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- None of these
- q1=±3μC ; q2=∓1μC
- q1=∓3μC ; q2=∓2μC
- q1=±5μC ; q2=∓2μC

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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=0.009

or (q1+q2)2=4×10−12

or 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=−1μC

B) From q1q2=−3×10−12 and q1+q2=−2×10−6, we get after solving q1=−3μC and q2=1μC

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