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 mm, 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. ltBrgt (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 seven remaining Cs atom?

Explanation

(i) Fig, Chiloine atom at the center of the cube is attracted equally by eight cesium atoms at the eight corners of the cube Symmetry shows that these forces would cancel out in pairs. Therefore, net electric field on CI atm due to eight Cs atoms is zero.
(ii) Removing a Cs atom at the corner A is equivalent to adding a singly charged negative Cs ion at A. Net force on the CI atom at A would be `F = (e^(2))/(4pi in_(0) r^(2))`, where r = distance between CI iion and Cs ion, i.e,
`r = sqrt((0.20)^(2) + (0.20)^(2) + (0.20)^(2)) xx 10^(-9) m = 0.346xx10^(-9) m`
`F = (e^(2))/(4pi in_(0) r^(2)) = (9xx10^(9) (1.6xx10^(-19))^(2))/((0.346xx10^(-9))^(2)) = 1.92xx10^(-9) N`