A) 0.9 N
B) 1.8 N
C) 2.72 N
D) 3.6 N
Correct Answer: D
Solution :
The situation is shown in figure. Force on C due to A \[{{F}_{AC}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{A}}{{q}_{C}}}{r_{AC}^{2}}\] \[=\frac{9\times {{10}^{9}}\times 1\times 2\times {{10}^{-12}}}{{{(0.1)}^{2}}}=1.8N\] Similarly, \[{{F}_{BC}}=1.8\,N\] Hence, net force on \[C={{F}_{AC}}+{{F}_{BC}}=3.6\,N\]You need to login to perform this action.
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