A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
B) zero
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\left( -q \right)}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
Correct Answer: B
Solution :
Potential at A \[=+\frac{q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}-\frac{q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] = zero
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