Question

Find the electric force between two small iron particles, each of mass $280mg$ placed $10cm$ apart if $0.01\%$ of the electrons of one particle are transferred to other. (atomic weight and atomic number of iron are $56g$ and $26$ respectively)

• A. $1.4\times 10^{10}N$
• B. $1.4\times 10^{12}N$
• C. $1.4\times 10^{13}N$
• D. $1.4\times 10^{11}N$

Answer 1

Answer: (B)

According to coulomb's law, $F=\frac{k{q}_1{q}_2}{r^2}$
The Avogadro's number is $6.022\times 10^{23}$
Number of atoms in $280mg$ of iron is
$\frac{6.022\times 10^{23}\times 280\times 10^{-3}}{56}=3.011\times 10^{21}\approx 3\times 10^{21}$
Therfore, the total number of electron in the iron particles are
$7.8\times 10^{22}$ (since in one iron atom 26 electrons are present) The charge transfered is $0.01\%$ of the total electrons which is
$\frac{0.01}{100}\times 7.8\times 10^{22}=7.8\times 10^{18}$ Hence, the net force is
$F=\frac{9\times 10^9\times {\left(7.8\times 10^{18}\times 1.6\times 10^{-19}\right)}^2}{0.1^2}=1401.75\times 10^9\approx 1.4\times 10^{12}$
(here,magnitude of $k=9\times 10^9$ )