Exercise 29 notes that more energy is required to ionize sodium that is retrieved by adding that electrons to an isolated chlorine atom, but the NaCl bond represents a lower energy because the attracting ions draw close together. Quantifying the energy-lowering effect of having alternating plus and minus charges can be rather involved for 3D lattice, but a one dimensional calculation is instructive. Consider an infinite line of point charges alternating between +e and –e, with a uniform spacing between adjacent opposite charge of a. (a) The electrostatic potential energy per ion is the same for a given positive ion as for a given negative ion. Why? (b) Calculate the electrostatic potential energy per ion for simplicity. Assume that a positive charge is at the origin. The following power series expansion will be helpful: $\mathbf{ln}\left(1+x\right)\mathbf{=}\mathbf{-}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\gamma }}\frac{{\left(-x\right)}^{\mathbf{n}}}{\mathbf{n}}$.

(a)

Note while choosing the potential chooseeither the positive of the negative charge at the origin such that the net potential is same at all the point such that it is the summation till infinity

(b)

Consider the equation for the positive charge at A from all the negative charges is:

$$\begin{gathered} U_{+,-}=\frac{\left(+e\right)\left(-e\right)}{4\pi {\in }_{0}a}+\frac{\left(+e\right)\left(-e\right)}{{\mu }_{0}3a}+\frac{\left(+e\right)\left(-e\right)}{{\mu }_{0}5a}+\dots +\frac{\left(+e\right)\left(-e\right)}{{\mu }_0\left(2n-1\right)a} \\ =\frac{-e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\frac{1}{2n-1} \end{gathered}$$

Consider the coulomb potential for all the positive charges as:

$$\begin{gathered} U_{++}=\frac{\left(+e\right)\left(+e\right)}{4\pi {\in }_{0}2a}+\frac{\left(+e\right)\left(+e\right)}{4\pi {\in }_{0}4a}+\dots +\frac{\left(+e\right)\left(+e\right)}{4\pi {\in }_0\left(2n\right)a} \\ =\frac{e^2}{4\pi {\in }_{0}a}\left(\sum _{n=1}^{\infty }\frac{1}{2n}\right) \end{gathered}$$

Solve for the net potential at the point B as:

$$\begin{gathered} U_A=U_{+}+U_{++} \\ =\left(\frac{-e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\frac{1}{2n-1}\right)+\left(\frac{e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\frac{1}{2n-1}\right) \\ =\frac{e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\left(\frac{1}{2n}-\frac{1}{2n-1}\right) \end{gathered}$$

Consider the formula as:

$U_A=\frac{-e^2}{4\pi {\in }_{0}a}{\log }_{e}2\left\{{\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots \right\}$

Consider the equation for the negative charges at B as:

$$\begin{gathered} U_{-,-}=\frac{\left(+e\right)\left(-e\right)}{4\pi {\in }_{0}a}+\frac{\left(+e\right)\left(-e\right)}{4\pi {\in }_{0}3a}+\dots +\frac{\left(+e\right)\left(-e\right)}{4\pi {\mu }_0\left(2n-1\right)a} \\ =\frac{-e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\frac{1}{2n-1} \end{gathered}$$

Consider the coulomb potential for all the positive charges as:

$$\begin{gathered} U_{++}=\frac{\left(+e\right)\left(+e\right)}{4\pi {\in }_{0}2a}+\frac{\left(+e\right)\left(+e\right)}{4\pi {\in }_{0}4a}+\dots +\frac{\left(+e\right)\left(+e\right)}{4\pi {\in }_0\left(2n\right)a} \\ =\frac{e^2}{4\pi {\in }_{0}a}\left(\sum _{n=1}^{\infty }\frac{1}{2n}\right) \end{gathered}$$

Solve for the net potential at the point B as:

$$\begin{gathered} U_B=U_{--}+U_{-+} \\ =\left(\frac{-e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\frac{1}{2n-1}\right)+\left(\frac{e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\frac{1}{2n-1}\right) \\ =\frac{e^2}{4\pi {\in }_{0}a}\sum _{n=1}^{\infty }\left(\frac{1}{2n}-\frac{1}{2n-1}\right) \\ =\frac{e^2}{4\pi {\in }_{0}a}{\log }_{e}2 \end{gathered}$$