Mathematical Physics Dilemma

Hello everyone.

I came across a physics question as follows,

A set of positive point charges are kept at the points on the x-axis, such that


And a set of negative point charges are placed on the x-axis, such that,


If the magnitude of the positive point charge is +q+q and the magnitude of the negative point charge is q-q, assuming ε0\varepsilon_0 is the absolute permittivity of free space, then prove that the electric potential at the origin is


This is how I proved it.

We know by the definition of electric potential due to a point charge,


where x0x_{0} is the separation between the origin and the point where the point charge is placed. Also it is known that the electric potential due to a system of point charges is the algebraic sum of the electric potentials due to the individual point charges.

Hence by applying this to every single point charge given, we get

V=q4πε0(x0)+q4πε0(2x0)+q4πε0(3x0)+q4πε0(4x0)+............. V=\frac{q}{4\pi\varepsilon_0(x_{0})} + \frac{-q}{4\pi\varepsilon_0(2x_{0})} + \frac{q}{4\pi\varepsilon_0(3x_{0})} + \frac{-q}{4\pi\varepsilon_0(4x_{0})}+.............\infty

Factoring out q4πε0x0\frac{q}{4\pi\varepsilon_0x_{0}}, we get

V=q4πε0(x0)[112+1314+.............] V=\frac{q}{4\pi\varepsilon_0(x_{0})}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.............\infty]

We know that the infinite summation in the square brackets is the Maclaurin's series expansion of ln(1+x)\ln(1+x) where x=1x=1

Thus we get,


But instead of substituting for the logarithm expansion if we did as I have done in the attached picture, then we get answer as zero. Please tell me where I have gone wrong.

Note by Anirudh Chandramouli
5 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

First of all, you cannot apply the Maclaurin's series of ln(1+x)\ln(1 + x) outside its radius of convergence, which is 1<x<1 -1 < x < 1 .
There are, however, other methods to show that ln2=k=1(1)k+1k \ln 2 = \sum\limits_{k = 1}^\infty \frac{(-1) ^{k + 1}}{k}

Your mistake comes from the subtraction of two divergent series as, k=11k\sum\limits_{k = 1}^\infty \frac{1}{k} does not converge!

Ameya Daigavane - 5 years, 1 month ago

Log in to reply

Okay now I get it. Thank you so much

Anirudh Chandramouli - 5 years, 1 month ago

Log in to reply

Actually, if you change the order of summation for non-absolutely converging series, the sum changes. In fact, by a suitable rearangement of terms, the summation given above can be made equal to any real number. Plus, the Maclaurin's series of ln(1+x)\ln(1+x) has the following domain: (1,1](-1, 1]

Deeparaj Bhat - 5 years, 1 month ago

Log in to reply

And secondly when you rearrange the order of the terms of a series that goes to infinite you are changing the value of the sum

Prit Savani - 5 years, 1 month ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...