Question

Two copper balls, each weighing $10\, g$, 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 $63.5$ )

• 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,
$Q=9.48 \times 10^{16} \times 1.6 \times 10^{-19}=0.015 C$
Force of attraction between the balls
$F=9 \times 10^{9} \times \frac{(0.015)^{2}}{(0.1)^{2}}=2 \times 10^{8} N$