A) \[-10N\]
B) \[+10N\]
C) \[+20N\]
D) \[-20N\]
Correct Answer: A
Solution :
In second case, charges will be \[-\,2\mu C\] and \[+3\mu C\] Since \[F\propto {{Q}_{1}}{{Q}_{2}}\] i.e. \[\frac{F}{{{F}'}}=\frac{{{Q}_{1}}{{Q}_{2}}}{Q{{'}_{1}}Q{{'}_{2}}}\] \ \[\frac{40}{{{F}'}}=\frac{3\times 8}{-2\times 3}=-4\] Þ \[{F}'=10\,N\](Attractive)
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