Two copper balls, each weighing 10g are kept in air 10 cm apart. If one electron from every $10^{6}$ atoms is transferred from one ball to the other, the coulomb force between them is (atomic weight of copper is 63.5)
Text Solution
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Number of atoms in given mass $= \frac{10}{63.5} \times 6.02 \times 10^{23}$
= 9.48 × × 10 22 
Transfer of electron between balls $= \frac{9.48 \times 10^{22}}{10^{6}}$
= 9.48 × × 10
16
Hence magnitude of charge gained by each ball.
Q = 9.48 × × 10 16 × × 1.6 × × 10 –19 = 0.015 C
Force of attraction between the balls $F = 9 \times 10^{9} \times \frac{(0.015)^{2}}{(0.1)^{2}} = 2 \times 10^{8} \mathrm{N}.$
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