Two copper balls, each weighing 10g, are kept in air 10 cm apart. If one electron from every 10⁶atoms is transferred from one ball to the other, then the coulomb force between them is (atomic weight of copper is 63.5) .

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$2.0 \times 10^{10} N$

$2.0 \times 10^{4} N$

$2.0 \times 10^{8} N$

$2.0 \times 10^{6} N$

answer is C.

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Detailed Solution

63.5g for 1 mole of atom

10g for x

Number of electrons,

$n = \frac{6 \times 10^{23}}{63.5} \times 10 \times \frac{1}{10^{6}} = \frac{6 \times 10^{18}}{63.5}$

As q = ne

$q = \frac{6 \times 10^{18} \times 1.6 \times 10^{- 19}}{63.5}$

or $q = 1.5 \times 10^{- 2} C$

$∴ F = \frac{9 \times 10^{9} \times 1.5 \times 10^{- 2} \times 1.5 \times 10^{- 2}}{{(\frac{10}{100})}^{2}}$

$= 2.0 \times 10^{8} N$

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