A paisa coin is made up of $\mathrm{Al - Mg}$ alloy and weighs $0.75\, g$ It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8$ $\mathrm{kC}$. Suppose that these equal charges were concentrated in two point charges separated by :
$(i)$ $1$ $\mathrm{cm}$ $(\sim \frac{1}{2} \times $ diagonal of the one paisa coin $)$
$(ii)$ $100\,\mathrm{m}$ $(\sim $ length of a long building $)$
$(iii)$ $10^6$ $\mathrm{m}$ (radius of the earth).
Find the force on each such point charge in each of the three cases. What do you conclude from these results ?
Here, $r_{1}=1 \mathrm{~cm}=10^{-2} \mathrm{~m}$
$r_{2}=100 \mathrm{~m}$ $r_{3}=10^{6} \mathrm{~m}$ $\frac{1}{4 \pi \epsilon_{0}}=k=9 \times 10^{9}$
$(i)$ $\mathrm{F}_{1}=\frac{k|q|^{2}}{r_{1}^{2}}=\frac{9 \times 10^{9} \times\left(3.48 \times 10^{4}\right)^{2}}{\left(10^{-2}\right)^{2}}=1.0899 \times 10^{23} \mathrm{~N}$ $=1.09 \times 10^{23} \mathrm{~N}$
$(ii)$ $\mathrm{F}_{2}=\frac{k|q|^{2}}{r_{2}^{2}}=\frac{9 \times 10^{9} \times\left(3.48 \times 10^{4}\right)^{2}}{(100)^{2}}=1.09 \times 10^{15} \mathrm{~N}$
$(iii)$ $\mathrm{F}_{3}=\frac{k|q|^{2}}{r_{3}^{2}}=\frac{9 \times 10^{9} \times\left(3.48 \times 10^{4}\right)^{2}}{\left(10^{6}\right)^{2}}=1.09 \times 10^{7} \mathrm{~N}$
Here, the force between charges is much more hence, it is difficult to disturb electrical neutrality of matter.
The law, governing the force between electric charges is known as
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