$A$ paisa coin is made up of $Al-Mg$ alloy and weighs $0.75\,g$. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\,kC$. Suppose that these equal charges were concentrated in two point charges separated by:
$(i)$ $1\,cm$ $(\sim \frac{1}{2} \times \text{diagonal of the one paisa coin})$
$(ii)$ $100\,m$ $(\sim \text{length of a long building})$
$(iii)$ $10^6\,m$ $(\text{radius of the earth})$.
Find the force on each such point charge in each of the three cases. What do you conclude from these results?

(N/A) Given: Charge $q = 34.8\,kC = 3.48 \times 10^4\,C$. Coulomb's constant $k = 9 \times 10^9\,N\cdot m^2/C^2$.
The force between two point charges is given by $F = \frac{k|q|^2}{r^2}$.
$(i)$ For $r_1 = 1\,cm = 10^{-2}\,m$:
$F_1 = \frac{9 \times 10^9 \times (3.48 \times 10^4)^2}{(10^{-2})^2} = \frac{9 \times 10^9 \times 12.11 \times 10^8}{10^{-4}} = 1.09 \times 10^{23}\,N$.
$(ii)$ For $r_2 = 100\,m$:
$F_2 = \frac{9 \times 10^9 \times (3.48 \times 10^4)^2}{(100)^2} = \frac{109 \times 10^{21}}{10^4} = 1.09 \times 10^{15}\,N$.
$(iii)$ For $r_3 = 10^6\,m$:
$F_3 = \frac{9 \times 10^9 \times (3.48 \times 10^4)^2}{(10^6)^2} = \frac{109 \times 10^{21}}{10^{12}} = 1.09 \times 10^7\,N$.
Conclusion: The calculated forces are extremely large. This indicates that it is nearly impossible to separate the positive and negative charges in a neutral object,which explains why matter is generally electrically neutral.