A) 0.02 V
B) 0.04 V
C) 0.06 V
D) 1.02 V
E) None of the above
Correct Answer: E
Solution :
The given, \[\lambda ={{10}^{-12}}C/m,\,l=50\,cm\], \[r=1\times {{10}^{-2}}m\] and \[{{\varepsilon }_{0}}=8.8\times {{10}^{-12}}F/m\] We know that, \[\lambda =\frac{Q}{l}\] \[{{10}^{-12}}=\frac{Q}{50\,cm}\] or \[{{10}^{-12}}=\frac{2Q}{2\times 50\,cm}\] \[{{10}^{-12}}=\frac{2Q}{100\,cm},{{10}^{-12}}=\frac{2Q}{1m},Q=\frac{1}{2}\times {{10}^{-12}}C\], Electric potential \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{r}\] \[=\frac{1}{4\times 3.14\times 8.8\times {{10}^{-12}}}.\frac{\frac{1}{2}\times {{10}^{-12}}}{1\times {{10}^{-2}}}\] \[=\frac{100}{221.05}=0.4\,V\]
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