Irodov 3.128: A Logarithm in a Charge Chain – Amit & Deepali Sinha Foundation

Problem Irodov 3.128

Point charges $q$ and $-q$ alternate along an infinite straight line, neighbouring charges being a distance $a$ apart. Find the interaction energy of any one charge with all the others. (Hint: use the power series of $\ln(1 + \alpha)$ in $\alpha$ .)

A stretch of the infinite chain: charges of alternating sign a distance a apart (drawn for q > 0), continuing without end in both directions. — A stretch of the infinite chain: charges of alternating sign a distance $a$ apart (drawn for $q > 0$ ), continuing without end in both directions.

Given

infinite chain of alternating charges $q$ , $-q$
spacing $a$

Find

the interaction energy

W

of one charge with all the others

Solution

Step 1The energy of one charge in the field of the rest

Pick any charge of the chain, say a $q$ , and call it the reference charge. Its interaction energy with all the others is $q$ times the potential $\varphi$ that the others produce at its position:

W = q\varphi \tag{1}

Step 2The potential as an alternating series

On each side the charge at distance $na$ carries $(-1)^n q$ : the nearest neighbours are $-q$ , the next ones $+q$ , and so on. The two sides contribute equally, so

\begin{aligned} \varphi &= 2\cdot\frac{1}{4\pi\varepsilon_0}\sum_{n=1}^{\infty}\frac{(-1)^n q}{na}\\ &= -\frac{2q}{4\pi\varepsilon_0 a}\left(1 - \frac12 + \frac13 - \frac14 + \cdots\right) \end{aligned}

The series converges only because the signs alternate; a chain of like charges would have an infinite potential.

Step 3Summing the series with the logarithm

The power series of the logarithm, valid for $-1 < \alpha \le 1$ , is

\ln(1 + \alpha) = \alpha - \frac{\alpha^2}{2} + \frac{\alpha^3}{3} - \frac{\alpha^4}{4} + \cdots

At the end point $\alpha = 1$ it is exactly the bracket above:

1 - \frac12 + \frac13 - \frac14 + \cdots = \ln 2

Hence

\varphi = -\frac{2q\ln 2}{4\pi\varepsilon_0 a} \tag{2}

Step 4The interaction energy

Substituting (2) into (1):

W = -\frac{2\ln 2}{4\pi\varepsilon_0}\,\frac{q^2}{a}

(3)

A reference charge $-q$ sits in the potential $+2q\ln 2/(4\pi\varepsilon_0 a)$ , so its energy is the same; every charge of the chain is bound equally.

Check

Dimensions: $q^2/(\varepsilon_0 a)$ is charge times potential, an energy. Bounds: the partial sums $1$ , $1/2$ , $5/6$ of the series alternately overshoot and undershoot, and $\ln 2 = 0.693$ lies between each consecutive pair, so $\lvert W\rvert$ lies between $q^2/(4\pi\varepsilon_0 a)$ and $2q^2/(4\pi\varepsilon_0 a)$ . So $W$ is negative, as it must be with unlike charges nearest.

Remark

The number $2\ln 2 \approx 1.386$ is the Madelung constant of a one-dimensional ionic crystal: the binding of an ion in an alternating lattice is the energy of a single nearest-neighbour pair times this constant. In three dimensions the same alternating sum, for rock salt, gives about $1.748$ ; there too the convergence depends on the order in which the terms are added, which is why a lattice sum is taken over neutral shells.