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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.
- What is the net electric field on the Cl atom due to eight Cs atoms?
- 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?
Solution
- Zero, from symmetry.
- 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–
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