A) \[9\,N{{C}^{-1}}\]
B) \[0.9\,N{{C}^{-1}}\]
C) \[90\,N{{C}^{-1}}\]
D) \[0.09\,N{{C}^{-1}}\]
Correct Answer: D
Solution :
(d) \[0.09\,N{{C}^{-1}}\] The point lies on equatorial line of a short dipole. \[\therefore \,\,E=\frac{2ql}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}=\frac{9\times {{10}^{9}}\times {{10}^{-6}}\times {{10}^{-8}}}{{{\left( {{10}^{-1}} \right)}^{3}}}\]You need to login to perform this action.
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