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="12-18\\over 2" =-3nC
Using Coulomb's law, F="\\frac{1\u00d7q^2}{4\\pi \\epsilon _o\u00d7r^2}" ="(9\u00d710 \n^9\n )\u00d7 \n\\frac{\n \n(3\u00d710 \n\u22129\n ) ^\n2}{(0.3)^2}\n \n\u200b\t\n =9\u00d710 \n^{\u22127}\n N"
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