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 0.40nm, whereas a CI atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the CI atom carries an excess electron.
(i) What is the net electric field on the CI 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 CI atom due to seven remaining Cs atoms ?

Explanation

(i) The cesium atoms are situated at the corners of a cube and 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 Cs atoms, on CI atom will cancel out.
Hence, `E = F/q, F=0`
(ii) We define force on a charge particle due to external electric field as F = qE. If eight cesium atoms are situated at the corners of a cube, the net force on CI alom 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 `Cl^(-)` ion by each Cs+ ion,
`F =(ke^(2))/r^(2)`........(1)
From Pythagoras theorem,
`r = sqrt((0.2 xx 10^(-9))^(2) + (0.2 xx 10^(-9))^(2) + (0.2 xx 10^(-9))^(2))`
`=sqrt(4+4+4) xx 10^(-10)`
`=sqrt(12) xx 10^(-10)`
`=3.46 xx 10^(-10)`m
`therefore` From equation (1),
`F = (9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(3.46 xx 10^(-10))^(2)`
`therefore F = 1.92 xx 10^(-9)` N