Question

Two copper balls, each weighing $10g$, are kept in air $10cm$ 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 $63.5$ )

• A. $2.0\times 10^{10}N$
• B. $2.0\times 10^{4}N$
• C. $2.0\times 10^{8}N$
• D. $2.0\times 10^{6}N$

Answer 1

Answer: (C)

Number of atoms in given mass $=\frac{10}{63.5}\times 6.02\times 10^{23}$
$=9.48\times 10^{22}$

Transfer of electron between balls $=\frac{9.48\times 10^{22}}{10^6}$
$=9.48\times 10^{16}$
Hence, magnitude of charge gained by each ball,
$Q=9.48\times 10^{16}\times 1.6\times 10^{-19}=0.015C$
Force of attraction between the balls
$F=9\times 10^9\times \frac{(0.015{)}^2}{(0.1{)}^2}=2\times 10^{8}N$