When they are connected by a conducting wire, finally charge on each will be half of total charge on both. Let q be the final charge on each, then
q=12−18212-18\over 2212−18 =-3nC
Using Coulomb's law, F=1×q24πϵo×r2\frac{1×q^2}{4\pi \epsilon _o×r^2}4πϵo×r21×q2 =(9×109)×(3×10−9)2(0.3)2=9×10−7N(9×10 ^9 )× \frac{ (3×10 −9 ) ^ 2}{(0.3)^2} =9×10 ^{−7} N(9×109)×(0.3)2(3×10−9)2=9×10−7N
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