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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
1 year, 7 months 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!

- 1 year, 7 months ago

Okay now I get it. Thank you so much

- 1 year, 7 months ago

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: $$(-1, 1]$$

- 1 year, 6 months ago

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

- 1 year, 7 months ago