A) 9 mm
B) 4 mm
C) 8 mm
D) 2 mm
Correct Answer: B
Solution :
As, \[\frac{{{\varepsilon }_{0}}A}{d}=4\pi {{\varepsilon }_{0}}R\] or \[\frac{{{\varepsilon }_{0}}\pi {{D}^{2}}}{4d}=4\pi {{\varepsilon }_{0}}R\] or \[d=\frac{{{D}^{2}}}{16R}=\frac{{{\left( 0\,.\,08 \right)}^{2}}}{16\times 0\,.\,10}\] \[=4\,\times {{10}^{-3}}\,m\,=\,4\,mm\]
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