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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

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

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

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

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

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

This is how I proved it.

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

\(V=\frac{q}{4\pi\varepsilon_0x_{0}}\)

where \(x_{0}\) 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

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

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

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

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

Thus we get,

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

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
7 months, 4 weeks ago

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

Your mistake comes from the subtraction of two divergent series as, \[\sum\limits_{k = 1}^\infty \frac{1}{k}\] does not converge! Ameya Daigavane · 7 months, 3 weeks ago

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@Ameya Daigavane Okay now I get it. Thank you so much Anirudh Chandramouli · 7 months, 3 weeks ago

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@Anirudh Chandramouli 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 \(\ln(1+x)\) has the following domain: \((-1, 1]\) Deeparaj Bhat · 7 months, 1 week 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 · 7 months, 3 weeks ago

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