Question

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.

1. What is the net electric field on the Cl atom due to eight Cs atoms?
2. 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 atoms?

Answer 1

1. Zero, from symmetry.
2. Removing a +ve Cs ion is equivalent to adding singly charged –ve Cs ion at that location.

Net force then is F = `e^2/(4pi ε_0 r^2)`

Where r = distance between the Cl ion and a Cs ion.

= `sqrt((0.20)^2 + (0.20)^2 + (0.20)^2) xx 10^-9` m

= `sqrt(3(0.20)^2) xx 10^-9` m

= 0.346 × 10^–9 m

Hence, F = `((8.99 xx 10^9)(1.6 xx 10^-19)^2)/(0.346 xx 10^-9)^2` = 192 × 10^–11

= 1.92 × 10^–9 N, directed from A to Cl^–