A) 1.8 nC
B) 2 nC
C) 1 nC
D) 1.5 Nc
Correct Answer: B
Solution :
For a conducting charged sphere, the electric field outside the sphere\[(E)=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}}\] \[1.8\times {{10}^{3}}=9\times {{10}^{9}}\times \frac{Q}{{{(0.1)}^{2}}}\] \[Q=\frac{1.8\times {{10}^{3}}\times {{10}^{-2}}}{9\times {{10}^{9}}}\] \[=2\times {{10}^{-9}}C=2\,nC\]
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