question_answer

Two point charges \[-q\] and \[+q\] are located at points (0, 0, \[-a\]) and (0, 0, \[a\]) respectively. The electric potential at a point (0, 0, \[z\]), where \[z<a\] is

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 \[V={{V}_{1}}+{{V}_{2}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{(z-a)}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{(z+a)}\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\frac{(z+a-z+a)}{(z-a)(z+a)}\] \[\Rightarrow \] \[V=\frac{2qa}{4\pi {{\varepsilon }_{0}}({{z}^{2}}-{{a}^{2}})}\]