A) \[150\,\,N{{C}^{-1}}\]
B) \[250\,\,N{{C}^{-1}}\]
C) \[60\,\,N{{C}^{-1}}\]
D) \[35\,\,N{{C}^{-1}}\]
Correct Answer: B
Solution :
Electric field \[\overset{\to }{\mathop{\mathbf{E}}}\,\] in broad side-on position is \[\overset{\to }{\mathop{\mathbf{E}}}\,=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{p}{{{r}^{3}}}\] where \[p\] is dipole moment given by \[p=q\times 2l=(3\times {{10}^{-6}}C)(2\times {{10}^{-3}}m)\] \[=6\times {{10}^{-9}}C\text{-}m\] Hence,\[\overset{\to }{\mathop{\mathbf{E}}}\,=9\times {{10}^{9}}\times \frac{6\times {{10}^{-9}}}{{{(0.6)}^{3}}}=250\,\,N{{C}^{-1}}\]You need to login to perform this action.
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