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

x=x0x=x_{0},x=3x0x=3x_{0},x=5x0x=5x_{0}........\infty

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

x=2x0x=2x_{0},x=4x0x=4x_{0},x=6x0x=6x_{0}........\infty

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

V=qln24πε0x0V=\frac{q\ln2}{4\pi\varepsilon_0x_{0}}

This is how I proved it.

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

V=q4πε0x0V=\frac{q}{4\pi\varepsilon_0x_{0}}

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,

V=qln24πε0x0V=\frac{q\ln2}{4\pi\varepsilon_0x_{0}}

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
3 years, 6 months ago

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1 vote

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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 - 3 years, 6 months ago

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Okay now I get it. Thank you so much

Anirudh Chandramouli - 3 years, 6 months ago

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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 - 3 years, 5 months ago

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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 - 3 years, 6 months ago

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