A) \[2.5\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}\]
B) \[4.5\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}\]
C) \[4.5\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}\]
D) \[2.5\times {{10}^{6}}N{{m}^{2}}{{C}^{-1}}\]
Correct Answer: B
Solution :
(b) \[4.5\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}\] Here, \[q=4\mu C=4\times {{10}^{-6}}\,C.\,l=10\,cm\] \[\phi =\frac{q}{{{\varepsilon }_{0}}}=\frac{4\times {{10}^{-6}}}{8.85\times {{10}^{-12}}}\]
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