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Two identical particles are charged and held at a distance of $$1 m$$ from each other. They are found to be attracting each other with a force of $$0.027 N$$. 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 $$0.009 N$$. Find the initial charge on each particle :


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

The correct option is B
$$q_1 = \pm3 \mu C \ ; \ q_2= \mp1 \mu C$$

Let the initial charges are $$q_1$$ and $$q_2$$.
In first case , $$F=\dfrac{kq_1q_2}{r^2}=-0.027 $$ (minus sign due to attraction force)
or $$\dfrac{(9\times 10^9)q_1q_2}{1^2}=-0.027 \Rightarrow q_1q_2=-3\times 10^{-12} . . (1)$$
When a conducting wire are connected them, the charge on each particle will be half of the total charge i.e $$(q_1+q_2)/2$$. 
For repulsion, $$F=\dfrac{(9\times 10^9(\dfrac{q_1+q_2}{2})^2}{1^2}=0.009$$
or $$(q_1+q_2)^2=4\times 10^{-12}$$
or $$q_1+q_2=\pm 2\times 10^{-6}  ...(2)$$
A) From $$q_1q_2=-3\times 10^{-12}$$ and $$q_1+q_2=+ 2\times 10^{-6} $$,  we get after solving $$q_1=3 \mu C$$ and $$q_2=-1 \mu C$$
B)  From $$q_1q_2=-3\times 10^{-12}$$ and $$q_1+q_2=- 2\times 10^{-6} $$,  we get after solving $$q_1=-3 \mu C$$ and $$q_2=1 \mu C$$

Physics

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