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# Two identical particles are charged and held at a distance of 1m from each other. They are found to be attracting each other with a force of 0.027N. Now, they are connected by a conducting wire so that charge flows between them. When the charge flow stops, they are found to be repelling each other with a force of 0.009N. Find the initial charge on each particle :None of theseq1=±3μC ; q2=∓1μCq1=∓3μC ; q2=∓2μCq1=±5μC ; q2=∓2μC

A
q1=±3μC ; q2=1μC
B
None of these
C
q1=±5μC ; q2=2μC
D
q1=3μC ; q2=2μC
Solution
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#### Let the initial charges are q1 and q2.In first case , F=kq1q2r2=−0.027 (minus sign due to attraction force)or (9×109)q1q212=−0.027⇒q1q2=−3×10−12..(1)When a conducting wire are connected them, the charge on each particle will be half of the total charge i.e (q1+q2)/2. For repulsion, F=(9×109(q1+q22)212=0.009or (q1+q2)2=4×10−12or q1+q2=±2×10−6...(2)A) From q1q2=−3×10−12 and q1+q2=+2×10−6, we get after solving q1=3μC and q2=−1μCB) From q1q2=−3×10−12 and q1+q2=−2×10−6, we get after solving q1=−3μC and q2=1μC

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