A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2a}{L}\left( 1+\frac{1}{\sqrt{5}} \right)\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2a}{L}\left( 1-\frac{1}{\sqrt{5}} \right)\]
C) zero
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q}{L}\left( 1+\sqrt{5} \right)\]
Correct Answer: B
Solution :
\[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}\] Here, \[V=2{{V}_{+ve}}+2{{V}_{-ve}}\] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{2q}{L}-\frac{2q}{L\sqrt{5}} \right]\] \[V=\frac{2q}{4\pi {{\varepsilon }_{0}}L}\left( 1-\frac{1}{\sqrt{5}} \right)\]
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