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