The figure represents a crystal unit of cesium chloride, $CsCl$. The cesium atoms, represented by open circles, are situated at the corners of a cube of side $0.40 \, nm$, whereas a $Cl$ atom is situated at the centre of the cube. The $Cs$ atoms are deficient in one electron while the $Cl$ atom carries an excess electron.
$(i)$ What is the net electric field on the $Cl$ atom due to eight $Cs$ atoms?
$(ii)$ Suppose that the $Cs$ atom at the corner $A$ is missing. What is the net force now on the $Cl$ atom due to the seven remaining $Cs$ atoms?

(N/A) $(i)$ The $Cs^{+}$ ions are situated at the corners of a cube and the $Cl^{-}$ ion is at the centre. Due to the symmetry of the cube, the electric field produced by each $Cs^{+}$ ion at the centre is equal in magnitude and directed away from the corner towards the opposite corner. These fields cancel each other out in pairs. Therefore, the net electric field at the centre is $0 \, N/C$.
$(ii)$ Let $\vec{E}_{total}$ be the electric field due to all eight $Cs^{+}$ ions at the centre, which is $0$. If one $Cs^{+}$ ion at corner $A$ is removed, the new electric field $\vec{E}'$ at the centre is given by $\vec{E}' + \vec{E}_{A} = 0$, where $\vec{E}_{A}$ is the field due to the $Cs^{+}$ ion at $A$. Thus, $\vec{E}' = -\vec{E}_{A}$.
The distance $r$ from a corner to the centre of a cube of side $a = 0.40 \, nm$ is half the body diagonal: $r = \frac{\sqrt{3}a}{2} = \frac{\sqrt{3} \times 0.40 \times 10^{-9}}{2} = 0.20\sqrt{3} \times 10^{-9} \, m \approx 3.464 \times 10^{-10} \, m$.
The magnitude of the electric field due to one $Cs^{+}$ ion at the centre is $E_{A} = \frac{k e}{r^2} = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{(0.20\sqrt{3} \times 10^{-9})^2} = \frac{14.4 \times 10^{-10}}{12 \times 10^{-20}} = 1.2 \times 10^{10} \, N/C$.
The force on the $Cl^{-}$ ion (charge $-e$) due to the remaining seven $Cs^{+}$ ions is $\vec{F} = (-e) \vec{E}' = (-e) (-\vec{E}_{A}) = e \vec{E}_{A}$.
The magnitude of the force is $F = e E_{A} = (1.6 \times 10^{-19}) \times (1.2 \times 10^{10}) = 1.92 \times 10^{-9} \, N$, directed towards the empty corner $A$.