Net electric field at point A as shown in figure is at an angle of 60^◦ with x-axis. Find P₂ /P₁ .

The correct answer is:
$2\sqrt{3}$

Step-by-Step Explanation:

1. Identify the dipoles and point A from the image:
From the given diagram, we have two short dipoles placed at the origin:
- Dipole 1 has a dipole moment vector ${\overrightarrow{P}}_1$ pointing in the negative x-direction (leftward), so:
${\overrightarrow{P}}_1=-P_1\hat{i}$
- Dipole 2 has a dipole moment vector ${\overrightarrow{P}}_2$ pointing in the positive y-direction (upward), so:
${\overrightarrow{P}}_2=P_2\hat{j}$
Point A lies on the positive x-axis at a distance $r$ from the origin.

2. Electric field due to Dipole 1 at point A:
Since point A lies along the axis of dipole 1, the electric field is in the axial position. The axial electric field of a dipole is directed parallel to the dipole moment vector:
${\overrightarrow{E}}_1=\frac{1}{4\pi {\epsilon }_0}\frac{2{\overrightarrow{P}}_1}{r^3}=-\frac{2k{P}_1}{r^3}\hat{i}$
where $k=\frac{1}{4\pi {\epsilon }_0}$.

3. Electric field due to Dipole 2 at point A:
Since point A lies on the perpendicular bisector (equatorial line) of dipole 2, the electric field is in the equatorial position. The equatorial electric field of a dipole is directed opposite to the dipole moment vector:
${\overrightarrow{E}}_2=-\frac{1}{4\pi {\epsilon }_0}\frac{{\overrightarrow{P}}_2}{r^3}=-\frac{k{P}_2}{r^3}\hat{j}$

4. Net electric field at point A:
The net electric field vector ${\overrightarrow{E}}_{\text{net}}$ is the vector sum of both fields:
${\overrightarrow{E}}_{\text{net}}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=-\frac{2k{P}_1}{r^3}\hat{i}-\frac{k{P}_2}{r^3}\hat{j}$

5. Angle of the net electric field with the x-axis:
Since both the x and y components of the electric field are negative, the net field points into the third quadrant. The magnitude of the angle $\theta$ it makes with the x-axis is given by:
$tan\theta =\left|\frac{E_y}{E_x}\right|$
Substituting the values of $E_x$ and $E_y$:
$tan\left({60}^{\circ }\right)=\frac{\left(\frac{k{P}_2}{r^3}\right)}{\left(\frac{2k{P}_1}{r^3}\right)}$
Simplifying the ratio:
$\sqrt{3}=\frac{P_2}{2P_1}$
Solving for the ratio $\frac{P_2}{P_1}$:
$\frac{P_2}{P_1}=2\sqrt{3}$