Question:

Infinite numbers of positive and singly negative charged ions are placed in a straight line alternatively with spacing of \(s\). Calculate the potential energy per ion.

Answer:

As shown in the figure, the energy of a particle is gained from all the other particles. The electric potential energy is derived from integrating the electric force.

\[ U = -\int F \cdot dr = \frac{k q q’}{r} \]

where \(k = \frac{1}{4\pi \epsilon_0}\).

One particle is affected by both-side particles, so it should be multiplied by a factor of two. Otherwise, we can just sum all the other energy produced by each ion.

\[ U_{\mathrm{total}} = \sum U = 2k   \]

\[   \left(-\frac{e^2}{s}+\frac{e^2}{2s}-\frac{e^2}{3s}+\cdots\right)  \]

\[     = =-\frac{2ke^2}{s}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right) \]

Recall that

\[ \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots \]

Substituting \(x=1\), we have

\[ \ln(1+1)=1-\frac{1}{2}+\frac{1}{3}+\cdots \]

Therefore,

\[ U_{\mathrm{total}} = -\frac{2ke^2}{s}\ln(2) \]

or

\[ U_{\mathrm{total}} = -\frac{2e^2}{4\pi \epsilon_0 s}\ln(2) \]

This is the answer!