What is the angle between the electric dipole moment and the electric…

The correct answer is Option (1) → \( 0^\circ \)

The dipole moment \( \vec{p} \) points along the axis from \( -q \) to \( +q \). On the axis, the field of the nearer charge is always the stronger one:

• Beyond \( +q \), the field of the nearer charge \( +q \) points away from it. It is stronger than the field of \( -q \), so the total field points along \( \vec{p} \).

• Beyond \( -q \), the field of the nearer charge \( -q \) points towards it. This is again the direction of \( \vec{p} \).

For \( r\gg a \) the axial field is

    \( \displaystyle \vec{E}=\frac{2\vec{p}}{4\pi\varepsilon_0r^{3}} \)

From the above, \( \vec{E} \) is a positive multiple of \( \vec{p} \). Hence the field on the axial line points the same way as the dipole moment, and the angle between them is \( 0^\circ \).

NCERT (2026-27):
The Field of an Electric Dipole – “… The dipole moment vector p of an electric dipole is defined by

    \( \mathbf{p}=q\times2a\,\hat{\mathbf{p}} \qquad (1.19) \)

that is, it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q, –q) and the direction is along the line from –q to q. In terms of p, the electric field of a dipole at large distances takes simple forms:

At a point on the dipole axis

    \( \displaystyle \mathbf{E}=\frac{2\mathbf{p}}{4\pi\varepsilon_0r^{3}} \)\( \displaystyle \quad(r\gg a) \qquad (1.20) \)”