Two short electric dipoles \(A\) and \(B\) having dipole moments \(p_1\) and \(p_2\) respectively are placed with their axes mutually perpendicular as shown in the figure. The resultant electric field at a point \(x\) is making an angle of \(60^\circ\) with the line joining points \(O\) and \(x\). The ratio of dipole moments \( \dfrac{p_2}{p_1} \) is:
Step 1: Understanding the Concept:
For a short dipole, the electric field at an axial point (distance \(r\)) is \(E_a = \frac{2kp}{r^3}\) and at an equatorial point (distance \(r\)) is \(E_e = \frac{kp}{r^3}\).
In the given diagram, the point \(x\) lies on the axial line of dipole \(B\) (with moment \(p_2\)) and on the equatorial line of dipole \(A\) (with moment \(p_1\)). Step 2: Key Formula or Approach:
1. Electric field from dipole B (Axial): \(E_B = \frac{2kp_2}{r^3}\) (along the \(x\)-axis).
2. Electric field from dipole A (Equatorial): \(E_A = \frac{kp_1}{r^3}\) (perpendicular to the \(x\)-axis).
3. The angle \(\theta\) made by the resultant field with the \(x\)-axis is \(\tan \theta = \frac{E_B}{E_A}\) or \(\frac{E_A}{E_B}\) depending on orientation. Step 3: Detailed Explanation:
According to the figure, the line joining O and x is the horizontal axis.
The resultant field makes \(60^\circ\) with the line \(Ox\).
The components of the electric field at point \(x\) are:
\[ E_{parallel} = E_A = \frac{kp_1}{r^3} \]
\[ E_{perpendicular} = E_B = \frac{2kp_2}{r^3} \]
Given \(\theta = 60^\circ\) with the horizontal line:
\[ \tan 60^\circ = \frac{E_B}{E_A} \]
\[ \sqrt{3} = \frac{\frac{2kp_2}{r^3}}{\frac{kp_1}{r^3}} = \frac{2p_2}{p_1} \]
\[ \frac{p_2}{p_1} = \frac{\sqrt{3}}{2} \] Step 4: Final Answer:
The ratio of the dipole moments \(p_2/p_1\) is \(\frac{\sqrt{3}}{2}\).
