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{ kq _1 q _2}{ r ^2} $
The Avogadro's number is $6.022 \times 10^{23}$
Number of atoms in $280 mg$ 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 $k=9 \times 10^9$ )