Ratio of Dipole Moments for Resultant Electric Field Angle

To find the ratio of the dipole moments $\frac{p_2}{p_1}$, we analyze the electric fields created by dipoles $A$ and $B$ at point $x$, which is located at a distance $r$ from the origin $O$ along the horizontal line $Ox$.

1. Electric Field due to Dipole $A$ ($p_1$): Point $x$ lies on the axial line of dipole $A$. The magnitude of the electric field $E_A$ due to dipole $A$ at point $x$ is: $E_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2 p_1}{r^3}$ This field $E_A$ is directed along the line joining $O$ and $x$ (horizontal direction).

2. Electric Field due to Dipole $B$ ($p_2$): Point $x$ lies on the equatorial line (broadside-on position) of dipole $B$. The magnitude of the electric field $E_B$ due to dipole $B$ at point $x$ is: $E_B = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p_2}{r^3}$ This field $E_B$ is directed perpendicular to the line joining $O$ and $x$ (vertical direction).

3. Angle of Resultant Electric Field: The resultant electric field $\vec{E} = \vec{E}_A + \vec{E}_B$ makes an angle of $60^\circ$ with the line joining points $O$ and $x$. Therefore, we have: $\tan 60^\circ = \frac{E_B}{E_A}$

4. Calculating the Ratio: Substitute $E_A$ and $E_B$ into the equation: $\tan 60^\circ = \frac{\frac{1}{4\pi\varepsilon_0} \cdot \frac{p_2}{r^3}}{\frac{1}{4\pi\varepsilon_0} \cdot \frac{2 p_1}{r^3}} = \frac{p_2}{2 p_1}$

   Since $\tan 60^\circ = \sqrt{3}$, we get: $\sqrt{3} = \frac{p_2}{2 p_1}$

   $\frac{p_2}{p_1} = 2\sqrt{3}$

Thus, the correct option is B.