Two copper balls, each weighing 10g are kept in air 10 cm apart. If one electron from every 106 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

Answers (1)

We know, Charges can be transferred from one body to another.

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

Number of electrons transferred between the balls = \frac{9.48\times 10^{22}}{10^{6}}=9.48\times10^{16}

\therefore Magnitude of charge gained by each ball =

Q=(9.48\times 10^{16})\times(1.6\times10^{-19})=0.015\: C

\therefore Force of attraction between the balls =  F=9\times 10^{9}\times\frac{(0.015)^{2}}{(0.1)^{2}}=2\times10^{8}N

Option C. is correct

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