A) zero
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{a}\left( 1+\frac{1}{\sqrt{5}} \right)\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{a}\left( 1-\frac{2}{\sqrt{5}} \right)\]
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{a}\left( 1-\frac{1}{\sqrt{5}} \right)\]
Correct Answer: D
Solution :
[d] Potential at point A, \[{{V}_{A}}=\frac{2Kq}{a}-\frac{2Kg}{a\sqrt{5}}\] |
Potential at point B,\[{{V}_{B}}=0\] |
\[\therefore \]Using work energy theorem, |
\[{{W}_{AB}}{{)}_{electric}}=Q({{V}_{A}}-{{V}_{B}})\] |
\[=\frac{2KqQ}{a}\left[ 1-\frac{1}{\sqrt{5}} \right]\]\[=\left( \frac{1}{4\pi {{\in }_{0}}} \right)\frac{2Qq}{a}\left[ 1-\frac{1}{\sqrt{5}} \right]\] |
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