A) zero
B) \[\left( \frac{-qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right)\sqrt{2}a\]
C) \[\left( \frac{-qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right).\frac{a}{\sqrt{2}}\]
D) \[\left( \frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}} \right).\sqrt{2}a\]
Correct Answer: A
Solution :
[a] We know that potential energy of two charge system is given by \[U=\frac{1}{4\pi {{\in }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}\] According to question, \[{{U}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( +q \right)\left( -Q \right)}{a}=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{a}\] \[\text{and }{{U}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( +q \right)\left( -Q \right)}{a}=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{a}\] \[\Delta U={{U}_{B}}-{{U}_{A}}=0\] We know that for conservative force, \[W=-\Delta U=0\]
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