A) \[0.01mm\]
B) 0.1 \[mm\]
C) 1.0 \[mm\]
D) 10 \[mm\]
Correct Answer: B
Solution :
\[4\pi {{\varepsilon }_{0}}r=\frac{{{\varepsilon }_{0}}A}{d}\] Þ \[d=\frac{A}{4\pi r}=\frac{\pi {{(20\times {{10}^{-3}})}^{2}}}{4\pi \times 1}=0.1\,mm\]
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