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A point dipole $\mathop p\limits^ \to = {p_0}\mathop x\limits^ \wedge $ is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Take V= 0 at infinity)
A. $\dfrac{{\left| {\mathop p\limits^ \to } \right|}}{{4\pi {\varepsilon _0}{d^2}}},\dfrac{{ - \mathop p\limits^ \to }}{{4\pi {\varepsilon _0}{d^3}}}$
B. $0,\dfrac{{\mathop p\limits^ \to }}{{4\pi {\varepsilon _0}{d^3}}}$
C. $\dfrac{{\left| {\mathop p\limits^ \to } \right|}}{{4\pi {\varepsilon _0}{d^2}}},\dfrac{{\mathop p\limits^ \to }}{{4\pi {\varepsilon _0}{d^3}}}$
D. $0,\dfrac{{ - \mathop p\limits^ \to }}{{4\pi {\varepsilon _0}{d^3}}}$

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: Electric dipoles will be having two charges definitely. Just like magnetic dipoles we can consider electric dipoles. In a magnetic dipole, the magnetic moment vector goes from south pole to north pole inside the magnet. We resolve that along axial and equatorial to get the magnetic fields. Similarly in electric dipole, the dipole moment vector goes from negative charge to positive charge.
Formula used:
$\eqalign{
  & {E_{axial}} = \dfrac{{2k\mathop p\limits^ \to }}{{{r^3}}} \cr
  & {E_{equitorial}} = \dfrac{{ - k\mathop p\limits^ \to }}{{{r^3}}} \cr
  & {V_\theta } = \dfrac{{kp\cos \theta }}{{{r^2}}} \cr} $

Complete answer:
Electric field can be produced by electric dipole and due to that field potential will also be present
The position of the point determines the direction of electric field and magnitude of potential
If the point is located on the axis then we have one formula to calculate the electric field over there and if the point is located on the equatorial point then we have another formula for the electric field.
For axial point the electric field will be
${E_{axial}} = \dfrac{{2k\mathop p\limits^ \to }}{{{r^3}}}$
‘r’ is the distance between the center of dipole and the point P.
So the electric field will be along the direction of electric moment vector($\mathop p\limits^ \to $)
For equatorial point the electric field will be
${E_{equitorial}} = \dfrac{{ - k\mathop p\limits^ \to }}{{{r^3}}}$
So the equatorial electric field will be anti parallel to the direction of electric dipole moment vector($\mathop p\limits^ \to $)
The point P on the y axis at a distance d will be equatorial point for the dipole and the filed will be
$\eqalign{
  & {E_{equitorial}} = \dfrac{{ - k\mathop p\limits^ \to }}{{{r^3}}} \cr
  & \therefore {E_P} = \dfrac{{ - \mathop p\limits^ \to }}{{4\pi {\varepsilon _0}{d^3}}} \cr} $
Potential at point P is given as
${V_\theta } = \dfrac{{kp\cos \theta }}{{{r^2}}}$
Where theta is the angle between the dipole moment vector and the line joining the dipole and point. For all equatorial points theta will be 90 degrees. So the cosine of 90 degrees will be zero. So potential will be zero.
${V_\theta } = \dfrac{{kp\cos \theta }}{{{r^2}}}$
$\eqalign{
  & \Rightarrow {V_P} = \dfrac{{kp\cos {{90}^0}}}{{{r^2}}} \cr
  & \therefore {V_P} = 0 \cr} $
All these formulas are valid for short dipoles only.

Hence option D is the answer.

Note:
If the point given by them is not a perfect axial point nor perfect equatorial point i.e an oblique point then we will resolve the electric dipole moment vector into axial direction and along the equatorial direction. After resolving we will find out the direction of electric field components at that point and finally we will find the resultant.