A) \[{{E}_{3}}={{E}_{0}}\]
B) \[{{E}_{3}}<{{E}_{0}}\]
C) \[{{E}_{3}}>{{E}_{0}}\]
D) \[{{E}_{3}}\ge {{E}_{0}}\]
Correct Answer: C
Solution :
\[{{E}_{1}}=\frac{\eta q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}},\,\,{{E}_{2}}=\frac{\eta q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\] Therefore \[E={{\vec{E}}_{1}}+{{\vec{E}}_{2}}\] \[=\,\,\,\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos \,\,60{}^\circ }\,\,=\,\,\frac{\sqrt{3}\,\eta q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\,\,\] Since \[{{\eta }^{-1}}<\sqrt{3},\,1\,<\,\sqrt{3}\eta ,\,\,\sqrt{3}\eta >1\] \[\Rightarrow \,\,\,\,\frac{\sqrt{3}\eta q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}>\frac{q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\] \[\Rightarrow \,\,\,\,\,\,\,{{E}_{3}}>{{E}_{0}}\,\left( {{E}_{0}}=\frac{q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}} \right)\]You need to login to perform this action.
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