Figure represents a crystal unit of cesium chloride, $\mathrm{CsCl}$. The cesium atoms, represented by open circles are situated at the corners of a cube of side $0.40\,\mathrm{nm}$, whereas a $\mathrm{Cl}$ atom is situated at the centre of the cube. The $\mathrm{Cs}$ atoms are deficient in one electron while the $\mathrm{Cl}$ atom carries an excess electron.

$(i)$ What is the net electric field on the $\mathrm{Cl}$ atom due to eight $\mathrm{Cs}$ atoms ?

$(ii)$ Suppose that the $\mathrm{Cs}$ atom at the corner $A$ is missing. What is the net force now on the $\mathrm{Cl}$ atom due to seven remaining $\mathrm{Cs}$ atoms ?

$(i)$ The cesium atoms are situated at the corners of a cube and $\mathrm{CI}$ atom is situated at the centre of the cube. From the given figure, we can analyses that the chlorine atom is at equal distance from all the eight corners of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all $\mathrm{Cs}$ atoms, on $\mathrm{Cl}$ atom will cancel out.

Hence $E=\frac{F}{q}, F=0$

$(ii)$ We define force on a charge particle due to external electric field as $\mathrm{F}=q \mathrm{E}$. If eight cesium atoms are situated at the corners of a cube, the net force on $\mathrm{Cl}$ atom is situated at the centre of the cube will be zero as net electric field at the centre of cube is zero. But, force acting on $\mathrm{Cl}^{-}$ion by each $\mathrm{Cs}^{+}$ion,

$\mathrm{F}=\frac{k e^{2}}{r^{2}} \quad \ldots \text { (1) }$

From Pythagoras theorem,

$r=\sqrt{\left(0.2 \times 10^{-9}\right)^{2}+\left(0.2 \times 10^{-9}\right)^{2}+\left(0.2 \times 10^{-9}\right)^{2}}$

$=\sqrt{4+4+4} \times 10^{-10}$

$=\sqrt{12} \times 10^{-10}$

$=3.46 \times 10^{-10} \mathrm{~m}$

$\therefore \text { From equation (1), }$

$\qquad$

$\mathrm{F} =\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{\left(3.46 \times 10^{-10}\right)^{2}}$

$\therefore \mathrm{F} =1.92 \times 10^{-9} \mathrm{~N}$