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 seven remaining Cs atom?

Explanation

(I )From the given figure, we can analy se the chlorine atom is at the centre of the cube i.e., at equal distance from all the eight coerners of cube where cesium atoms are placed. Thus due to symmetry the forces due to all Cs tons, on Cl atom will cancel out.
Hence, `E=(F)/(q)` where F=0
`thereforeE=0`
(ii) Thus, net force on Cla atom at A would be:
where r=distance between Cl ion and Cs ion.
Applying Pythagorous theorem we get
`r=sqrt((0.20)^(2)+(0.20)^(2)+(0.20)^(2))xx10^(-9)m`
`=0.346xx10^(9)m`
Now, `F=(q^(2))/(4piepsi_(0)r^(2))=(e^(2))/(4piepsi_(0)r_(2))`
`=(9xx10^(9)(1.6xx10^(-19))^(2))/((0.346xx10^(-9))^(2))=1.92xx10^(-9)N`