Ncert Exercises & Solutions

The figure represents a crystal unit of caesium chloride, CsCl. The caesium 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 \(E_{net}\) on the Cl-atom due to eight Cs-atoms, and suppose that the Cs-atom at corner A is missing. What is the net force \(F_{net}\) now on the Cl-atom due to seven remaining Cs-atoms?

1.	\(E_{net}=1.2 \times 10^9 ~\text{NC}^{-1},~F_{net} = 1.92 \times 10^{-9}~\text N\)
2.	\(E_{net}=0~\text{NC}^{-1},~F_{net} = 1.92 \times 10^{-9}~\text N\)
3.	\(E_{net}=0~\text{NC}^{-1},~F_{net} = 0~\text N\)
4.	\(E_{net}=1.2\times 10^9~\text{NC}^{-1},~F_{net} = 0~\text N\)

Hint: All Cs-atoms are at the same distance from Cl-atom and have the same charge.

Step 1: Find the net electric field at the Cl-atom.

(i) From the given figure, we can analyse that the chlorine atom is at the centre of the cube i.e., at an equal distance from all the eight corners of the cube where caesium atoms are placed. Thus, due to symmetry, the forces due to all Cs-atoms, on the Cl atom will cancel out.
Hence,

E = \frac{F}{q}

where F = 0

∴

E = 0

Step 2: Find the force on the Cl-atom in the second case.

(ii) Thus, the net force on Cl-atom at A would be,

F = \frac{e^{2}}{4 π ε_{0} r^{2}}

where r= distance between Cl-ion and Cs-ion.

Applying Pythagoras theorem, we get,

r = \frac{1}{2} \sqrt{[(0.40)^{2} + (0.40)^{2} + (0.40)^{2}]} \times 10^{- 9} m

= 0.346 \times 10^{- 9} m

Now,

F = \frac{q^{2}}{4 {πε}_{0} r^{2}} = \frac{e^{2}}{4 {πε}_{0} r^{2}}

= \frac{9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}}{{(0.346 \times 10^{- 9})}^{2}} = 1.92 \times 10^{- 9} N

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