The potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are respectively
A. \[{90^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\]
B. \[{0^0}\,and\,{\rm{ 9}}{{\rm{0}}^0}\]
C. \[{90^0}\,and\,{\rm{ }}{{\rm{0}}^0}\]
D. \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\]

Hint:The values of maximum and minimum of the potential at a point due to an electric dipole will be determined by the angles like if \[\theta = {0^0},\cos \theta = 1\] then the electric potential will be maximum at the dipole axis and if \[\theta = {180^0},\cos \theta = - 1\] then the electric potential will be minimum at the dipole axis

Formula used:
The potential due to this dipole is given as,
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
Where,
p is dipole moment,
r is distance from dipole,
k is coulomb’s constant,\[k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}\] and \[{\varepsilon _0}\] is permittivity.

Complete step by step solution:
The potential due to this dipole is:
\[V = \dfrac{{p\cos \theta }}{{4\pi {\varepsilon _0}{r^2}}}\]
If \[\theta = {0^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {0^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the maximum potential,
\[{V_{\max }} = \dfrac{{kp}}{{{r^2}}}\]

Now If \[\theta = {180^0}\], then we have
\[\begin{array}{l}V = \dfrac{{p\cos {{180}^0}}}{{4\pi {\varepsilon _0}{r^2}}}\\ \Rightarrow V{\rm{ = - }}\dfrac{{kp}}{{{r^2}}}\end{array}\]
This is the minimum potential,
\[{V_{\min }} = - \dfrac{{kp}}{{{r^2}}}\]
Therefore the potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are \[{0^0}\,and\,{\rm{ 18}}{{\rm{0}}^0}\].

Hence option D is the correct answer.

Note:The electric potential will be zero at the perpendicular axis of the dipole when \[\theta = {90^0}\] . As we know electric dipoles consist of two charges equal in magnitude (q) but opposite in nature one is a positive charge and other is a negative charge. Electric potential obeys superposition principle due to electric dipole as a whole can be sum of potential due to both the charges positive and negative.