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Question

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 :
  1. None of these
  2. q1=±3μC ; q2=1μC
  3. q1=3μC ; q2=2μC
  4. q1=±5μC ; q2=2μC

A
q1=±3μC ; q2=1μC
B
None of these
C
q1=±5μC ; q2=2μC
D
q1=3μC ; q2=2μC
Solution
Verified by Toppr

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=0.027q1q2=3×1012..(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=0.009
or (q1+q2)2=4×1012
or q1+q2=±2×106...(2)
A) From q1q2=3×1012 and q1+q2=+2×106, we get after solving q1=3μC and q2=1μC
B) From q1q2=3×1012 and q1+q2=2×106, we get after solving q1=3μC and q2=1μC

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