Question:
Two ideal electric dipoles A and B, having their dipole moment \(p_1\) and \(p_2\) respectively are placed on a plane with their centres at O as shown in the figure. At point C on the axis of dipole A, the resultant electric field is making an angle of \(37^\circ\) with the axis. The ratio of the dipole moment of A and B, \(\dfrac{p_1}{p_2}\) is : (take \(\sin 37^\circ = \dfrac{3}{5}\))
Two ideal electric dipoles A and B, having their dipole moment \(p_1\) and \(p_2\) respectively are placed on a plane with their centres at O as shown in the figure. At point C on the axis of dipole A, the resultant electric field is making an angle of \(37^\circ\) with the axis. The ratio of the dipole moment of A and B, \(\dfrac{p_1}{p_2}\) is : (take \(\sin 37^\circ = \dfrac{3}{5}\))
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C lies on the axial line of A and on the equatorial line of B at the same distance r from O. Use E_axial = 2kp/r^3 and E_equatorial = kp/r^3; these two fields are perpendicular.
Updated On: Nov 4, 2026
- \(\frac{3}{8}\)
- \(\frac{3}{2}\)
- \(\frac{2}{3}\)
- \(\frac{4}{3}\)
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The Correct Option is C
Solution and Explanation
Let \(OC = r\). Point C is on the axis of A and on the equatorial line of B.
Axial field of A: \(E_A = \dfrac{2kp_1}{r^3}\), directed along the axis.
Equatorial field of B: \(E_B = \dfrac{kp_2}{r^3}\), perpendicular to the axis.
The resultant makes \(37^\circ\) with the axis, so \(\tan 37^\circ = \dfrac{E_B}{E_A} = \dfrac{3}{4}\).
\(\dfrac{kp_2/r^3}{2kp_1/r^3} = \dfrac{p_2}{2p_1} = \dfrac34\) \(\Rightarrow\) \(\dfrac{p_1}{p_2} = \dfrac{4}{6} = \dfrac{2}{3}\).
Answer: (3) \(\frac{2}{3}\)
Axial field of A: \(E_A = \dfrac{2kp_1}{r^3}\), directed along the axis.
Equatorial field of B: \(E_B = \dfrac{kp_2}{r^3}\), perpendicular to the axis.
The resultant makes \(37^\circ\) with the axis, so \(\tan 37^\circ = \dfrac{E_B}{E_A} = \dfrac{3}{4}\).
\(\dfrac{kp_2/r^3}{2kp_1/r^3} = \dfrac{p_2}{2p_1} = \dfrac34\) \(\Rightarrow\) \(\dfrac{p_1}{p_2} = \dfrac{4}{6} = \dfrac{2}{3}\).
Answer: (3) \(\frac{2}{3}\)
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