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 \[E={{\overrightarrow{E}}_{1}}+{{\overrightarrow{E}}_{2}}\] \[=\,\,\,\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\,\cos \,60{}^\circ }\,\,=\,\,\frac{\sqrt{3}\eta q}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\] 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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