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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 $63.5$)
$2.0 \times {10^{10}}\,N$
$2.0 \times {10^4}\,N$
$2.0 \times {10^8}\,N$
$2.0 \times {10^6}\,N$
Solution
(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 $F = 9 \times {10^9} \times \frac{{{{(0.015)}^2}}}{{{{(0.1)}^2}}} = 2 \times {10^8}\,N.$
