A) \[\frac{q}{4\pi {{\varepsilon }_{0}}d}\]
B) \[\frac{-2q}{4\pi {{\varepsilon }_{0}}d}\]
C) \[\frac{q}{4\pi {{\varepsilon }_{0}}d}\]
D) \[\frac{-q}{4\pi {{\varepsilon }_{0}}d}\]
Correct Answer: C
Solution :
Net potential at point P is given by \[V=\left[ \frac{2q}{4\pi {{\varepsilon }_{0}}d}+\frac{2q}{4\pi {{\varepsilon }_{0}}d}-\frac{2q}{4\pi {{\varepsilon }_{0}}d}-\frac{2q}{4\pi {{\varepsilon }_{0}}(2d)} \right]\] \[=\,\,\frac{2q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{2d} \right]=\frac{2q}{8\pi {{\varepsilon }_{0}}d}=\frac{q}{4\pi {{\varepsilon }_{0}}d}\]
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