A) \[2.0\times {{10}^{4}}N\]
B) \[2.0\times {{10}^{10}}N\]
C) \[2.0\text{ }{{10}^{6}}N\]
D) \[2.0\times {{10}^{8}}N\]
Correct Answer: D
Solution :
No. of atoms in 10 gm of copper \[=\frac{6\times {{10}^{23}}\times 10}{63.5}=9.4\times {{10}^{22}}\]No- of \[\overline{e}\] transferred from one piece to other \[=\frac{9.4\times {{10}^{22}}}{{{10}^{6}}}=9.4\times {{10}^{16}}\]Charge on these electron \[\overline{e}=9.4\times {{10}^{16}}\times 1.6\times {{10}^{-19}}\] \[=15\times {{10}^{-3}}C\]equal but opposite charge will be on other piece \[\therefore \]\[\text{Fattraction}=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[=\frac{{{(1.5\times {{10}^{-2}})}^{2}}\times 9\times {{10}^{9}}}{0.01}\]\[=2\times {{10}^{7}}N\]
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