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Answer #1

+q -9 tq +q 9

Let consider the names of the charges as shown in the figure.

The potential energy exists in pairs therefore

$U_{12}=\frac{-kq^2}{d}\ ,U_{23}=\frac{-kq^2}{d}\ ,U_{34}=\frac{-kq^2}{d}\ , U_{14}=\frac{-kq^2}{d}\ ,U_{13}=\frac{kq^2}{\sqrt{2}d}\ ,U_{24}=\frac{kq^2}{\sqrt{2}d}$

Now considering the upper part of the cube,

$U_{56}=\frac{-kq^2}{d}\ ,U_{67}=\frac{-kq^2}{d}\ ,U_{78}=\frac{-kq^2}{d}\ , U_{58}=\frac{-kq^2}{d}\ ,U_{57}=\frac{kq^2}{\sqrt{2}d}\ ,U_{68}=\frac{kq^2}{\sqrt{2}d}$

Considering the pairs of upper and lower parts,

$U_{15}=\frac{-kq^2}{d}\ ,U_{26}=\frac{-kq^2}{d}\ ,U_{37}=\frac{-kq^2}{d}\ , U_{48}=\frac{-kq^2}{d}$

Now considering the body diagonals,

$U_{17}=\frac{-kq^2}{\sqrt{3}d}\ ,U_{28}=\frac{-kq^2}{\sqrt{3}d}\ ,U_{35}=\frac{-kq^2}{\sqrt{3}d}\ , U_{46}=\frac{-kq^2}{\sqrt{3}d}$

The total potential energy is the sum of all the potential energies due to the pairs.

$U=\frac{-4kq^2}{d}+\frac{2kq^2}{\sqrt{2}d}+\frac{-4kq^2}{d}+\frac{2kq^2}{\sqrt{2}d}+\frac{-4kq^2}{d}+\frac{-4kq^2}{\sqrt{3}d}$

$U=-\frac{12kq^2}{d}+\frac{4kq^2}{\sqrt{2}d}-\frac{4kq^2}{\sqrt{3}d}$

$U=\frac{4kq^2}{d}\left [ -4 +\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right ]$

Here it is clear from the above expression that the total potential energy of the system of charges is negative. The negative potential energy leads to stability. This shows these types of ionic crystal exists in nature.

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Answer 1

+q -9 tq +q 9

Let consider the names of the charges as shown in the figure.

The potential energy exists in pairs therefore

$U_{12}=\frac{-kq^2}{d}\ ,U_{23}=\frac{-kq^2}{d}\ ,U_{34}=\frac{-kq^2}{d}\ , U_{14}=\frac{-kq^2}{d}\ ,U_{13}=\frac{kq^2}{\sqrt{2}d}\ ,U_{24}=\frac{kq^2}{\sqrt{2}d}$

Now considering the upper part of the cube,

$U_{56}=\frac{-kq^2}{d}\ ,U_{67}=\frac{-kq^2}{d}\ ,U_{78}=\frac{-kq^2}{d}\ , U_{58}=\frac{-kq^2}{d}\ ,U_{57}=\frac{kq^2}{\sqrt{2}d}\ ,U_{68}=\frac{kq^2}{\sqrt{2}d}$

Considering the pairs of upper and lower parts,

$U_{15}=\frac{-kq^2}{d}\ ,U_{26}=\frac{-kq^2}{d}\ ,U_{37}=\frac{-kq^2}{d}\ , U_{48}=\frac{-kq^2}{d}$

Now considering the body diagonals,

$U_{17}=\frac{-kq^2}{\sqrt{3}d}\ ,U_{28}=\frac{-kq^2}{\sqrt{3}d}\ ,U_{35}=\frac{-kq^2}{\sqrt{3}d}\ , U_{46}=\frac{-kq^2}{\sqrt{3}d}$

The total potential energy is the sum of all the potential energies due to the pairs.

$U=\frac{-4kq^2}{d}+\frac{2kq^2}{\sqrt{2}d}+\frac{-4kq^2}{d}+\frac{2kq^2}{\sqrt{2}d}+\frac{-4kq^2}{d}+\frac{-4kq^2}{\sqrt{3}d}$

$U=-\frac{12kq^2}{d}+\frac{4kq^2}{\sqrt{2}d}-\frac{4kq^2}{\sqrt{3}d}$

$U=\frac{4kq^2}{d}\left [ -4 +\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right ]$

Here it is clear from the above expression that the total potential energy of the system of charges is negative. The negative potential energy leads to stability. This shows these types of ionic crystal exists in nature.

You can comment below for any further inquiry. Please provide positive feedback.