A) \[18\times {{10}^{-3}}m\]
B) \[9\times {{10}^{-3}}m\]
C) \[4.5\times {{10}^{-3}}m\]
D) None of these
Correct Answer: A
Solution :
Capacitance of conductor \[C=1\,pF\] \[=1\times {{10}^{-12}}F\] The capacitance of a spherical conductor is \[C=4\pi {{\varepsilon }_{0}}r\] or \[1\times {{10}^{-12}}=4\pi {{\varepsilon }_{0}}r\] so, \[r=\frac{1\times {{10}^{-12}}}{4\pi {{\varepsilon }_{0}}}\] \[=9\times {{10}^{9}}\times 1\times {{10}^{-12}}\] \[=9\times {{10}^{-3}}m\] so, diameter \[=2r=2\times 9\times {{10}^{-3}}\] \[=18\times {{10}^{-3}}m\] (where\[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\])
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