A) 6N
B) 4N
C) 3N
D) 5N
Correct Answer: A
Solution :
[a] 6 N \[{{F}_{1}}=\left( \frac{1}{4\pi {{\varepsilon }_{0}}} \right)\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}\times 4\times {{10}^{-6}}}{{{\left( 20\times {{10}^{-2}} \right)}^{2}}}=\frac{108\times {{10}^{-3}}}{4\times {{10}^{-2}}}=2.7\,N\]\[{{F}_{2}}=\frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}\times 5\times {{10}^{-6}}}{{{\left( 20\times {{10}^{-2}} \right)}^{2}}}\] \[=\frac{105\times {{10}^{-3}}}{4\times {{10}^{-2}}}=3.375\,N\] \[{{F}_{net}}={{F}_{1}}+{{F}_{2}}\] \[=2.7+3.375=6.075\,N\cong 6N\]You need to login to perform this action.
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