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Correct option is A)

Let the initial charges are $q_{1}$ and $q_{2}$.

In first case , $F=r_{2}kq_{1}q_{2} =−0.027$ (minus sign due to attraction force)

or $1_{2}(9×10_{9})q_{1}q_{2} =−0.027⇒q_{1}q_{2}=−3×10_{−12}..(1)$

When a conducting wire are connected them, the charge on each particle will be half of the total charge i.e $(q_{1}+q_{2})/2$.

For repulsion, $F=1_{2}(9×10_{9}(2q_{1}+q_{2} )_{2} =0.009$

or $(q_{1}+q_{2})_{2}=4×10_{−12}$

or $q_{1}+q_{2}=±2×10_{−6}...(2)$

A) From $q_{1}q_{2}=−3×10_{−12}$ and $q_{1}+q_{2}=+2×10_{−6}$, we get after solving $q_{1}=3μC$ and $q_{2}=−1μC$

B) From $q_{1}q_{2}=−3×10_{−12}$ and $q_{1}+q_{2}=−2×10_{−6}$, we get after solving $q_{1}=−3μC$ and $q_{2}=1μC$

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