A) 0.9 N
B) 1.8 N
C) 2.7 N
D) 3.6 N
Correct Answer: B
Solution :
FA = force on C due to charge placed at A \[=9\times {{10}^{9}}\times \frac{{{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(10\times {{10}^{-2}})}^{2}}}=1.8\,N\] FB = force on C due to charge placed at B \[=9\times {{10}^{9}}\times \frac{{{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(0.1)}^{2}}}=1.8N\] Net force on C \[{{F}_{net}}=\sqrt{{{({{F}_{A}})}^{2}}+{{({{F}_{B}})}^{2}}+2{{F}_{A}}{{F}_{B}}\cos {{120}^{o}}}=1.8\,N\]You need to login to perform this action.
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