An Ionic Crystal. Figure P23.57 shows eight-point charges arranged at the corners of a cube with sides of length d. The values of the charges are +q and -q, as shown. This is a model of one cell of a cubic ionic crystal. In sodium chloride (NaCl), for instance, the positive ions are ${\mathit{N}}^{\mathbf{a}\mathbf{+}}$ and the negative ions are $\mathit{C}\mathit{I}$. (a) Calculate the potential energy $\mathit{U}$of this arrangement. (Take as zero the potential energy of the eight charges when they are infinitely far apart.) (b) In part (a), you should have found that $\mathit{U}\mathbf{<}\mathbf{0}$. Explain the relationship between this result and the observation that such ionic crystals exist in nature.

In each square face, the diagonal of the square and radius$\sqrt{2d}$ and each square has two connection and the cube have a total of 12 connection and their potential is positive.

The cube has 4 diagonals so from the cube rules each diagonal is$\sqrt{3d}$and as the charges are opposite so the potential is negative.

Taking charges at the distance (d) far from the each other and has 12 sides and since charges are not the same therefore potential is negative.

Let the voltage is positive and its radius is $\sqrt{2d}$and charges are (q)

So, potential it is given by

$V_1=\frac{12kq}{\sqrt{2d}}$

Now taking voltage as negative and radius is$\sqrt{2d}$and charges (q) from 4 connection

$V_2=\frac{-4kq}{\sqrt{3d}}$

As the voltage is negative, the charge is (q) and radius (r)

$V_3=\frac{-12kq}{d}$

So, the total voltage is

$$\begin{gathered} v_t=\frac{12kq}{\sqrt{2d}}-\frac{4kq}{\sqrt{3d}}-\frac{12kq}{d} \\ =\frac{-1.45q^2}{\pi {\epsilon }_{0}d} \end{gathered}$$

Therefore, the Potential energy of this arrangement isrole="math" localid="1664271395597" $V_t\frac{-1.45q^2}{\pi {\epsilon }_{0}d}$

Hence, As the crystal is solid, thus due to attraction force our result is negative.