A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
B) \[\text{zero}\]
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 point on the broad side due to dipole is zero. Thus, potential at A will be zero due to dipole BC.
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