Two copper balls,each weighing 10,g,are kept | vedclass

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(C) $1$. Calculate the number of atoms in $10\,g$ of copper: $N = \frac{	ext{mass}}{	ext{atomic weight}} 	imes N_A = \frac{10}{63.5} 	imes 6.022 \

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$2.0 	imes 10^8\,N$

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(C) $1$. Calculate the number of atoms in $10\,g$ of copper: $N = \frac{	ext{mass}}{	ext{atomic weight}} 	imes N_A = \frac{10}{63.5} 	imes 6.022 	imes 10^{23} \approx 9.48 	imes 10^{22}$ atoms.$2$. Calculate the number of electrons transferred: $n = \frac{N}{10^6} = \frac{9.48 	imes 10^{22}}{10^6} = 9.48 	imes 10^{16}$ electrons.$3$. Calculate the charge on each ball: $Q = n 	imes e = 9.48 	imes 10^{16} 	imes 1.6 	imes 10^{-19} \approx 0.015\,C$.$4$. Calculate the Coulomb force using $F = k \frac{Q_1 Q_2}{r^2}$,where $k = 9 	imes 10^9\,N\cdot m^2/C^2$,$Q_1 = Q_2 = 0.015\,C$,and $r = 0.1\,m$: $F = 9 	imes 10^9 	imes \frac{(0.015)^2}{(0.1)^2} = 9 	imes 10^9 	imes \frac{2.25 	imes 10^{-4}}{0.01} = 9 	imes 10^9 	imes 2.25 	imes 10^{-2} = 20.25 	imes 10^7 \approx 2.0 	imes 10^8\,N$.

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 $63.5$).

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