A copper atom consists of copper nucleus surrounded by 29 electrons. The atomic weight of copper is $63.5gmo{l}^{-1}$. Let us now take two pieces of copper each weighing 10 g. Let us transfer one electron from one piece to another for every 1000 atoms in a piece.

 The force of attraction between them when they are separated by a distance of 1cm is $\mathrm{X}.057\times {10}^{16}\mathrm{N}$ Find $\mathrm{X}$  

63.5g copper contains $N_A=6x{10}^{23 }$copper atoms, therefore the number of electrons transferred, $N=\frac{6\times {10}^{23}\times 10}{63.5\times 1000}$

Since only one electrons is transferred,  $N=\frac{6\times {10}^{23}}{63.5}\times 10$

$\Rightarrow q=\frac{1920}{127}C$

Separation between pieces is $r =1 cm = {10}^{-2 }m$ One piece of copper has positive charge and the other negative charge, so force of attraction between the pieces is 

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

$\Rightarrow F=\frac{9\times {10}^9\left(\frac{1920}{127}\right)\times \left(\frac{1920}{127}\right)}{{\left({10}^{-2}\right)}^2}N$

$\Rightarrow F=2.057\times {10}^{16}N$

So, x =2