  • #9
jbriggs444
Science Advisor
Homework Helper
11,690
6,372
so if I do work between two points, there should be a potential difference.

What method would I use to find this potential difference?
Change the order of the summation.

Rather than taking the difference between two divergent sums, take the sum of the termwise differences.

To take this a step farther, since the potential difference to a reference at infinity diverges, define the reference potential at the origin instead. The stated problem is then solved by definition.
 
Last edited:
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
39,567
8,830
We were told there was a finite solution to the potential by the professor.
So this should be in a homework forum?

As @BvU notes, potential is relative. Without a specified zero potential the question cannot be answered. Most often, the potential is taken to be zero at infinity, but here that is problematic. At the least, it will depend on the direction in which one goes to infinity.
Even if we take infinity along the negative x axis, the sum is not going to converge. This opens the possibility that we could consider the difference in potential between the origin and the point -x, then let x tend to infinity, but my guess is this still will not converge.
So I have to wonder whether the problem has been stated accurately. Are the charges perhaps alternating in sign?
 
  • #11
SherlockHolmie
14
1
So this should be in a homework forum?

As @BvU notes, potential is relative. Without a specified zero potential the question cannot be answered. Most often, the potential is taken to be zero at infinity, but here that is problematic. At the least, it will depend on the direction in which one goes to infinity.
Even if we take infinity along the negative x axis, the sum is not going to converge. This opens the possibility that we could consider the difference in potential between the origin and the point -x, then let x tend to infinity, but my guess is this still will not converge.
So I have to wonder whether the problem has been stated accurately. Are the charges perhaps alternating in sign?
There's no obligation for us to turn this in. I went talking to the professor after the test trying to figure out how to do it, but I'm the only one working on this problem still.
 
  • #12
BvU
Science Advisor
Homework Helper
15,369
4,356
professor after the test trying to figure out how to do it
I'm interested in prof's intentions with this exercise. Why he thinks it's doable, for one.
 
  • #13
SherlockHolmie
14
1
I'm interested in prof's intentions with this exercise. Why he thinks it's doable, for one.
He said he mostly wanted us to compare it to the potential of an infinite wire, where reference point at infinity can't be taken. I'm guessing there's a natural log somewhere in the real potential function.
 
  • #14
jbriggs444
Science Advisor
Homework Helper
11,690
6,372
Potential at origin
We were told there was a finite solution to the potential by the professor
The potential difference between the origin and a reference at infinity diverges. If the question is as stated, the professor is simply incorrect.

So we must ask: Was the question: "Potential at origin" stated correctly and completely?

compare it to the potential of an infinite wire, where reference point at infinity can't be taken
This suggests that the professor is well aware that the reference at infinity cannot be used. Which, in turn, suggests that another reference be used instead. Which, again, means that we need to know what question you are actually being asked to answer.

Like "what is the potential difference between x=-1 and x=0". Or between x and y.
 
  • #15
SherlockHolmie
14
1
The potential difference between the origin and a reference at infinity diverges. If the question is as stated, the professor is simply incorrect.

So we must ask: Was the question: "Potential at origin" stated correctly and completely?


This suggests that the professor is well aware that the reference at infinity cannot be used. Which, in turn, suggests that another reference be used instead. Which, again, means that we need to know what question you are actually being asked to answer.

Like "what is the potential difference between x=-1 and x=0". Or between x and y.
243028
 
  • #16
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
Interesting little headcracker...

The interesting thing is, if I have done the math correctly, that the potential at any other finite point ##r_i## of the x-axis is also infinite. The only way to I can think to "renormalize" the infinity is to set the potential reference point at the origin as the one with potential zero as others have already suggested.
 
  • #17
36,295
13,371
You can define the potential at the origin to be zero, then the answer is zero - but that is trivial, of course. The field is finite everywhere outside the point charges, with this assignment we get a finite potential everywhere.
 

Suggested for: Potential at the origin due to an infinite set of point charges

Replies
7
Views
178
Replies
2
Views
812
Replies
2
Views
148
Replies
7
Views
864
Replies
16
Views
1K
  • Last Post
Replies
1
Views
116
Replies
1
Views
661
Replies
23
Views
301
Top