A) +10 N
B) +20 N
C) -20 N
D) -10 N
Correct Answer: D
Solution :
Two charges \[+\,3\mu C\] and \[8\mu C\] are given which repel each other when a charge \[-5\mu C\] is added then in second case changes becomes \[-2\mu C\] and \[3\mu C\]. From the relation \[F\propto {{q}_{1}}{{q}_{2}}\] Hence \[\frac{{{F}_{1}}}{{{F}_{1}}}=\frac{{{q}_{1}}\times {{q}_{2}}}{{{q}_{1}}{{q}_{2}}}\] or \[\frac{40}{{{F}_{1}}}=\frac{3\times 8}{-2\times 3}=-\,4\] so, \[{{F}_{1}}=-\frac{40}{4}=10\,\,\text{attractive}\]
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