represents a crystal unit of cesium chloride,

(i) From the given figure, we can observe that Cl^‒ atom is at the centre whereas Cs⁺ atoms are at eight corners at equal distance. So, by symmetry net force on Cl^‒ atom due to other Cs⁺ atoms is zero. Due to which, net electric field due to other Cs⁺ atoms is also zero (as, |E| = F/q)

(ii) In the given question, removing Cs atom at the corner A is equivalent to adding a singly charged negative Cs ion at point A.

So,

Net force = q²/(4πε_or²)

Here, q = charge on electron, r = distance between Cl and Cs atoms

Applying Pythagoras theorem, we have

$r=\sqrt{(0.02)^{2}+(0.20)^{2}+(0.20)^{2}} \times 10^{-9} \mathrm{~m}=0.346 \times 10^{-9} \mathrm{~m}$

Now,

Net force $=\frac{q^{2}}{4 \pi \varepsilon_{0} r^{2}}=\frac{9 \times 10^{9}\left(1.6 \times 10^{-19}\right)^{2}}{\left(0.346 \times 10^{-9}\right)^{2}}=1.92 \times 10^{-9} \mathrm{~N}$

The direction of force is from A to Cl^‒