A paisa coin is electrically neutral and contains equal amounts of positive and negative charge of magnitude \(34.8~\text{kC}\)34.8 kC. Suppose that these equal charges were concentrated in two-point charges separated by 1 cm, 100 m and 106 m . Find the force on each such point charge in each of the three cases respectively.
1. \(1.09\times10^{23}~\text{N},~1.09\times10^{15}~\text{N}~\text{and}~1.09\times10^{7}~\text{N}\)
2. \(1.09\times10^{22}~\text{N},~1.09\times10^{15}~\text{N}~\text{and}~1.09\times10^{7}~\text{N}\)
3. \(1.09\times10^{23}~\text{N},~1.09\times10^{14}~\text{N}~\text{and}~1.09\times10^{8}~\text{N}\)
4. \(1.09\times10^{22}~\text{N},~1.09\times10^{15}~\text{N}~\text{and}~1.09\times10^{7}~\text{N}\)

Hint: Use the coulomb's law.
Step 1: Find the force between the charges in each case.
Here, 
q=±34.8kC=±3.48×104C
r1=1cm=10-2m,r2=100m,r3=106mand14πε0=9×109
F1=|q|24πε0r12=9×109×3.48×104210-22=1.09×1023N
F2=|q|24πε0r22=9×109×3.48×1042(100)2=1.09×1015N
F3=|q|24πε0r32=9×109×348×10421062=1.09×107N
Step 2: Find the conclusion by observing the force between the charges.
When charges in an electrically neutral matter are separated by a distance, they exert a large force on each other and hence, it is no possible to separate the charges in an electrically neutral matter.