A) 2 mC
B) \[2\times {{10}^{-7}}mC\]
C) \[2\,nC\]
D) \[2\mu C\]
Correct Answer: D
Solution :
(d)\[2\mu C\] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{{{q}_{1}}{{q}_{2}}}{{{d}^{2}}}\] \[\therefore \,\,\left( 10\times {{10}^{-3}} \right)\times 10=\frac{\left( 9\times {{10}^{9}} \right)\times {{q}^{2}}}{{{\left( 0.6 \right)}^{2}}}\] Or \[{{q}^{2}}=\frac{{{10}^{-1}}\times 3.6}{9\times {{10}^{9}}}=4\times {{10}^{-12}}\] Or \[q=2\times {{10}^{-6}}C=2\mu C\]
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