Q. Two copper balls, each weighting $10g$ 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)
A) $2.0 \times 10^{10}N$
B) $2.0 \times 10^{4}N$
C) $2.0 \times 10^{8}N$
D) $2.0 \times 10^{6}N$
Explanation
(c) Number of atoms in given mass $= \ \frac{10}{63.5} \times 6.02 \times 10^{23}$
$= 9.48 \times 10^{22}$
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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
$F = 9 \times 10^{9} \times \frac{(0.015)^{2}}{(0.1)^{2}} = 2 \times 10^{8}N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.