A) \[\frac{qa}{4\pi {{\varepsilon }_{_{0}}}{{z}^{2}}}\]
B) \[\frac{q}{4\pi {{\varepsilon }_{_{0}}}a}\]
C) \[\frac{2qa}{4\pi {{\varepsilon }_{_{0}}}a\left( {{z}^{2}}-{{a}^{2}} \right)}\]
D) \[\frac{2qa}{4\pi {{\varepsilon }_{_{0}}}a\left( {{z}^{2}}+{{a}^{2}} \right)}\]
Correct Answer: C
Solution :
Potential at P due to \[(+q)\] charge \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{(z-a)}\] Potential at P due to \[(-q)\] charge \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{-q}{(z+a)}\] Total potential at P due to (AB) electric dipole \[V={{V}_{1}}+{{V}_{2}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{(z-a)}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{(z+a)}\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\frac{(z+a-z+a)}{(z-a)(z+a)}\] \[\Rightarrow \] \[V=\frac{2qa}{4\pi {{\varepsilon }_{0}}({{z}^{2}}-{{a}^{2}})}\]You need to login to perform this action.
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