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Q. What is the Coulomb's force between two $\alpha$ -particles separated by a distance of $3.2 \times 10^{-15} m ?$

Electric Charges and Fields

Solution:

Here, $r=3.2 \times 10^{-15} m$
Charge on $\alpha$ -particles, $q_{1}=q_{2}=2 \times 1.6 \times 10^{-19} C$
Now, $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}$
$=9 \times 10^{9} \times \frac{\left(2 \times 1.6 \times 10^{-19}\right)^{2}}{\left(3.2 \times 10^{-15}\right)^{2}}=90 N$