Q61 / 75 · JEE Main 2026 (02 April Shift 1) · Physics
Two short electric dipoles $A$ and $B$ having dipole moment $p_1$ and $p_2$ respectively are placed with their axis mutually perpendicular as shown in the figure. The resultant electric field at a point $x$ is making an angle of $60°$ with the line joining points $O$ and $x$. The ratio of the dipole moments $p_2/p_1$ is _______.


- A$\dfrac{\sqrt{3}}{2}$
- Correct answerB$2\sqrt{3}$
- C$\dfrac{1}{\sqrt{3}}$
- D$\sqrt{3}$
Solution
Let the distance of point $x$ from the origin $O$ be $r$.
Point $x$ lies on the axial line of dipole A. The electric field at $x$ due to dipole A is directed along the horizontal axis and its magnitude is given by:
$E_A = \dfrac{1}{4\pi\epsilon_0} \dfrac{2p_1}{r^3}$
Point $x$ lies on the equatorial line of dipole B. The electric field at $x$ due to dipole B is directed perpendicular to the horizontal axis and its magnitude is given by:
$E_B = \dfrac{1}{4\pi\epsilon_0} \dfrac{p_2}{r^3}$
The resultant electric field makes an angle of $60^{\circ}$ with the line $Ox$ (the horizontal axis). Therefore, the tangent of this angle is the ratio of the perpendicular component of the electric field to the parallel component:
$\tan 60^{\circ} = \dfrac{E_B}{E_A}$
Substituting the expressions for $E_A$ and $E_B$:
$\sqrt{3} = \dfrac{\dfrac{1}{4\pi\epsilon_0} \dfrac{p_2}{r^3}}{\dfrac{1}{4\pi\epsilon_0} \dfrac{2p_1}{r^3}}$
$\sqrt{3} = \dfrac{p_2}{2p_1}$
$\dfrac{p_2}{p_1} = 2\sqrt{3}$
Answer: $2\sqrt{3}$
Point $x$ lies on the axial line of dipole A. The electric field at $x$ due to dipole A is directed along the horizontal axis and its magnitude is given by:
$E_A = \dfrac{1}{4\pi\epsilon_0} \dfrac{2p_1}{r^3}$
Point $x$ lies on the equatorial line of dipole B. The electric field at $x$ due to dipole B is directed perpendicular to the horizontal axis and its magnitude is given by:
$E_B = \dfrac{1}{4\pi\epsilon_0} \dfrac{p_2}{r^3}$
The resultant electric field makes an angle of $60^{\circ}$ with the line $Ox$ (the horizontal axis). Therefore, the tangent of this angle is the ratio of the perpendicular component of the electric field to the parallel component:
$\tan 60^{\circ} = \dfrac{E_B}{E_A}$
Substituting the expressions for $E_A$ and $E_B$:
$\sqrt{3} = \dfrac{\dfrac{1}{4\pi\epsilon_0} \dfrac{p_2}{r^3}}{\dfrac{1}{4\pi\epsilon_0} \dfrac{2p_1}{r^3}}$
$\sqrt{3} = \dfrac{p_2}{2p_1}$
$\dfrac{p_2}{p_1} = 2\sqrt{3}$
Answer: $2\sqrt{3}$
Hence, the correct answer is option: 'B'.