question_answer

Two equal charges \[q\] of opposite sign separated by a distance \[2a\] constitute an electric dipole of dipole moment \[p\]. If \[P\] is a point at a distance \[r\] from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle \[\theta \] with the axis of the dipole, then the potential at \[P\] is given by \[(r>>2a)\] (Where\[p=2qa\]) [MP PET 1997]

A)                    \[V=\frac{p\cos \theta }{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]

B)                                       \[V=\frac{p\cos \theta }{4\pi {{\varepsilon }_{0}}r}\]

C)            \[V=\frac{p\sin \theta }{4\pi {{\varepsilon }_{0}}r}\]           

D)            \[V=\frac{p\cos \theta }{2\pi {{\varepsilon }_{0}}{{r}^{2}}}\]

Correct Answer: A

Solution :

           For the given situation, diagram can be drawn as follows As shown in figure component of dipole moment along the line OP will be \[p'=p\cos \theta \]. Hence electric potential at point P will be \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{p\cos \theta }{{{r}^{2}}}\]